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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)

• Java
• Python
• C++
• Go
• TypeScript

• 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).

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