Leetcode 2295 Replace Elements in an Array Solution in c++ | Hindi Coding Community

 

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.

vector<int> arrayChange(vector<int>& A, vector<vector<int>>& op) {

unordered_map<int,int> store;

for(int i=0;i<A.size();i++) store[A[i]]=i;

for(auto i:op){

A[store[i[0]]]= i[1];

store[i[1]]= store[i[0]];

}

return A;

}