Classic binary search on the answer, basically the same shape as the 'eating bananas' problem.
Recognize that the minimum k can be found using binary search on the answer, since feasibility is monotonic: if a given k works, any larger k also works. For a fixed k, compute the total hours needed by summing ceil(vault/k) for each vault, and check if it's ≤ h. Binary search k in the range [1, max(vault)] to find the smallest feasible k.
Pro tip: Clarify that k must be an integer and that each hour you can remove up to k units from a single vault, so you can't split an hour across vaults. Also, mention that if h is less than the number of vaults, it's impossible, so return -1 or handle appropriately.
Restate the problem: given an array of positive integers and an integer h, find the minimum integer k such that all vaults can be emptied in at most h hours, where each hour you can remove at most k units from one vault. Note that k must be at least 1 and at most the maximum vault value.
For a given k, compute the total hours required: sum over all vaults of ceil(vault / k). If this sum is ≤ h, then k is feasible; otherwise, it's not.
Since feasibility is monotonic (if k works, any larger k also works), binary search for the smallest feasible k in the range [1, max(vault)]. Initialize low=1, high=max(vault), and while low < high, compute mid and adjust based on feasibility.
The feasibility check takes O(n) time, and binary search runs O(log(max(vault))) iterations, so overall time complexity is O(n log(max(vault))). Space complexity is O(1).
If h < number of vaults, it's impossible to empty all vaults because each vault requires at least one hour, so return -1 or indicate impossibility. Also, ensure integer division and ceiling are handled correctly.
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DP table, O(n * target) time and space, pretty textbook.
Start by clarifying the problem constraints (e.g., number of integers, target size) to choose between dynamic programming and meet-in-the-middle. Then present a DP solution using a boolean table or bitset, and explain how to reconstruct the subset by backtracking through the DP table. Finally, discuss time and space complexity, and mention optimizations like bitset or pruning.
Pro tip: Mention that if the target is large but the number of elements is small (≤40), meet-in-the-middle is more efficient; this shows you consider trade-offs. Also, emphasize that reconstruction requires storing parent pointers or using a 2D DP table, and discuss how to handle memory constraints.
Ask about input size, target range, and whether negative numbers or zeros are allowed. This determines the best algorithm and avoids over-engineering.
For small n and target, use DP with a boolean table; for large target but small n, use meet-in-the-middle. Explain the trade-offs.
Define dp[i][s] = true if a subset of first i elements sums to s. Transition: dp[i][s] = dp[i-1][s] or dp[i-1][s - nums[i]]. Base case dp[0][0] = true.
If dp[n][target] is true, backtrack from (n, target): if dp[i-1][target] is true, skip element i; else include it and move to (i-1, target - nums[i]).
Time O(n * target), space O(n * target) or O(target) with bitset. Mention meet-in-the-middle O(2^(n/2)) time and space, and pruning techniques.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.