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3432. Count Partitions with Even Sum Difference 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int countPartitions(std::vector<int>& nums) {
    // If we add the same number in the left subarray and remove it from the
    // right subarray, then the difference remains the same parity. So, just
    // return the number of ways to partition the array into two subarrays when
    // the array sum is even.
    return accumulate(nums.begin(), nums.end(), 0) % 2 == 0 ? nums.size() - 1
                                                            : 0;
  }
};

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class Solution {
  public int countPartitions(int[] nums) {
    // If we add the same number in the left subarray and remove it from the
    // right subarray, then the difference remains the same parity. So, just
    // return the number of ways to partition the array into two subarrays when
    // the array sum is even.
    return Arrays.stream(nums).sum() % 2 == 0 ? nums.length - 1 : 0;
  }
}

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class Solution:
  def countPartitions(self, nums: list[int]) -> int:
    # If we add the same number in the left subarray and remove it from the
    # right subarray, then the difference remains the same parity. So, just
    # return the number of ways to partition the array into two subarrays when
    # the array sum is even.
    return len(nums) - 1 if sum(nums) % 2 == 0 else 0