LeetCode 2809: Minimum Time to Make Array Sum At Most x Solution

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2809. Minimum Time to Make Array Sum At Most x

Hard

Array Dynamic Programming Sorting

Problem Explanation

Explanation

To solve this problem, we can use a binary search approach to find the minimum time required to make the sum of elements in nums1 less than or equal to x. We will iterate through possible time values, and for each time value, we will simulate the process of incrementing elements based on nums2 and selecting an index to make it zero. We will keep track of the sum of nums1 at each step and adjust our binary search based on whether the sum is less than x or not.

Algorithm:
1. Initialize low as 0 and high as a large value, which is the maximum possible time needed.
2. Perform binary search on the possible time values.
3. In each iteration of binary search, calculate the sum of nums1 after mid seconds.
4. If the sum is less than or equal to x, update high to mid - 1.
5. If the sum is greater than x, update low to mid + 1.
6. Return the final low if it is within the bounds, else return -1.
Time Complexity: O(n log x), where n is the length of nums1 and x is the maximum possible time needed.
Space Complexity: O(1)

Solution Code

class Solution {
    public int minOperations(int[] nums1, int[] nums2, int x) {
        int n = nums1.length;
        int sum = 0;
        for (int i = 0; i < n; i++) {
            sum += nums1[i];
        }

        int low = 0, high = n;
        int ans = Integer.MAX_VALUE;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            int target = x - sum;

            if (target == 0) {
                ans = Math.min(ans, mid);
            }

            for (int i = 0; i < mid; i++) {
                target -= (i < n) ? nums1[i] : 0;
                target -= (i < n) ? nums2[i] : 0;
                ans = (target >= 0) ? Math.min(ans, mid) : ans;
            }

            if (target > 0) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }

        return (ans == Integer.MAX_VALUE) ? -1 : ans;
    }
}

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