2295. Replace Elements in an Array

Problem

You are given a 0-indexed array nums that consists of n distinct positive integers. Apply m operations to this array, where in the ith operation you replace the number operations[i][0] with operations[i][1].

It is guaranteed that in the ith operation:

• operations[i][0] exists in nums.

• operations[i][1] does not exist in nums.

Return the array obtained after applying all the operations.

  Example 1:

Input: nums = [1,2,4,6], operations = [[1,3],[4,7],[6,1]]
Output: [3,2,7,1]
Explanation: We perform the following operations on nums:
- Replace the number 1 with 3. nums becomes [3,2,4,6].
- Replace the number 4 with 7. nums becomes [3,2,7,6].
- Replace the number 6 with 1. nums becomes [3,2,7,1].
We return the final array [3,2,7,1].

Example 2:

Input: nums = [1,2], operations = [[1,3],[2,1],[3,2]]
Output: [2,1]
Explanation: We perform the following operations to nums:
- Replace the number 1 with 3. nums becomes [3,2].
- Replace the number 2 with 1. nums becomes [3,1].
- Replace the number 3 with 2. nums becomes [2,1].
We return the array [2,1].

  Constraints:

• n == nums.length

• m == operations.length

• 1 <= n, m <= 10^5

• All the values of nums are distinct.

• operations[i].length == 2

• 1 <= nums[i], operations[i][0], operations[i][1] <= 10^6

• operations[i][0] will exist in nums when applying the ith operation.

• operations[i][1] will not exist in nums when applying the ith operation.

Solution (Java)

class Solution {
    public int[] arrayChange(int[] nums, int[][] operations) {
        HashMap<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < nums.length; i++) {
            map.put(nums[i], i);
        }
        for (int[] operation : operations) {
            int index = map.get(operation[0]);
            nums[index] = operation[1];
            map.put(operation[1], index);
        }
        return nums;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).