README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2200-2299/2295.Replace%20Elements%20in%20an%20Array/README_EN.md	1445	Weekly Contest 296 Q3	Array Hash Table Simulation

2295. Replace Elements in an Array

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Description

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.

Solutions

Solution 1: Hash Table

First, we use a hash table $d$ to record the indices of each number in the array $\textit{nums}$. Then, we iterate through the operation array $\textit{operations}$. For each operation $[x, y]$, we replace the number at index $d[x]$ in $\textit{nums}$ with $y$, and update the index of $y$ in $d$ to $d[x]$.

Finally, we return $\textit{nums}$.

The time complexity is $O(n + m)$, and the space complexity is $O(n)$. Here, $n$ and $m$ are the lengths of the array $\textit{nums}$ and the operation array $\textit{operations}$, respectively.

Python3

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

Java

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

C++

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 x = op[0], y = op[1];
            nums[d[x]] = y;
            d[y] = d[x];
        }
        return nums;
    }
};

Go

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

TypeScript

function arrayChange(nums: number[], operations: number[][]): number[] {
    const d: Map<number, number> = new Map(nums.map((x, i) => [x, i]));
    for (const [x, y] of operations) {
        nums[d.get(x)!] = y;
        d.set(y, d.get(x)!);
    }
    return nums;
}