← Capital One Interview Insights
This was the warmup and I almost tripped on the k=1 case.
Start by clarifying the problem and edge cases, especially k=1. Then outline an efficient algorithm: for k>1, repeatedly divide each element by k until it's no longer divisible, and check if the result is 1; for k=1, count elements equal to 1. Finally, discuss time and space complexity and potential optimizations.
Pro tip: Mention that for k>1, you can precompute powers of k up to the maximum array value and use a hash set for O(1) lookups, which is efficient for multiple queries. Also, explicitly handle k=1 as a special case to avoid infinite loops.
Confirm that the array contains positive integers and k >= 1. Discuss the special case when k=1, where only elements equal to 1 are powers of 1.
For each element, repeatedly divide by k while divisible, then check if the final value is 1. Alternatively, precompute powers of k up to the maximum element and use a set for O(1) membership checks.
When k=1, count the number of elements equal to 1, since 1^n = 1 for any n, and no other number is a power of 1.
Discuss time complexity: O(n log_k(max)) for division method, or O(n + log_k(max)) for precomputation. Space complexity: O(1) for division, O(log_k(max)) for precomputation. Consider trade-offs.
Walk through examples like k=2, array=[1,2,4,8,16,3] and k=1, array=[1,1,2,3] to verify correctness. Also test with large numbers and k=1.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, compute the total number of days elapsed from day 1 of month 1 to the target date by summing the lengths of all preceding months and adding the target day minus 1. Then, add the starting phase index and take the result modulo 8 to get the phase index on the target date. Handle edge cases like invalid dates and ensure the calculation is efficient.
Pro tip: Clarify whether the starting phase is for day 1 (index 0) and confirm that month lengths are given in order. Also, mention that you can precompute prefix sums of month lengths for O(1) date queries if multiple queries are expected.
Restate the problem: given a starting phase on day 1, month lengths, and a target date, find the phase index. Confirm the indexing: day 1 corresponds to the starting phase, and phases cycle every 8 days.
Calculate the number of days from day 1 of month 1 to the target date. Sum the lengths of all months before the target month, then add (target day - 1).
Add the starting phase index to the days elapsed, then take modulo 8 to get the phase index on the target date. Ensure the result is in the range 0-7.
Check for invalid dates (e.g., day exceeding month length, month out of range). Consider if the target date is day 1 of month 1: days elapsed = 0, phase = starting phase.
If multiple queries are expected, precompute prefix sums of month lengths to answer each query in O(1) time after O(n) preprocessing.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the grid as a 2D array and simulate the process in rounds: each round, scan for all groups of size >= T using BFS/DFS, mark them for removal, then apply gravity to each column. Repeat until a round removes nothing, then return the grid. Emphasize correctness, complexity, and edge cases.
Pro tip: Mention that you can optimize by only re-checking cells adjacent to removed groups in subsequent rounds, and clarify that gravity is applied after all removals in a round to match typical game mechanics.
Confirm whether removal happens simultaneously for all groups in a round, whether gravity is applied after each removal or after all removals, and how to handle T <= 1 or empty grids.
Use BFS/DFS to find 4-directionally connected components of the same color. Track visited cells to avoid re-processing, and collect groups with size >= T.
Set all cells in qualifying groups to 0, then for each column, compact non-zero cells downward (stable order). This can be done in-place with a write pointer.
Repeat detection and removal/gravity until a full pass finds no removable groups. Return the final grid.
Discuss time complexity O(R*C) per round, worst-case O(R*C * rounds). Suggest optimizations like only re-checking affected columns/cells or using union-find for dynamic connectivity.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The problem statement asked me to also define the transformation rule and state necessary conditions, which felt unusual for a coding problem.
Use a hash map to group strings by a canonical signature that is invariant under up to two swaps, then count pairs within each group. For each string, generate all possible strings reachable by 0, 1, or 2 swaps and use the lexicographically smallest as the key. This reduces the problem to counting pairs in groups, which can be done in O(N) time after O(N * L^2) preprocessing.
Pro tip: Mention that generating all 2-swap variants naively is O(L^4), but you can optimize to O(L^2) by considering only swaps that affect mismatched positions relative to a reference string. Also, clarify that unordered pairs mean each pair counted once, so use combinations.
Clarify that strings are of equal length L, leading zeros allowed, and we need unordered pairs where one can be transformed into the other with at most two swaps. Confirm that swaps are within a single string.
For each string, generate all strings reachable by 0, 1, or 2 swaps and choose the lexicographically smallest as its canonical key. This ensures two strings are in the same group iff they are transformable into each other.
Instead of generating all O(L^4) swap combinations, note that only swaps involving mismatched positions relative to a reference string matter. This reduces to O(L^2) per string by considering at most two swaps among mismatched indices.
Use a hash map to count frequencies of each canonical key. For each group of size k, add k*(k-1)/2 to the total count of unordered pairs.
Time: O(N * L^2) for generating keys (assuming optimized swap generation) plus O(N) for hashing and counting. Space: O(N * L) for storing keys and the hash map. Mention that L is typically small, so this is efficient.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.