LeetCode 2604: Minimum Time to Eat All Grains Solution

Master LeetCode problem 2604 (Minimum Time to Eat All Grains), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2604. Minimum Time to Eat All Grains

Hard

Array Two Pointers Binary Search Sorting

Problem Explanation

Explanation:

To solve this problem, we can use binary search to find the minimum time required to eat all the grains. We need to find the minimum time t such that all the grains can be eaten within t hours. We can start with a lower bound of 1 hour and an upper bound of the maximum number of grains multiplied by the maximum eating time. We then perform a binary search within this range to find the minimum time required.

In each iteration of the binary search, we calculate the total time required to eat all grains with the given time limit. If the total time is less than or equal to the time limit, we update the upper bound. Otherwise, we update the lower bound. We continue this process until the lower bound is equal to the upper bound, at which point we have found the minimum time required.

Solution Code

class Solution {
    public int minTime(int[] grains, int time) {
        int low = 1, high = 0;
        for (int g : grains) {
            high = Math.max(high, g);
        }
        
        while (low < high) {
            int mid = low + (high - low) / 2;
            int total = 0;
            for (int g : grains) {
                total += (g + mid - 1) / mid;
            }
            if (total <= time) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        
        return low;
    }
}

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