LeetCode-in-Java

3810. Minimum Operations to Reach Target Array

Medium

You are given two integer arrays nums and target, each of length n, where nums[i] is the current value at index i and target[i] is the desired value at index i.

You may perform the following operation any number of times (including zero):

• Choose an integer value x
• Find all maximal contiguous segments where nums[i] == x (a segment is maximal if it cannot be extended to the left or right while keeping all values equal to x)
• For each such segment [l, r], update simultaneously:
  • nums[l] = target[l], nums[l + 1] = target[l + 1], ..., nums[r] = target[r]

Return the minimum number of operations required to make nums equal to target.

Example 1:

Input: nums = [1,2,3], target = [2,1,3]

Output: 2

Explanation:

• Choose x = 1: maximal segment [0, 0] updated -> nums becomes [2, 2, 3]
• Choose x = 2: maximal segment [0, 1] updated (nums[0] stays 2, nums[1] becomes 1) -> nums becomes [2, 1, 3]
• Thus, 2 operations are required to convert nums to target.

Example 2:

Input: nums = [4,1,4], target = [5,1,4]

Output: 1

Explanation:

• Choose x = 4: maximal segments [0, 0] and [2, 2] updated (nums[2] stays 4) -> nums becomes [5, 1, 4]
• Thus, 1 operation is required to convert nums to target.

Example 3:

Input: nums = [7,3,7], target = [5,5,9]

Output: 2

Explanation:

• Choose x = 7: maximal segments [0, 0] and [2, 2] updated -> nums becomes [5, 3, 9]
• Choose x = 3: maximal segment [1, 1] updated -> nums becomes [5, 5, 9]
• Thus, 2 operations are required to convert nums to target.

Constraints:

• 1 <= n == nums.length == target.length <= 105
• 1 <= nums[i], target[i] <= 105

Solution

import java.util.HashSet;
import java.util.Set;

public class Solution {
    public int minOperations(int[] nums, int[] target) {
        Set<Integer> virelantos = new HashSet<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != target[i]) {
                virelantos.add(nums[i]);
            }
        }
        return virelantos.size();
    }
}

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