3432. Count Partitions with Even Sum Difference LeetCode Solution
In this guide, you will get 3432. Count Partitions with Even Sum Difference LeetCode Solution with the best time and space complexity. The solution to Count Partitions with Even Sum Difference problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of 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.
[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.
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
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
3432. Count Partitions with Even Sum Difference LeetCode Solution in C++
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;
}
}; /* code provided by PROGIEZ */
3432. Count Partitions with Even Sum Difference LeetCode Solution in Java
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;
}
} // code provided by PROGIEZ
3432. Count Partitions with Even Sum Difference LeetCode Solution in Python
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 # code by PROGIEZ
