LeetCode 2449: Minimum Number of Operations to Make Arrays Similar Solution

Master LeetCode problem 2449 (Minimum Number of Operations to Make Arrays Similar), 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.

2449. Minimum Number of Operations to Make Arrays Similar

Hard

Array Greedy Sorting

Problem Explanation

Explanation

To solve this problem, we need to determine the minimum number of operations required to make the given array nums similar to the target array. We can achieve this by calculating the frequency of elements in both arrays and then comparing these frequencies. The idea is to match the frequency of each element in nums with the frequency of the same element in the target array.

We can iterate through the arrays and count the occurrences of each element. Then, by comparing the frequencies of the elements in both arrays, we can determine the minimum number of operations required to make the arrays similar.

The steps involved in the algorithm are as follows:

Create a map to store the frequency of elements in both arrays.
Iterate through both arrays and count the occurrences of each element.
Compare the frequencies of elements in both arrays to find the minimum number of operations required.
Return the minimum number of operations.

The time complexity of this algorithm is O(n) where n is the length of the arrays. The space complexity is also O(n) to store the frequency of elements.

Solution Code

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

class Solution {
    public int minOperations(int[] nums, int[] target) {
        Map<Integer, Integer> freq = new HashMap<>();

        for (int num : nums) {
            freq.put(num, freq.getOrDefault(num, 0) + 1);
        }
        
        for (int num : target) {
            freq.put(num, freq.getOrDefault(num, 0) - 1);
        }

        int minOps = 0;
        for (int count : freq.values()) {
            minOps += Math.abs(count);
        }

        return minOps / 2;
    }
}

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