2464. Minimum Subarrays in a Valid Split

Explanation

To solve this problem, we can iterate through the input array and maintain two pointers - one at the start and one at the end. We will split the array into two subarrays at each possible split point and check if the split is valid. A split is considered valid if the left subarray contains at least one element and the sum of elements in the left subarray is less than or equal to the sum of elements in the right subarray.

We will keep track of the minimum number of subarrays we can get for a valid split. To do this, we will iterate through all possible split points and update the minimum count of subarrays whenever we find a valid split.

View Leetcode Problem

class Solution {
    public int minSubarrays(int[] nums) {
        int n = nums.length;
        int minSubarrays = Integer.MAX_VALUE;

        for (int i = 1; i < n; i++) {
            int leftSum = 0;
            int rightSum = 0;
            int leftCount = 0;
            int rightCount = 0;

            for (int j = 0; j < i; j++) {
                leftSum += nums[j];
                leftCount++;
            }

            for (int j = i; j < n; j++) {
                rightSum += nums[j];
                rightCount++;

                if (leftSum <= rightSum) {
                    minSubarrays = Math.min(minSubarrays, leftCount + 1);
                    break;
                }
            }
        }

        return minSubarrays == Integer.MAX_VALUE ? -1 : minSubarrays;
    }
}

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