LeetCode 2025: Maximum Number of Ways to Partition an Array Solution

Master LeetCode problem 2025 (Maximum Number of Ways to Partition an Array), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2025. Maximum Number of Ways to Partition an Array

Hard

Array Hash Table Counting Enumeration Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible pivot indices and calculate the sum of elements on the left side and the right side of the pivot. By keeping track of the frequency of each difference between the two sums, we can determine the maximum number of ways to partition the array after changing at most one element. We will try changing each element to k and update the counts accordingly.

The algorithm can be summarized as follows:

Initialize a hashmap to store the frequencies of differences between the left and right sums.
Initialize a variable result to store the maximum number of ways to partition the array.
Iterate through all possible pivot indices.
Calculate the sum of elements on the left and right sides of the pivot.
Update the hashmap with the difference between the two sums.
If the pivot is not the last element, calculate the number of ways for the current pivot and update the result.
Try changing each element to k, update the sums, and update the counts in the hashmap.
Update the result with the maximum number of ways obtained.

The time complexity of this algorithm is O(n^2) where n is the length of the input array. The space complexity is O(n) to store the hashmap.

Solution Code

class Solution {
    public int maxNumOfSubarrays(int[] nums, int k) {
        int n = nums.length;
        int result = 0;
        Map<Integer, Integer> count = new HashMap<>();
        
        for (int i = 1; i < n; i++) {
            int leftSum = 0, rightSum = 0;
            for (int j = 0; j < i; j++) {
                leftSum += nums[j];
            }
            for (int j = i; j < n; j++) {
                rightSum += nums[j];
            }
            int diff = leftSum - rightSum;
            count.put(diff, count.getOrDefault(diff, 0) + 1);
            if (i < n - 1) {
                result = Math.max(result, count.get(diff) + 1);
            }
        }
        
        for (int i = 0; i < n; i++) {
            int original = nums[i];
            nums[i] = k;
            int leftSum = 0, rightSum = 0;
            for (int j = 0; j < i; j++) {
                leftSum += nums[j];
            }
            for (int j = i; j < n; j++) {
                rightSum += nums[j];
            }
            int diff = leftSum - rightSum;
            count.put(diff, count.getOrDefault(diff, 0) + 1);
            result = Math.max(result, count.get(diff));
            nums[i] = original;
        }
        
        return result;
    }
}

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