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Formatted question description: https://leetcode.ca/all/2295.html

# 2295. Replace Elements in an Array

• Difficulty: Medium.
• Related Topics: Array, Hash Table, Simulation.
• Similar Questions: Find All Numbers Disappeared in an Array.

## Problem

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 (Java, C++, Python)

• class Solution {
public int[] arrayChange(int[] nums, int[][] operations) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
map.put(nums[i], i);
}
for (int[] operation : operations) {
int index = map.get(operation[0]);
nums[index] = operation[1];
map.put(operation[1], index);
}
return nums;
}
}

############

class Solution {
public int[] arrayChange(int[] nums, int[][] operations) {
Map<Integer, Integer> d = new HashMap<>();
for (int i = 0; i < nums.length; ++i) {
d.put(nums[i], i);
}
for (var op : operations) {
int a = op[0], b = op[1];
nums[d.get(a)] = b;
d.put(b, d.get(a));
}
return nums;
}
}

• class Solution:
def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
d = {v: i for i, v in enumerate(nums)}
for a, b in operations:
nums[d[a]] = b
d[b] = d[a]
return nums

############

# 2295. Replace Elements in an Array
# https://leetcode.com/problems/replace-elements-in-an-array/

class Solution:
def arrayChange(self, nums: List[int], operations: List[List[int]]) -> List[int]:
mp = {x:i for i, x in enumerate(nums)}

for a, b in operations:
currIndex = mp[a]
nums[currIndex] = b
mp[b] = currIndex

return nums


• class Solution {
public:
vector<int> arrayChange(vector<int>& nums, vector<vector<int>>& operations) {
unordered_map<int, int> d;
for (int i = 0; i < nums.size(); ++i) {
d[nums[i]] = i;
}
for (auto& op : operations) {
int a = op[0], b = op[1];
nums[d[a]] = b;
d[b] = d[a];
}
return nums;
}
};

• func arrayChange(nums []int, operations [][]int) []int {
d := map[int]int{}
for i, v := range nums {
d[v] = i
}
for _, op := range operations {
a, b := op[0], op[1]
nums[d[a]] = b
d[b] = d[a]
}
return nums
}

• function arrayChange(nums: number[], operations: number[][]): number[] {
const d = new Map(nums.map((v, i) => [v, i]));
for (const [a, b] of operations) {
nums[d.get(a)] = b;
d.set(b, d.get(a));
}
return nums;
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).