LeetCode-in-Java
3432. Count Partitions with Even Sum Difference
Easy
You are given an integer array nums of length n.
A partition is defined as an index i where 0 <= i < n - 1, splitting the array into two non-empty subarrays such that:
- Left subarray contains indices
[0, i]. - Right subarray contains indices
[i + 1, n - 1].
Return the number of partitions where the difference between the sum of the left and right subarrays is even.
Example 1:
Input: nums = [10,10,3,7,6]
Output: 4
Explanation:
The 4 partitions are:
[10],[10, 3, 7, 6]with a sum difference of10 - 26 = -16, which is even.[10, 10],[3, 7, 6]with a sum difference of20 - 16 = 4, which is even.[10, 10, 3],[7, 6]with a sum difference of23 - 13 = 10, which is even.[10, 10, 3, 7],[6]with a sum difference of30 - 6 = 24, which is even.
Example 2:
Input: nums = [1,2,2]
Output: 0
Explanation:
No partition results in an even sum difference.
Example 3:
Input: nums = [2,4,6,8]
Output: 3
Explanation:
All partitions result in an even sum difference.
Constraints:
2 <= n == nums.length <= 1001 <= nums[i] <= 100
Solution
public class Solution {
public int countPartitions(int[] nums) {
int ct = 0;
int n = nums.length;
for (int i = 0; i < n - 1; i++) {
int sum1 = 0;
int sum2 = 0;
for (int j = 0; j <= i; j++) {
sum1 += nums[j];
}
for (int k = i + 1; k < n; k++) {
sum2 += nums[k];
}
if (Math.abs(sum1 - sum2) % 2 == 0) {
ct++;
}
}
return ct;
}
}