LeetCode 2333: Minimum Sum of Squared Difference Solution

Master LeetCode problem 2333 (Minimum Sum of Squared Difference), a medium 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.

2333. Minimum Sum of Squared Difference

Medium

Array Binary Search Greedy Sorting Heap (Priority Queue)

Problem Explanation

Explanation

To solve this problem, we can approach it in the following way:

Calculate the initial sum of squared differences between nums1 and nums2.
Iterate through each pair of corresponding elements in nums1 and nums2, and calculate the potential reduction in the sum of squared differences if we adjust the elements by +1 or -1.
Use dynamic programming to keep track of the minimum sum of squared differences achievable after making the allowed adjustments.

The time complexity of this solution is O(n) where n is the length of the arrays nums1 and nums2. The space complexity is also O(n) for storing the dynamic programming table.

Solution Code

class Solution {
    public int minSumOfSquaredDifference(int[] nums1, int[] nums2, int k1, int k2) {
        int n = nums1.length;
        int sumDiff = 0;

        for (int i = 0; i < n; i++) {
            sumDiff += (nums1[i] - nums2[i]) * (nums1[i] - nums2[i]);
        }

        int[][] dp = new int[n][k1 + 1];

        for (int i = 0; i < n; i++) {
            for (int k = 0; k <= k1; k++) {
                dp[i][k] = Integer.MAX_VALUE;
            }
        }

        for (int i = 0; i < n; i++) {
            dp[i][0] = sumDiff;
        }

        for (int i = 0; i < n; i++) {
            for (int k = 1; k <= k1; k++) {
                for (int j = 0; j < n; j++) {
                    int diff = Math.abs(nums1[i] + k - nums2[j]);
                    int newDiff = (nums1[i] + k - nums2[j]) * (nums1[i] + k - nums2[j]);

                    if (j > 0) {
                        dp[i][k] = Math.min(dp[i][k], dp[j - 1][k - 1] + sumDiff - newDiff + diff * diff);
                    } else {
                        dp[i][k] = Math.min(dp[i][k], sumDiff - newDiff + diff * diff);
                    }
                }
            }
        }

        int result = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            result = Math.min(result, dp[i][k1]);
        }

        return result;
    }
}

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