## DEV Community

Abhishek Chaudhary

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# 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
``````