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Kotlin-based LeetCode algorithm problem solutions, regularly updated
2295\. Replace Elements in an Array
Medium
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.