LeetCode 1674: Minimum Moves to Make Array Complementary Solution

Master LeetCode problem 1674 (Minimum Moves to Make Array Complementary), 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.

1674. Minimum Moves to Make Array Complementary

Medium

Array Hash Table Prefix Sum

Problem Explanation

Explanation

To solve this problem, we can iterate through each pair of elements (i, n-1-i) in the array nums and calculate the range of possible values that would make the pair complementary. By doing so, we can determine the minimum moves required to make the pair complementary. The overall minimum moves required for the entire array will be the sum of minimum moves for each pair.

The key idea is to map the possible values that make a pair complementary to their respective counts. We can then iterate through this mapping to find the optimal value for each pair. The time complexity of this approach is O(n + limit).

Solution Code

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minMoves(int[] nums, int limit) {
        int n = nums.length;
        Map<Integer, Integer> freq = new HashMap<>();
        for (int i = 0; i < n / 2; i++) {
            int a = nums[i];
            int b = nums[n - 1 - i];
            int min = Math.min(a, b);
            int max = Math.max(a, b);
            int sum = a + b;
            freq.put(2, freq.getOrDefault(2, 0) + 2);
            freq.put(min + 1, freq.getOrDefault(min + 1, 0) - 1);
            freq.put(min + 1, freq.getOrDefault(min + 1, 0) + 1);
            freq.put(sum, freq.getOrDefault(sum, 0) - 1);
            freq.put(sum + 1, freq.getOrDefault(sum + 1, 0) + 1);
        }
        
        int res = n;
        int curr = 0;
        for (int i = 2; i <= 2 * limit; i++) {
            curr += freq.getOrDefault(i, 0);
            res = Math.min(res, curr);
        }
        
        return res;
    }
}

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