LeetCode 2780: Minimum Index of a Valid Split Solution

Master LeetCode problem 2780 (Minimum Index of a Valid Split), a medium 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.

2780. Minimum Index of a Valid Split

Medium

Array Hash Table Sorting

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array to find the dominant element and its count. Then, we iterate again to keep track of the prefix sums of the dominant element. Finally, we iterate one more time to check for valid splits by comparing the prefix sums.

Find the dominant element and its count in the array.
Calculate the prefix sums of the dominant element.
Check for valid splits by comparing prefix sums.
Return the minimum index of a valid split or -1 if no valid split exists.

Time complexity: O(n) where n is the length of the input array nums. Space complexity: O(1) since we are using only a constant amount of extra space.

Solution Code

class Solution {
    public int minDominoIdx(int[] nums) {
        int n = nums.length;
        int dominant = -1, count = 0;
        for (int num : nums) {
            if (count == 0) {
                dominant = num;
                count = 1;
            } else {
                count += num == dominant ? 1 : -1;
            }
        }
        if (count <= 0) return -1;
        
        int prefixSum = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] == dominant) prefixSum++;
        }
        
        int validSplitIdx = -1;
        int leftPrefixSum = 0;
        for (int i = 0; i < n - 1; i++) {
            if (nums[i] == dominant) leftPrefixSum++;
            if (leftPrefixSum * 2 == i + 1 && (prefixSum - leftPrefixSum) * 2 == n - i - 1) {
                validSplitIdx = i;
                break;
            }
        }
        
        return validSplitIdx;
    }
}

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