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Classic binary search on the answer problem.
Recognize that the minimum eating speed k lies between 1 and the maximum pile size. Use binary search to efficiently find the smallest k such that the total hours required to eat all piles (sum of ceil(pile/k)) is ≤ h. For each candidate k, compute the total hours in O(n) time, resulting in O(n log m) overall, where m is the maximum pile size.
Pro tip: During the interview, explicitly state the time and space complexity and discuss edge cases like when h is less than the number of piles (impossible) or when h is very large (k=1). Also, mention that the binary search is on the answer space, not the array indices.
Clarify that each hour you choose one pile and eat up to k bananas from it. If the pile has fewer than k bananas, you finish it and cannot eat from another pile that hour. The goal is to find the minimum integer k such that all piles are finished within h hours.
The minimum possible speed is 1 (if h is large enough) and the maximum needed speed is the size of the largest pile (since eating faster than that doesn't reduce hours further). So set low = 1, high = max(piles).
For a given k, compute the total hours required: sum over piles of ceil(pile / k). If this sum is ≤ h, then k is feasible; otherwise, it's not.
While low < high, compute mid = (low + high) // 2. If mid is feasible, set high = mid; else set low = mid + 1. At the end, low is the minimum feasible speed.
Time complexity: O(n log m) where n is number of piles and m is max pile size. Space: O(1). Discuss edge cases: if h < n, return -1 or indicate impossible; if h is very large, answer is 1.
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