LeetCode 3432: Count Partitions with Even Sum Difference Solution

Master LeetCode problem 3432 (Count Partitions with Even Sum Difference), a easy 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.

3432. Count Partitions with Even Sum Difference

Easy

Array Math Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can iterate through the array and keep track of the running prefix sum. At each index, we can calculate the sum of elements on the left of that index and the sum of elements on the right of that index. The difference between these two sums will determine if the partition has an even sum difference. We can count the number of such partitions.

Initialize a variable count to keep track of the number of partitions with even sum difference.
Iterate through the array and calculate the prefix sum.
For each index i, calculate the sum of elements on the left and right of that index.
Check if the difference between the left sum and right sum is even. If it is, increment the count.
Return the count as the final result.

Time Complexity

The time complexity of this solution is O(n), where n is the number of elements in the array.

Space Complexity

The space complexity of this solution is O(1) as we are using a constant amount of extra space.

Solution Code

public int countPartitionsWithEvenSumDifference(int[] nums) {
    int count = 0;
    for (int i = 0; i < nums.length - 1; i++) {
        int leftSum = 0;
        int rightSum = 0;
        for (int j = 0; j <= i; j++) {
            leftSum += nums[j];
        }
        for (int j = i + 1; j < nums.length; j++) {
            rightSum += nums[j];
        }
        if ((leftSum - rightSum) % 2 == 0) {
            count++;
        }
    }
    return count;
}

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