2025-12-05 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's December LeetCoding Challenge with cpp.

December LeetCoding Challenge 5

Description

Count Partitions with Even Sum Difference

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 of 10 - 26 = -16, which is even.
  • [10, 10], [3, 7, 6] with a sum difference of 20 - 16 = 4, which is even.
  • [10, 10, 3], [7, 6] with a sum difference of 23 - 13 = 10, which is even.
  • [10, 10, 3, 7], [6] with a sum difference of 30 - 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 <= 100
  • 1 <= nums[i] <= 100

Solution

class Solution {
public:
  int countPartitions(vector<int>& nums) {
    int sum = accumulate(nums.begin(), nums.end(), 0);
    if(sum % 2) return 0;
    return nums.size() - 1;
  }
};

// Accepted
// 616/616 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 78.41 % of cpp submissions (22.3 MB)
Algorithm