LeetCode 2035: Partition Array Into Two Arrays to Minimize Sum Difference Solution

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2035. Partition Array Into Two Arrays to Minimize Sum Difference

Hard

Array Two Pointers Binary Search Dynamic Programming Bit Manipulation Ordered Set Bitmask

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. The goal is to partition the given array into two arrays such that the absolute difference of the sums of the two arrays is minimized. We can iterate through all possible partitions and calculate the absolute difference for each partition. We will maintain a 2D array to store the minimum absolute difference possible for each possible sum.

Let's define dp[i][j] as the minimum absolute difference that can be achieved by partitioning the first i elements into two arrays such that the sum of one array is j. We can calculate dp[i][j] using the following recurrence relation: dp[i][j] = min(dp[i-1][j], dp[i-1][j-nums[i]] + abs(nums[i]))

The final answer will be the minimum of dp[n][s] and dp[n][total_sum - s], where n is the total number of elements and s is the sum of one of the arrays.

Solution Code

class Solution {
    public int minDifference(int[] nums) {
        int n = nums.length / 2;
        int totalSum = Arrays.stream(nums).sum();
        int[][] dp = new int[n + 1][totalSum + 1];
        
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= totalSum; j++) {
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = 0; k < i * 2; k++) {
                    if (j - nums[k] >= 0) {
                        dp[i][j] = Math.min(dp[i][j], dp[i - 1][j - nums[k]] + Math.abs(nums[k]));
                    }
                }
            }
        }
        
        int minDiff = Integer.MAX_VALUE;
        for (int j = 0; j <= totalSum; j++) {
            minDiff = Math.min(minDiff, Math.min(dp[n][j], dp[n][totalSum - j]));
        }
        
        return minDiff;
    }
}

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