Class Solution
java.lang.Object
g3401_3500.s3432_count_partitions_with_even_sum_difference.Solution
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 <= 100
1 <= nums[i] <= 100
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Constructor Summary
Constructors -
Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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countPartitions
public int countPartitions(int[] nums)
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