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Problem solution · Java

Count Partitions with Even Sum Difference

Count Partitions with Even Sum Difference: a Java solution using sliding window or two pointers. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sliding window or two pointers
Source
walkccc LeetCode Solutions
Length
10 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sliding window or two pointers

For Count Partitions with Even Sum Difference, the implementation maintains a moving interval and updates only the information that enters or leaves the window.

  1. Choose the invariant that makes a window valid or useful.
  2. Advance the right boundary and add the new element.
  3. Move the left boundary only as needed while maintaining the invariant and updating the answer.

Code notes

  • 10 lines of Java from the credited upstream file 3432.java.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Confirm that neither pointer moves backwards; if so, the scan is usually linear apart from the window’s data-structure operations.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCount Partitions with Even Sum Difference · JavaJava
Use this to learn the idea, then write your own version.
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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