Replace Elements in an Array

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 <= 105
• All the values of nums are distinct.
• operations[i].length == 2
• 1 <= nums[i], operations[i][0], operations[i][1] <= 106
• 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:

class Solution:
    def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
        n = len(nums)
        pos = {}
        for i in range(n):
            pos[nums[i]] = i
        for a, b in operations:
            i = pos[a]
            nums[i] = b
            pos[b] = i
        return nums