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