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# Replace Elements in an Array LeetCode Solution

## Problem – Replace Elements in an Array LeetCode Solution

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]` with `operations[i]`.

It is guaranteed that in the `ith` operation:

• `operations[i]` exists in `nums`.
• `operations[i]` 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], operations[i] <= 106`
• `operations[i]` will exist in `nums` when applying the `ith` operation.
• `operations[i]` will not exist in `nums` when applying the `ith` operation.

### Replace Elements in an Array LeetCode Solution in C++

``````class Solution {
public:
vector<int> arrayChange(vector<int>& nums, vector<vector<int>>& operations) {

unordered_map<int,int> hm;     //****store index of element in map
for(int i =0;i<nums.size();i++){

hm[nums[i]] = i;   //** use map to find the index
}

for(int i =0;i<operations.size();i++){

int a  = operations[i];
int b = operations[i];

nums[hm[a]] = b;

hm[b] = hm[a];  //**also store the changed element in hashmap

}
return nums;

}
};
``````

### Replace Elements in an Array LeetCode Solution in Java

``````public int[] arrayChange(int[] A, int[][] op) {
HashMap<Integer,Integer> store= new HashMap<>();
for(int i=0;i<A.length;i++) store.put(A[i],i);
for(var i:op){
A[store.get(i)]= i;              //replace value to its index
store.put(i,store.get(i));       //update new value with its index
}
return A;
}
``````

### Replace Elements in an Array LeetCode Solution in Python

``````class Solution:
def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
replacements = {}
for x, y in reversed(operations):
replacements[x] = replacements.get(y, y)
for idx, val in enumerate(nums):
if val in replacements:
nums[idx] = replacements[val]
return nums
``````
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