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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 0indexed 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 innums
. 
operations[i][1]
does not exist innums
.
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 innums
when applying theith
operation. 
operations[i][1]
will not exist innums
when applying theith
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; } }

Todo

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/replaceelementsinanarray/ 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
Explain:
nope.
Complexity:
 Time complexity : O(n).
 Space complexity : O(n).