← Capital One Interview Insights
First clarify edge cases and constraints, then outline a solution that iterates through the array, checks divisibility by 3, and counts digit occurrences of 7. Write clean code with helper functions and test with examples.
Pro tip: Mention that you can optimize by checking divisibility first (cheap operation) before counting digits, and use arithmetic digit extraction to avoid string conversion overhead.
Ask about input size, negative numbers, zero, and whether 'contains the digit 7 at least twice' means exactly two or at least two occurrences. Confirm return type and handling of non-integers.
Explain that you'll iterate through each number, check if divisible by 3, then count digit 7s. State time complexity O(n * d) where d is average digits, and space O(1).
Write a helper function that counts occurrences of digit 7 using modulo and division, handling negative numbers by taking absolute value. Alternatively, convert to string and count.
Loop through the array, for each number check if divisible by 3 and if digit count >= 2, increment a counter. Return the counter.
Walk through test cases: empty array, numbers with multiple 7s but not divisible by 3, numbers divisible by 3 but fewer than two 7s, negative numbers, and zero.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the rules and edge cases first, then propose a simulation using queues or two pointers to compare top cards each round. Discuss time and space complexity, and consider optimizations like early termination or cycle detection if decks can be infinite.
Pro tip: Mention that you would handle infinite loops by detecting repeated deck states, showing awareness of potential pitfalls beyond the basic simulation.
Ask about deck sizes, card values, tie-breaking rules, and whether decks can be infinite or cycle. Confirm that the game ends when one deck is empty.
Use queues (or deques) to represent each player's deck, allowing O(1) removal from the top and addition to the bottom. Alternatively, use two pointers if decks are arrays and we only need to track positions.
While both decks are non-empty, compare the top cards. The winner takes both cards and places them at the bottom of their deck. Increment the round counter.
Return the round count when one deck is empty. If infinite loops are possible, detect cycles by storing seen deck states or limit rounds.
Time complexity is O(R) where R is the number of rounds, and space is O(N) for the queues. Discuss potential optimizations if needed.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one is where I probably lost the most time.
Clarify the problem constraints and edge cases first, then propose an efficient algorithm that precomputes prefix sums and counts of non-negative values to evaluate all possible split points in O(R*C) time. Discuss how to handle partitions with no non-negative cells and how to compute floored averages, then compare candidate splits to find the minimum difference.
Pro tip: Mention that you would validate the solution with brute force on small random matrices to catch off-by-one errors in partition boundaries, and discuss time/space complexity trade-offs to show engineering maturity.
Ask about matrix dimensions, value ranges, and what to do if a partition has no non-negative cells (e.g., skip that split or treat average as 0). Confirm that split indices are 0-based and that partitions are non-empty.
Use 2D prefix sums for non-negative values and counts to compute the sum and count for any submatrix in O(1). This allows evaluating all O(R*C) split points efficiently.
For each candidate split (r, c), compute the four partition averages as floor(sum / count) for non-negative cells. If any partition has count 0, skip that split.
Track the minimum difference between the maximum and minimum partition averages across all valid splits. Return the row and column indices of the best split.
State time complexity O(R*C) and space O(R*C) for prefix arrays. Suggest testing with small random matrices against a brute-force implementation to verify correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use a difference array to efficiently mark the start and end of each lamp's coverage, then compute prefix sums to get coverage counts at all points. For each control point, look up the count in O(1) time. This handles large inputs well and avoids checking each lamp for each point.
Pro tip: Clarify whether the segments are inclusive of endpoints and whether control points can be outside the range of all lamps; handle these edge cases explicitly. Also, mention that if the number of control points is small, a brute-force approach might be acceptable, but the difference array solution scales better.
Confirm the meaning of 'includes' (inclusive/exclusive endpoints), the range of coordinates, and the number of lamps and control points. Ask about potential duplicates or unsorted input.
Decide between brute-force (O(L*Q)) and difference array with prefix sums (O(L + Q + R)), where R is the coordinate range. Explain why the latter is better for large inputs.
Create an array of size max_coordinate+2 (or use a hash map for sparse coordinates). For each lamp [start, end], increment diff[start] and decrement diff[end+1] (if inclusive).
Compute the running sum over the difference array to get coverage counts at each point. For each control point, return the count from the prefix sum array (or 0 if out of range).
Test with overlapping lamps, points at boundaries, points outside all segments, and large coordinate values. Discuss time and space complexity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.