"""
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.

hint:
1 Can you think of a data structure that will allow you to store the position of each number?
2 Use that data structure to instantly replace a number with its new value.

"""
from typing import List


class Solution:
    def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
        val_to_idx = {}
        for i, n in enumerate(nums):
            val_to_idx[n] = i
        for old, new in operations:
            idx = val_to_idx[old]
            val_to_idx[new] = idx
            val_to_idx.pop(old)
        res = [0] * len(nums)
        for key in val_to_idx:
            res[val_to_idx[key]] = key
        return res