Coding Interview Patterns Flashcards: Signals, Invariants & Complexity
Practice coding interview patterns with 210 English flashcards on problem signals, invariants, complexity, edge cases, and choosing between approaches. Language-agnostic DSA review to pair with hands-on coding.
Об этой колоде
Coding interview patterns become useful when you can explain why an approach fits. These 210 English flashcards practice recognizing problem signals, stating invariants, checking prerequisites, comparing approaches, and diagnosing a broken assumption.
The sequence starts with constraints and complexity, then moves through arrays, strings, hashing, two pointers, sliding windows, prefix sums, binary search, intervals, stacks, linked lists, trees, heaps, tries, graphs, union-find, topological sorting, backtracking, greedy choices, dynamic programming, and bit techniques. Each topic builds from its basic idea toward the conditions that make it work.
Cards map short scenarios to candidate patterns, named patterns to invariants and costs, and failed assumptions to explanations or alternatives. Selected contrast prompts distinguish neighboring ideas such as subarrays and subsequences, BFS and DFS, or greedy choice and dynamic programming. Mechanical reverse cards, problem-number recall, language-specific API memorization, and full solution listings are excluded because they add little to this reasoning-focused review. Answer each prompt before turning the card, then test the idea by coding a fresh example.
This is a language-agnostic companion to practical DSA exercises. It complements the Blind 75 Python solutions deck by teaching reusable reasoning across problems. It was authored independently in English, not translated or paraphrased from another catalog package.
The questions, answers, organization, metadata, and original generated cover are released under CC0 1.0 to the extent applicable rights exist. Algorithm facts are common knowledge; no commercial deck text or third-party media was copied. This is an independent study resource with no affiliation to an employer or interview platform.
Карточки в этой колоде
Карточка 1
Вопрос
Which input constraints should you clarify before choosing an interview algorithm?
Ответ
Input size, value range, ordering, duplicates, allowed mutations, and the exact output. These determine which operations are affordable and which assumptions are valid.
Карточка 2
Вопрос
What does a loop invariant describe?
Ответ
A property that holds at a defined point in every iteration. Show it holds initially, survives an iteration, and implies the result when the loop ends.
Карточка 3
Вопрос
When analyzing nested loops, why can multiplying their written bounds overestimate runtime?
Ответ
The loops may share progress. If an inner pointer only advances across the input once, its total work can be O(n), even inside an outer loop.
Карточка 4
Вопрос
What is the difference between auxiliary space and total space?
Ответ
Auxiliary space counts extra working storage. Total space also includes the input and, depending on the stated convention, the output. State which measure you report.
Карточка 5
Вопрос
What does amortized O(1) mean for an operation sequence?
Ответ
The total cost of m operations is O(m), even if some individual operations are expensive. It is a sequence-wide bound, not a probability claim.
Карточка 6
Вопрос
How does a counterexample help evaluate a proposed greedy rule?
Ответ
One valid input where the rule fails disproves it. Try small cases that force a locally attractive choice to block a better later choice.
Карточка 7
Вопрос
Why should an algorithm's correctness argument address termination separately?
Ответ
Preserving the right property is not enough if the loop never stops. Identify a bounded measure that moves strictly toward termination.
Карточка 8
Вопрос
What runtime lower bound follows from returning k separate results?
Ответ
At least Ω(k) time to emit them, or more if each result has multiple elements. Include output size when analyzing enumeration algorithms.
Карточка 9
Вопрос
What should you say when quoting expected O(1) hash-table lookup?
Ответ
It assumes a suitable hash distribution and controlled load factor. It is not a worst-case guarantee; hashing a long key can also cost time.
Карточка 10
Вопрос
Which edge cases best expose index and boundary errors?
Ответ
Empty input, one element, all equal elements, and answers at either end. Also check the smallest input that enters each branch.
Карточка 11
Вопрос
Why can sorting be an invalid optimization even when it reduces later search work?
Ответ
Sorting may destroy required order or original indices. Preserve the needed information or choose an approach that respects the input contract.
Карточка 12
Вопрос
What does an exchange argument establish in a greedy proof?
Ответ
That an optimal solution can be changed to include the greedy choice without making its objective worse. The remaining problem must still fit the same reasoning.
Карточка 13
Вопрос
Why can recursion use O(n) space even without an explicit collection?
Ответ
Each active call occupies a stack frame. A chain of n calls can therefore need O(n) auxiliary space.
Карточка 14
Вопрос
What evidence should accompany a faster solution after presenting brute force?
Ответ
Name the repeated work or discarded search space, explain why the shortcut is safe, and give the resulting time and space bounds.
Карточка 15
Вопрос
What makes an array useful when a problem repeatedly accesses positions by index?
Ответ
Constant-time indexed access in the usual RAM model. Inserting or removing near the front can still require shifting O(n) elements.
Карточка 16
Вопрос
What should 'one character' mean before solving a string problem?
Ответ
Clarify whether the unit is a byte, code unit, Unicode code point, or user-perceived character. Indexing and length depend on that choice.
Карточка 17
Вопрос
An unsorted array needs a duplicate-existence check. Which structure fits?
Ответ
A hash set of values seen so far. Stop when a value is already present; expected O(n) time and O(n) space.
Карточка 18
Вопрос
For two-sum on an unsorted array, what should a hash map store?
Ответ
Previously seen values mapped to their indices. For each value x, look for target − x before inserting x, so the same position is not reused.
Карточка 19
Вопрос
When does a frequency array beat a hash map for counting?
Ответ
When keys come from a small known integer or character range. Direct indexing gives predictable access with space proportional to that range.
Карточка 20
Вопрос
Why is repeated concatenation risky when constructing a long immutable string?
Ответ
Each append may copy the existing prefix, producing quadratic total work. Collect pieces and join them, or use an appropriate mutable builder.
Карточка 21
Вопрос
How can you group anagrams without comparing every pair of words?
Ответ
Map a canonical character signature to a group. Sorted characters work; a frequency tuple works when the alphabet and character rules are fixed.
Карточка 22
Вопрос
What information does a set lose compared with a frequency map?
Ответ
Multiplicity. A set can answer whether a value exists but cannot distinguish one occurrence from several.
Карточка 23
Вопрос
How can a hash set support finding the longest consecutive integer run in expected O(n) time?
Ответ
Start scanning a run only at values whose predecessor is absent. Each distinct value then belongs to one forward scan; iterate distinct values.
Карточка 24
Вопрос
Why can a mutable object be a dangerous hash-map key?
Ответ
Changing fields used by its hash or equality can make the stored entry unreachable by ordinary lookup. Use immutable keys or stable key values.
Карточка 25
Вопрос
An array contains only integers from 0 through k. When is counting sort attractive?
Ответ
When k is small enough: count each value and reconstruct the order in O(n + k) time with O(k) count storage. General arbitrary keys need another approach.
Карточка 26
Вопрос
How can a single scan find both the minimum value and its earliest index?
Ответ
Keep the current minimum and index; update only on a strictly smaller value. Updating on equality would select a later occurrence.
Карточка 27
Вопрос
How do you compare two strings as multisets of characters?
Ответ
Compare character counts under the same character interpretation. Order does not matter, but every character's multiplicity does.
Карточка 28
Вопрос
Why does a hash collision not imply that two keys are equal?
Ответ
A hash compresses many possible keys into fewer codes. A correct table also checks key equality when resolving collisions.
Карточка 29
Вопрос
When is sorting a useful preprocessing step for detecting duplicate values?
Ответ
When reordering is allowed and O(n log n) time is acceptable. Equal values become adjacent, so a scan finds duplicates without a separate hash set.
Карточка 30
Вопрос
What is the key distinction between a subarray and a subsequence?
Ответ
A subarray is contiguous. A subsequence preserves relative order but may skip positions. Sliding-window methods usually depend on contiguity.
Карточка 31
Вопрос
For an unsorted two-sum query, how do hashing and sorting trade off?
Ответ
Hashing gives expected O(n) time with O(n) extra storage. Sorting plus two pointers costs O(n log n) time and requires care with original indices and mutation.
Карточка 32
Вопрос
How can a frequency map detect whether any permutation of a string can be a palindrome?
Ответ
Count odd frequencies. At most one character may have an odd count; all other occurrences must form mirrored pairs.
Карточка 33
Вопрос
Why must compound hash keys encode boundaries unambiguously?
Ответ
Naive concatenation can merge different tuples into the same key, such as (1, 23) and (12, 3). Use tuples or a length-aware encoding.
Карточка 34
Вопрос
What does coordinate compression preserve about numeric values?
Ответ
Their relative order and equality, by replacing distinct sorted values with ranks. It does not preserve numeric distances or sums.
Карточка 35
Вопрос
How can you compute products except self without division?
Ответ
Combine the product strictly before each position with the product strictly after it. Prefix and suffix passes handle zeros; use a numeric type large enough for the products.
Карточка 36
Вопрос
Why can a count of matching pairs overflow even when every input value fits in an integer?
Ответ
The number of matching pairs can grow quadratically with the input length. Size the result type for the count, not just the input values.
Карточка 37
Вопрос
A sorted array needs a pair with a target sum. Which search pattern fits?
Ответ
Opposite-end two pointers. Compare the endpoint sum with the target and move the endpoint that can move the sum in the needed direction.
Карточка 38
Вопрос
What invariant supports in-place removal of unwanted array values with read and write pointers?
Ответ
The prefix before write contains exactly the retained values from the processed input, in order. Read scans every input position once.
Карточка 39
Вопрос
How can two pointers check a palindrome without constructing a reversed string?
Ответ
Compare matching elements from opposite ends and move inward. Any mismatch rejects it; the pointers meeting or crossing completes the check.
Карточка 40
Вопрос
When does a fixed-size sliding window apply?
Ответ
When every candidate is a contiguous block of the same length and its summary can be updated as one element enters and one leaves.
Карточка 41
Вопрос
What invariant should a longest-window algorithm restore after adding a new rightmost element?
Ответ
The current window satisfies the required constraint. Move the left boundary and update state until validity returns, then consider its length.
Карточка 42
Вопрос
Why is moving the left pointer safe when a sorted-array endpoint sum is too small?
Ответ
With that left value, every candidate at or before the current right endpoint gives an equally small or smaller sum. The left position cannot form a valid pair.
Карточка 43
Вопрос
How does a three-way partition maintain separate regions?
Ответ
Track a low region, a middle region, an unclassified region, and a high region. Each step shrinks the unclassified region by placing one element correctly.
Карточка 44
Вопрос
How do you update the sum when a fixed-size window moves one position?
Ответ
Subtract the outgoing value and add the incoming value. After initializing the first window, all moves together take O(n) time.
Карточка 45
Вопрос
Why can a variable sliding window run in O(n) despite a nested shrink loop?
Ответ
Each boundary advances at most n times. With constant-time state updates, the total number of additions and removals is linear.
Карточка 46
Вопрос
For the longest substring without repeated characters, what window state is useful?
Ответ
Character counts or last-seen positions. Move the left boundary past the conflicting occurrence while ensuring it never moves backward.
Карточка 47
Вопрос
Why does opposite-end target-sum search fail on a generally unsorted array?
Ответ
Moving an endpoint no longer changes the sum predictably. The algorithm can discard a valid pair because the order-based elimination proof is missing.
Карточка 48
Вопрос
A positive-number array needs the shortest nonempty subarray with sum at least a positive T. When should the left boundary move?
Ответ
While the current sum is at least T, record the length and remove the leftmost value. Positive values make further shrinking reduce the sum predictably.
Карточка 49
Вопрос
For windows with at most k distinct values, what must happen when an outgoing count becomes zero?
Ответ
Remove that value from the active distinct set or decrement the distinct counter. A zero count must not still count as present.
Карточка 50
Вопрос
What changes when a fixed-window length exceeds the input length?
Ответ
There is no complete window. Return the contract's empty or missing-result value instead of treating a partial block as a valid candidate.
Карточка 51
Вопрос
How do you merge two sorted arrays with forward pointers?
Ответ
Repeatedly take the smaller unconsumed element, then append the remaining suffix. Each element is consumed once, giving O(m + n) time.
Карточка 52
Вопрос
Why must a minimum-cover substring track multiplicities of required characters?
Ответ
A target can require the same character more than once. A set of required characters would mark an undersupplied window as complete.
Карточка 53
Вопрос
How can counting subarrays with exactly k distinct values, for k ≥ 1, use an at-most helper?
Ответ
Compute atMost(k) − atMost(k − 1). The helper counts all valid subarrays ending at each right boundary after restoring the distinct-value limit.
Карточка 54
Вопрос
Why can a negative value break the usual shortest-sum sliding-window argument?
Ответ
Removing it increases the sum, and adding one can decrease the sum. The monotonic relation between window size and sum no longer holds.
Карточка 55
Вопрос
What does 'fast and slow pointers' mean when removing duplicates from a sorted array?
Ответ
A read pointer scans candidates while a write pointer marks the next unique slot. Sorting makes equal values adjacent, so the retained prefix can stay compact.
Карточка 56
Вопрос
What invariant prevents overwriting unread data during a backward merge into spare array capacity?
Ответ
The suffix after the write pointer already contains the largest merged elements. Filling from the end leaves the remaining source elements unread and intact.
Карточка 57
Вопрос
Why is 'find a contiguous range' alone insufficient to justify a variable sliding window?
Ответ
You also need a safe boundary-movement rule. Check how adding and removing elements affect the condition; contiguity by itself gives no such guarantee.
Карточка 58
Вопрос
After restoring an at-most window's validity, why are there right − left + 1 valid subarrays ending at right?
Ответ
Every suffix starting between left and right is valid when removing elements cannot violate the constraint. Count those starts, including the one-element suffix.
Карточка 59
Вопрос
How can you avoid duplicate value pairs in a sorted two-pointer enumeration?
Ответ
After emitting a pair, skip equal values on both sides. First confirm the output wants unique value pairs, since index-pair counting needs different handling.
Карточка 60
Вопрос
When does a character-frequency sliding window detect an anagram of a pattern?
Ответ
When the window has the pattern's length and identical character counts. Maintain count differences or a mismatch counter as the window moves.
Карточка 61
Вопрос
What does a prefix-sum array P mean when P[0] = 0?
Ответ
P[i] is the sum of the first i input elements. The extra zero represents the empty prefix and makes ranges beginning at index 0 work uniformly.
Карточка 62
Вопрос
When is binary search valid on a Boolean predicate over ordered candidates?
Ответ
When the predicate changes at most once, such as false then true. The search uses that monotonic boundary to discard a whole interval.
Карточка 63
Вопрос
Which workload favors a difference array?
Ответ
Many range additions followed by final value reconstruction. Mark each range's start and end changes, then take one prefix sum.
Карточка 64
Вопрос
In a lower-bound search over [lo, hi), what does hi initially equal for an n-element array?
Ответ
n, an exclusive boundary. The result can equal n when no element is at least the target, so do not index the array without checking.
Карточка 65
Вопрос
For a static array, how do prefix sums answer the half-open range [l, r)?
Ответ
Return P[r] − P[l]. Building P takes O(n) time and space; each range-sum query then takes O(1).
Карточка 66
Вопрос
How can prefix sums count subarrays whose sum equals k when negative values are allowed?
Ответ
For each current prefix p, add the number of earlier prefixes equal to p − k, then record p. A frequency map gives expected O(n) time.
Карточка 67
Вопрос
Why is binary-searching an answer different from binary-searching an input array?
Ответ
The candidates are possible result values. A feasibility check tells which side contains the boundary, even if the original input is unsorted.
Карточка 68
Вопрос
How does a difference array encode an addition of v to [l, r)?
Ответ
Add v at l and subtract v at r, using a boundary slot when needed. The reconstructed prefix totals apply v only within that range.
Карточка 69
Вопрос
For lower bound, how should equality with the target move the search boundary?
Ответ
Move hi to mid. An equal element is a candidate, but an earlier equal or qualifying element may still exist.
Карточка 70
Вопрос
Why initialize the prefix-frequency map with zero appearing once?
Ответ
It represents the empty prefix before the array. This lets a subarray starting at index 0 contribute to the count.
Карточка 71
Вопрос
Why are plain prefix sums inconvenient for many interleaved point updates and range-sum queries?
Ответ
Changing one value can invalidate a long suffix of prefix sums. A Fenwick tree or segment tree can support both operations in O(log n).
Карточка 72
Вопрос
How can you binary-search the minimum capacity needed to finish ordered work within a deadline?
Ответ
Define whether a capacity suffices, prove larger capacities remain feasible, bracket a feasible answer, and search for the first feasible capacity.
Карточка 73
Вопрос
What must every binary-search iteration do to guarantee termination?
Ответ
Strictly shrink the candidate interval while preserving the boundary invariant. Mixing inclusive and exclusive update rules can leave the same interval unchanged.
Карточка 74
Вопрос
For the longest subarray with a specified sum, which occurrence of each prefix sum should you retain?
Ответ
The earliest index. For a later endpoint, it gives the longest matching span; overwriting it with a later occurrence can shorten the answer.
Карточка 75
Вопрос
What runtime should you report for binary search with a nonconstant feasibility check?
Ответ
O(C log R), where C is one check's cost and R is the number of discrete candidates. Include any preprocessing separately.
Карточка 76
Вопрос
How can prefix sums turn a longest balanced binary subarray into an equal-prefix problem?
Ответ
Map one symbol to +1 and the other to −1. Equal prefix sums enclose a zero-sum range with equal counts of the two symbols.
Карточка 77
Вопрос
What is upper bound in a sorted array?
Ответ
The first position whose value is strictly greater than the target, or n if none exists. Lower bound instead finds the first value at least the target.
Карточка 78
Вопрос
Why must a prefix-sum counting algorithm query before recording the current prefix?
Ответ
Recording first can count the empty subarray ending at the current boundary, especially for target zero. Query only earlier prefixes for nonempty ranges.
Карточка 79
Вопрос
How do you safely compute a midpoint in a fixed-width integer search?
Ответ
Use lo + (hi − lo) / 2 with integer division when the nonnegative difference fits the type. Choose bounds or a wider type that also keep the subtraction safe.
Карточка 80
Вопрос
Why can duplicate values degrade searching a rotated sorted array to O(n)?
Ответ
Equal endpoints and midpoint can hide which side is sorted. Some cases permit discarding only one boundary element at a time.
Карточка 81
Вопрос
What preprocessing usually simplifies merging overlapping intervals?
Ответ
Sort by start coordinate. Keep the current merged interval and either extend its end or emit it when the next interval starts beyond it.
Карточка 82
Вопрос
Which data structure matches nested bracket validation?
Ответ
A stack of unmatched opening brackets. Each closing bracket must match the most recent unmatched opener, and the stack must be empty at the end.
Карточка 83
Вопрос
A problem asks for each element's next greater element. Which pattern is promising?
Ответ
A monotonic stack of unresolved positions. A new larger value resolves the smaller pending values it overtakes.
Карточка 84
Вопрос
Why must interval endpoint conventions be explicit?
Ответ
Touching endpoints overlap for closed intervals, but adjacent half-open intervals do not. The convention changes merge tests and event ordering.
Карточка 85
Вопрос
How can a sweep line find the maximum number of simultaneous intervals?
Ответ
Turn starts and ends into signed events, sort by coordinate, and track the running active count. Handle same-coordinate ties according to the endpoint convention.
Карточка 86
Вопрос
Why is one pass after sorting enough to merge intervals?
Ответ
No later interval starts earlier than the next one being inspected. Once that start is beyond the current end, future intervals cannot bridge the gap.
Карточка 87
Вопрос
What information should a stack store for next-greater distances?
Ответ
Indices, so the distance is currentIndex − previousIndex. Values alone do not identify positions or distinguish repeated occurrences.
Карточка 88
Вопрос
Why is counting opening and closing brackets insufficient to validate their sequence?
Ответ
Counts ignore order and nesting. A closing bracket may appear before its opener, or bracket types may cross despite balanced totals.
Карточка 89
Вопрос
For half-open intervals [start, end), how should equal-time starts and ends affect room counts?
Ответ
Process ends before starts, or aggregate their net change before evaluating the active count for the next segment. A room freed at time t can be reused at t.
Карточка 90
Вопрос
Why is a monotonic-stack algorithm often O(n) even though one step can pop many items?
Ответ
Each item is pushed once and popped at most once. Summed across the scan, stack operations are linear.
Карточка 91
Вопрос
How can a stack help simplify an absolute filesystem path lexically?
Ответ
Process components: ignore empty components and '.', pop for '..' when possible, and push ordinary names. This lexical result does not resolve symbolic links.
Карточка 92
Вопрос
How can a monotonic deque find each sliding-window maximum?
Ответ
Keep candidate indices in decreasing value order. Remove expired indices from the front and dominated values from the back; the front gives the maximum.
Карточка 93
Вопрос
What mistake can lose coverage when merging an interval contained inside the current one?
Ответ
Replacing the current end with the new end. Use the larger end so a nested interval cannot shrink the merged coverage.
Карточка 94
Вопрос
For a strictly next-greater query, what should happen to an equal-valued stack entry?
Ответ
Do not resolve it with the equal value. With a decreasing stack of unresolved indices, pop only when the new value is strictly greater.
Карточка 95
Вопрос
What event allows a monotonic stack to finalize a rectangle in a histogram?
Ответ
A shorter bar supplies a right limit for taller bars being popped. A popped bar's candidate span starts after the new stack top, or at index 0 if the stack is empty. Handle equal heights consistently.
Карточка 96
Вопрос
What comparison detects overlap between two nonempty half-open intervals?
Ответ
max(start1, start2) < min(end1, end2). A strict comparison excludes intervals that only touch.
Карточка 97
Вопрос
How does a stack support evaluating a postfix arithmetic expression?
Ответ
Push operands; for an operator, pop its right operand and then its left operand, compute, and push the result. Preserve order for subtraction and division.
Карточка 98
Вопрос
Why can a newer value dominate an older value in a sliding-window maximum deque?
Ответ
If the newer value is at least as large, it expires no earlier and is never worse as a future maximum. The older candidate can be removed.
Карточка 99
Вопрос
How can you merge two already sorted lists of disjoint intervals to find their intersections?
Ответ
Compare one interval from each list, emit any overlap, then advance the one with the earlier end. Total time is O(m + n).
Карточка 100
Вопрос
Why must a histogram stack algorithm handle bars still pending after the scan?
Ответ
Those bars may extend to the array's end and contain the largest rectangle. Flush them using the end boundary or a suitable sentinel.
Карточка 101
Вопрос
What must you save before reversing a singly linked list node's next pointer?
Ответ
Its original next node. Otherwise rewiring can lose access to the remaining list.
Карточка 102
Вопрос
Why does a dummy head simplify linked-list insertion and deletion?
Ответ
It supplies a predecessor even when the real head changes. The same pointer update can handle both the first node and interior nodes.
Карточка 103
Вопрос
How do fast and slow pointers detect a cycle in a singly linked list?
Ответ
Advance one pointer one step and the other two. A meeting implies a cycle; reaching null with the fast pointer means the list terminates.
Карточка 104
Вопрос
What does a recursive binary-tree traversal use for auxiliary space?
Ответ
O(h) call-stack space, where h is tree height. This is O(log n) for a balanced tree but O(n) for a chain.
Карточка 105
Вопрос
During iterative list reversal, what do prev and current represent?
Ответ
prev heads the reversed processed prefix; current heads the unprocessed suffix. Rewire one node while preserving access to the suffix.
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Coding Interview Patterns Flashcards: Signals, Invariants & Complexity
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Карточка 106
Вопрос
How can you remove the nth node from the end of a list in one pass?
Ответ
Start lead at the head and lag at a dummy head. Advance lead n nodes, then move both until lead is null; remove lag.next. Reject invalid n according to the input contract.
Карточка 107
Вопрос
Which tree traversal naturally computes a value that depends on both children's results?
Ответ
Postorder. Process left and right subtrees before combining their results at the parent.
Карточка 108
Вопрос
How can you locate a cycle's entry after Floyd's two-speed pointers meet?
Ответ
Reset one pointer to the head and move both one step at a time. Their next meeting is the entry; this follows from the distances modulo the cycle length.
Карточка 109
Вопрос
What is the difference between tree depth and tree height?
Ответ
Depth measures distance from the root to a node; height measures the longest downward distance to a leaf. State whether distances count edges or nodes.
Карточка 110
Вопрос
How can you merge two sorted linked lists using O(1) auxiliary node storage?
Ответ
Relink the smaller current node onto the result tail, advancing that list. Attach the remaining suffix when one list ends; existing nodes are reused.
Карточка 111
Вопрос
When is breadth-first traversal more natural than depth-first traversal on a tree?
Ответ
When results are grouped by depth or you need the nearest qualifying node by edge count. A queue processes one distance layer before the next.
Карточка 112
Вопрос
Why must linked-list intersection compare node identity rather than node value?
Ответ
Intersection means sharing the same node object and suffix. Separate nodes can hold equal values without the lists intersecting.
Карточка 113
Вопрос
Why is checking only immediate children insufficient to validate a binary search tree?
Ответ
A descendant can satisfy its parent yet violate an ancestor's constraint. Carry inherited lower and upper bounds, with an explicit duplicate policy.
Карточка 114
Вопрос
What is the lowest common ancestor of two nodes in a rooted tree?
Ответ
The deepest node that is an ancestor of both, allowing a node to be its own ancestor.
Карточка 115
Вопрос
What property makes inorder traversal useful in a binary search tree?
Ответ
It visits keys in sorted order under the tree's duplicate policy. In a strict BST, each visited key must be greater than the previous one.
Карточка 116
Вопрос
Why must maximum tree-path sum separate its returned value from its global candidate?
Ответ
The parent can extend only one downward branch. A complete path considered at the current node may join both children, but that fork cannot be extended upward.
Карточка 117
Вопрос
How can two pointers find the intersection of two acyclic singly linked lists without measuring lengths?
Ответ
After reaching a list's end, switch that pointer to the other head. Each traverses both lengths, so they meet at the shared node or at null.
Карточка 118
Вопрос
What extra information makes preorder serialization unambiguous for an arbitrary binary tree?
Ответ
Explicit null-child markers or another equivalent shape encoding. Values in preorder alone do not determine the structure.
Карточка 119
Вопрос
Why can repeated subtree-height calculations make a tree-balance check O(n²)?
Ответ
The same descendants may be scanned from many ancestors. Return height and balance together in one postorder traversal to visit each node once.
Карточка 120
Вопрос
How does BST ordering guide a lowest-common-ancestor search for two existing distinct keys?
Ответ
Move left if both keys are smaller and right if both are larger. The first split, or a node matching one key, is their lowest common ancestor.
Карточка 121
Вопрос
What runtime does a search in an ordinary unbalanced BST guarantee?
Ответ
O(h), where h is its height, and O(n) in the worst case. Logarithmic search requires a balance guarantee or a stated expected-shape assumption.
Карточка 122
Вопрос
When computing a root-to-leaf path sum, why is reaching a null child insufficient for success?
Ответ
A valid endpoint must be a leaf with no children. A missing child beside an existing child does not finish a root-to-leaf path.
Карточка 123
Вопрос
What does a min-heap guarantee about its root and children?
Ответ
The root is a minimum element, and each parent is no larger than its children. The whole array representation is not sorted.
Карточка 124
Вопрос
A stream needs the k largest values seen so far. Which heap should you maintain?
Ответ
A min-heap of at most k values. Its root is the smallest retained value, so a larger arrival can replace it.
Карточка 125
Вопрос
What shared structure does a trie store?
Ответ
Prefixes of keys. Following one edge per symbol reaches a key's prefix node, while a terminal marker distinguishes a complete stored key.
Карточка 126
Вопрос
What are the usual binary-heap costs for peek, insertion, and root removal?
Ответ
Peek is O(1); insertion and root removal are O(log n). Moving a changed element along one root-to-leaf path restores heap order.
Карточка 127
Вопрос
How can a heap merge k sorted input streams?
Ответ
Keep each nonempty stream's next element in a min-heap. Emit the minimum and replace it with that stream's next item. With k streams and N total elements, O(k + N log k) time includes initialization for k ≥ 2.
Карточка 128
Вопрос
Why is bottom-up heap construction O(n), not O(n log n)?
Ответ
Most nodes are near the leaves and can move only a short distance. Summing each node's possible sift-down work across all heights is linear.
Карточка 129
Вопрос
How do two heaps support a running median?
Ответ
Keep the lower half in a max-heap and the upper half in a min-heap, with sizes differing by at most one and every lower value no greater than every upper value.
Карточка 130
Вопрос
What is a trie's lookup cost for a key of length L?
Ответ
O(L) when each child transition is O(1). Child maps or ordered child containers can change that transition cost; space depends on stored prefixes and representation.
Карточка 131
Вопрос
Why does a trie node need a terminal marker even if it has children?
Ответ
A stored key can be a prefix of another key. The marker distinguishes a complete word from a prefix that merely leads to longer words.
Карточка 132
Вопрос
When does sorting make more sense than a top-k heap?
Ответ
When you need the entire sorted order or k is close to n and a full sort is acceptable. A heap's advantage is strongest when only a small retained subset is needed.
Карточка 133
Вопрос
Why does a priority queue not by itself support efficient arbitrary deletion?
Ответ
The heap efficiently exposes only its root. Removing another item needs its position, an indexed-heap design, or a lazy-deletion scheme with cleanup.
Карточка 134
Вопрос
What output cost remains after a trie reaches a requested prefix?
Ответ
Enumerating matching completions still costs time proportional to the visited subtree and emitted text. Prefix lookup does not make all autocomplete results free.
Карточка 135
Вопрос
How can a priority queue break tied priorities without comparing the payloads?
Ответ
Attach a unique increasing sequence number and compare priority first, then sequence number. Equal-priority items can then leave in insertion order without requiring an order on their payloads.
Карточка 136
Вопрос
Why can a trie use more memory than a hash set of complete strings?
Ответ
Nodes and child containers have overhead, especially for sparse branches. Shared prefixes save repeated symbols but do not guarantee a smaller representation.
Карточка 137
Вопрос
What should graph modeling identify before choosing a traversal?
Ответ
The states as vertices and legal transitions as edges, including direction and cost. A grid cell, word, or puzzle configuration can be a vertex.
Карточка 138
Вопрос
When does ordinary BFS find a shortest path?
Ответ
When every edge has the same nonnegative cost, including the unweighted case. Processing vertices by distance layer makes first discovery a shortest-edge-count path.
Карточка 139
Вопрос
What is the space cost of an adjacency list compared with an adjacency matrix?
Ответ
A list uses O(V + E) space; a matrix uses O(V²). A matrix gives constant-time edge lookup, while lists efficiently enumerate actual neighbors.
Карточка 140
Вопрос
Which pattern finds all vertices reachable from a start vertex?
Ответ
DFS or BFS with a visited set. Each reachable vertex and edge is processed a bounded number of times with adjacency lists.
Карточка 141
Вопрос
What role does a parent map play in shortest-path traversal?
Ответ
It records the predecessor used to reach each state. After reaching the target, follow parents backward and reverse the sequence to recover a path.
Карточка 142
Вопрос
Why should BFS mark a vertex visited when enqueuing it?
Ответ
To prevent several parents from adding it before its first removal. First enqueue already fixes its distance in an unweighted graph.
Карточка 143
Вопрос
When is Dijkstra's algorithm appropriate?
Ответ
For shortest paths with nonnegative edge weights. Its greedy finalization relies on no later path reducing a settled distance through a negative edge.
Карточка 144
Вопрос
How can a grid traversal avoid confusing physical cells with full search states?
Ответ
Include all information that changes future moves in the visited key, such as remaining obstacle removals or collected keys. Position alone may merge different states.
Карточка 145
Вопрос
How do you detect a directed cycle with DFS?
Ответ
Track unvisited, active, and finished vertices. An edge to an active vertex closes a cycle on the current recursion path.
Карточка 146
Вопрос
Why can DFS with a visited set fail to find a shortest unweighted path?
Ответ
Its first discovered route may follow a deep detour. DFS reachability order is not distance order; BFS provides that guarantee.
Карточка 147
Вопрос
What does a topological ordering guarantee?
Ответ
For every directed edge u → v, u appears before v. Such an ordering exists exactly when the directed graph is acyclic.
Карточка 148
Вопрос
Why should stale priority-queue entries be skipped in a common Dijkstra implementation?
Ответ
A vertex can receive a better distance after an older entry was pushed. Skip an entry whose stored distance differs from the current best distance.
Карточка 149
Вопрос
For an undirected simple graph, why does DFS ignore the edge back to its parent when detecting cycles?
Ответ
That edge is the same tree edge traversed in reverse, not a new cycle. A different already-visited neighbor indicates a cycle.
Карточка 150
Вопрос
What does union-find answer efficiently?
Ответ
Whether elements belong to the same connected component while components are merged. It does not store the actual connecting paths.
Карточка 151
Вопрос
How does Kahn's algorithm build a topological ordering?
Ответ
Enqueue all zero-indegree vertices, repeatedly remove one, and decrement its outgoing neighbors' indegrees. Enqueue each neighbor when its indegree becomes zero.
Карточка 152
Вопрос
Which shortest-path algorithm handles edges weighted only 0 or 1 without a heap?
Ответ
0–1 BFS with a deque. Push a relaxed zero-cost neighbor to the front and a one-cost neighbor to the back, preserving distance order.
Карточка 153
Вопрос
Why is a boolean visited flag usually wrong for Dijkstra at first enqueue?
Ответ
The first tentative distance need not be the shortest. Allow improvements; a vertex becomes settled when its smallest current distance is removed from the queue.
Карточка 154
Вопрос
How do path compression and union by size or rank affect union-find complexity?
Ответ
Together they give O(α(n)) amortized time per operation, where α is the inverse Ackermann function. The bound is effectively tiny for practical input sizes.
Карточка 155
Вопрос
What does processing fewer than V vertices in Kahn's algorithm reveal?
Ответ
A directed cycle remains. No vertex in the cyclic remainder can reach indegree zero after all removable dependencies are processed.
Карточка 156
Вопрос
How can BFS compute distance from every grid cell to the nearest source?
Ответ
Initialize the queue with all sources at distance zero. This multi-source BFS expands the nearest-source distance layers together.
Карточка 157
Вопрос
Why can a topological ordering be nonunique?
Ответ
Several vertices may currently have no remaining prerequisites. Choosing them in different orders can produce different valid orderings.
Карточка 158
Вопрос
What happens when union-find receives an edge whose endpoints already share a representative?
Ответ
The edge connects vertices already in one component. In incremental construction of an undirected forest, adding it creates a cycle.
Карточка 159
Вопрос
Which algorithm can handle negative edge weights and detect a reachable negative cycle?
Ответ
Bellman–Ford. Repeatedly relax all edges; an improvement after V − 1 full rounds indicates a negative cycle reachable from the source.
Карточка 160
Вопрос
Why does traversal need an outer loop to count every connected component of an undirected graph?
Ответ
One traversal reaches only one component. Start another traversal from each still-unvisited vertex and increment the component count.
Карточка 161
Вопрос
How can topological order simplify shortest paths in a weighted DAG?
Ответ
Relax each vertex's outgoing edges in topological order. Every predecessor is processed first, so negative weights are allowed and total time is O(V + E).
Карточка 162
Вопрос
When should you use BFS or DFS instead of union-find for connectivity?
Ответ
When the graph is static and you need traversal details such as paths or component members. Union-find is especially useful for repeated incremental edge additions and connectivity queries.
Карточка 163
Вопрос
What is the difference between a minimum spanning tree and a shortest-path tree?
Ответ
A minimum spanning tree minimizes total connecting edge weight. A shortest-path tree preserves shortest routes from a chosen source; neither objective implies the other.
Карточка 164
Вопрос
Why can stopping at the first meeting be unsafe in bidirectional BFS with arbitrary node-by-node expansion?
Ответ
A first meeting under an arbitrary expansion order may not minimize the combined distances. Use a layer-based stopping rule that accounts for both search depths.
Карточка 165
Вопрос
Which signal suggests backtracking rather than a single greedy choice?
Ответ
The task asks for all valid arrangements, or choices must be tried and undone because no safe local choice is known. Build a partial candidate and explore legal extensions.
Карточка 166
Вопрос
What belongs in a backtracking state?
Ответ
Enough information to determine legal next choices and recognize completion, such as the current position, chosen items, and remaining constraints.
Карточка 167
Вопрос
What makes a pruning condition safe?
Ответ
It proves that no completion of the current partial state can satisfy the goal or improve the required objective. A guess about likely failure is insufficient.
Карточка 168
Вопрос
How do combinations differ from permutations during generation?
Ответ
Combinations ignore order, so restrict future choices to later positions. Permutations care about order, so track which positions are already used.
Карточка 169
Вопрос
What should be true after a backtracking recursive call returns?
Ответ
The caller's mutable search state is restored exactly to its pre-choice state. Undo additions and constraint updates before trying a sibling choice.
Карточка 170
Вопрос
When generating unique subsets from sorted values, how do you skip duplicates safely?
Ответ
At one recursion depth, skip a value equal to the previous sibling candidate. Still allow equal values at deeper levels when the input provides multiple copies.
Карточка 171
Вопрос
Why can backtracking output alone require exponential time?
Ответ
A set with n distinct elements has 2^n subsets. Explicitly listing all subsets cannot be polynomial in n; copying their contents adds further cost.
Карточка 172
Вопрос
Why should a completed mutable candidate usually be copied before saving it?
Ответ
Later backtracking steps will modify the working candidate. Saving only a reference can make all recorded answers reflect subsequent changes.
Карточка 173
Вопрос
What is the main risk of memoizing backtracking solely by the current index?
Ответ
Different histories can leave different remaining choices or constraints. The memo key must include every part of the state that affects future results.
Карточка 174
Вопрос
For selecting the most nonoverlapping intervals, which greedy choice is justified?
Ответ
Choose the available interval with the earliest finishing time, then continue with compatible intervals. This leaves at least as much room for the remaining selections.
Карточка 175
Вопрос
What is the difference between greedy choice and dynamic programming?
Ответ
Greedy commits to a choice proven safe without exploring all alternatives. DP evaluates and combines subproblem alternatives when that local commitment is not justified.
Карточка 176
Вопрос
Why does choosing the largest coin repeatedly fail for some coin systems?
Ответ
The locally largest coin can leave an expensive remainder. With denominations 1, 3, 4 and amount 6, greedy uses 4 + 1 + 1, while 3 + 3 uses fewer coins.
Карточка 177
Вопрос
How does a farthest-reachable frontier solve reachability in a nonnegative jump-length array?
Ответ
Scan positions no farther than the current frontier and extend it with each reachable index plus its jump length. If the next position lies beyond the frontier, progress is impossible.
Карточка 178
Вопрос
Why does choosing the shortest interval not always maximize the number of nonoverlapping intervals?
Ответ
A short interval can cross the boundary between two compatible intervals and block both. Duration alone does not measure the future scheduling space it consumes.
Карточка 179
Вопрос
What must be proved before pruning a combination-sum branch because its sum exceeds the target?
Ответ
Remaining choices cannot reduce the sum. The pruning is safe for nonnegative additions under the stated goal, but negative numbers can make it invalid.
Карточка 180
Вопрос
How does branch and bound differ from ordinary feasibility pruning?
Ответ
It uses a bound on the best objective reachable from a partial state. Prune only when that bound cannot beat the best complete answer already found.
Карточка 181
Вопрос
What question distinguishes a greedy proof from evidence that a heuristic often works?
Ответ
Can every discarded alternative be ruled out for all valid inputs? Examples and benchmarks support a heuristic, but do not establish the safe-choice property.
Карточка 182
Вопрос
Why is earliest-finish interval scheduling insufficient when intervals have different rewards?
Ответ
Maximizing count and maximizing reward are different objectives. A single high-reward interval can beat several low-reward intervals, so weighted scheduling needs more information.
Карточка 183
Вопрос
Which combination of properties makes dynamic programming promising?
Ответ
Repeated subproblems and a recurrence that combines their results. Define a state whose answer is independent of the path used to reach it.
Карточка 184
Вопрос
What should a DP state definition say before you write a recurrence?
Ответ
Exactly what one table entry means, including its input boundary and any remaining resource or constraint. Ambiguous states lead to mismatched transitions.
Карточка 185
Вопрос
How do top-down memoization and bottom-up tabulation differ?
Ответ
Memoization computes states on demand through calls and caches them. Tabulation processes states in an explicit dependency order, often avoiding recursion overhead.
Карточка 186
Вопрос
What determines the runtime of a DP with a finite state table?
Ответ
The number of states actually evaluated times the work per state, plus preprocessing and output reconstruction. Count transitions rather than just table dimensions.
Карточка 187
Вопрос
Why are base cases part of a DP's meaning rather than convenient initial values?
Ответ
They encode valid empty or smallest subproblems. A wrong base value can invent impossible solutions or remove legitimate ones from every later transition.
Карточка 188
Вопрос
For 0/1 knapsack compressed to one capacity array, why iterate capacities downward?
Ответ
Each item must be used at most once. Descending order reads the previous item's state instead of reusing an update made for the current item.
Карточка 189
Вопрос
When can a DP table be compressed to a few rows or variables?
Ответ
When future states depend only on a bounded slice of earlier states. Keep those dependencies until their last use; reconstruction may need additional storage.
Карточка 190
Вопрос
What recurrence models choosing nonadjacent values for maximum sum?
Ответ
At each position, compare skipping it with taking it plus the best result before its neighbor. The base cases must specify whether choosing nothing is allowed.
Карточка 191
Вопрос
What DP state counts paths through a blocked grid when moves are only right or down?
Ответ
The number of ways to reach each cell from above or from the left. Blocked cells contribute zero; initialize an unblocked starting cell to one.
Карточка 192
Вопрос
For unbounded knapsack, why can capacities run upward within an item's pass?
Ответ
Reusing the current item's updated smaller-capacity result is allowed. Ascending order lets that item contribute more than once.
Карточка 193
Вопрос
How can loop order change coin-change counting from combinations to ordered sequences?
Ответ
Processing coin types outside amounts builds combinations without ordering them. Processing amounts outside all coin choices counts different last-coin sequences separately.
Карточка 194
Вопрос
What is the key distinction between longest common subsequence and longest common substring?
Ответ
A subsequence may skip characters; a substring must stay contiguous. Their DP transitions differ because a substring match cannot carry through a mismatch.
Карточка 195
Вопрос
Why is O(nW) knapsack called pseudopolynomial?
Ответ
It is polynomial in the numeric capacity W, but W can be exponential in the number of bits used to encode it. It is not polynomial in input bit length.
Карточка 196
Вопрос
For longest increasing subsequence, what does tails[length − 1] represent in the O(n log n) method?
Ответ
The smallest possible final value of an increasing subsequence of that length among processed values. The tails array itself need not be one actual subsequence.
Карточка 197
Вопрос
Why do counting and minimization DPs use different unreachable-state values?
Ответ
A count uses zero ways. A minimization state needs an explicit unreachable marker or infinity so an impossible predecessor cannot look like a cheap solution.
Карточка 198
Вопрос
What state supports edit distance between two strings?
Ответ
The minimum edits needed to transform one prefix into the other. Transitions account for insertion, deletion, and replacement or a matching final character.
Карточка 199
Вопрос
How can a DP recover one chosen solution instead of only its score?
Ответ
Store predecessor or choice information, or recompute choices from the full table. Walk backward from the final state to reconstruct the selected decisions.
Карточка 200
Вопрос
Why must an LIS implementation choose its binary-search boundary according to strictness?
Ответ
For a strictly increasing subsequence, replace the first tail at least equal to the value. A nondecreasing subsequence instead uses the first strictly greater tail.
Карточка 201
Вопрос
What operation tests whether bit i of a nonnegative integer mask is set?
Ответ
Check whether mask AND (1 shifted left by i) is nonzero. Ensure i is inside the integer representation's supported bit range.
Карточка 202
Вопрос
Why does XOR recover a unique value when every other value occurs exactly twice?
Ответ
Equal values cancel because x XOR x = 0, and XOR is associative and commutative. XORing all values leaves the single unpaired value.
Карточка 203
Вопрос
What does x AND (x − 1) do for a positive integer x?
Ответ
It clears x's lowest set bit. Repeating it counts set bits in time proportional to the number of set bits.
Карточка 204
Вопрос
When does a bitmask make a useful DP state?
Ответ
When a small set of items is either included or excluded and future choices depend on that subset. n items give 2^n possible masks, so n must be small.
Карточка 205
Вопрос
How do you set a bit and clear a bit without changing the others?
Ответ
Set bit i with mask OR (1 shifted left by i). Clear it with mask AND the bitwise complement of that single-bit mask, respecting the chosen word width.
Карточка 206
Вопрос
What makes a memoized recurrence invalid when it depends on mutable global state omitted from the key?
Ответ
The same key can have different answers under different global conditions. Include the relevant state in the key or remove that dependency.
Карточка 207
Вопрос
What condition recognizes a power of two among integers?
Ответ
x > 0 and x AND (x − 1) = 0. The positivity check excludes zero, which also makes the bitwise expression zero.
Карточка 208
Вопрос
Why does bitwise complement need a width convention in language-agnostic reasoning?
Ответ
Complement flips all bits in the representation. Fixed-width and arbitrary-precision signed integers can produce different-looking values; mask to the intended width when necessary.
Карточка 209
Вопрос
How can two unique values be recovered when every other value occurs twice?
Ответ
XOR all values, choose a set bit in that nonzero result, and partition by that bit. XOR within each group; the two unique values fall into different groups.
Карточка 210
Вопрос
Why is memoization alone insufficient to handle cyclic state dependencies?
Ответ
A call may revisit an unfinished state before any value is cached. Use cycle handling or a problem-specific iterative method; ordinary DAG-style DP assumes an acyclic dependency order.
210 карточек
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