LeetCode in Kotlin

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.

Solution

class Solution {
    fun arrayChange(nums: IntArray, operations: Array<IntArray>): IntArray {
        val map = HashMap<Int, Int>()
        for (i in nums.indices) {
            map[nums[i]] = i
        }
        for (operation in operations) {
            val index = map[operation[0]]!!
            nums[index] = operation[1]
            map[operation[1]] = index
        }
        return nums
    }
}

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