LeetCode in Kotlin

3379. Transformed Array

Easy

You are given an integer array nums that represents a circular array. Your task is to create a new array result of the same size, following these rules:

For each index i (where 0 <= i < nums.length), perform the following independent actions:

• If nums[i] > 0: Start at index i and move nums[i] steps to the right in the circular array. Set result[i] to the value of the index where you land.
• If nums[i] < 0: Start at index i and move abs(nums[i]) steps to the left in the circular array. Set result[i] to the value of the index where you land.
• If nums[i] == 0: Set result[i] to nums[i].

Return the new array result.

Note: Since nums is circular, moving past the last element wraps around to the beginning, and moving before the first element wraps back to the end.

Example 1:

Input: nums = [3,-2,1,1]

Output: [1,1,1,3]

Explanation:

• For nums[0] that is equal to 3, If we move 3 steps to right, we reach nums[3]. So result[0] should be 1.
• For nums[1] that is equal to -2, If we move 2 steps to left, we reach nums[3]. So result[1] should be 1.
• For nums[2] that is equal to 1, If we move 1 step to right, we reach nums[3]. So result[2] should be 1.
• For nums[3] that is equal to 1, If we move 1 step to right, we reach nums[0]. So result[3] should be 3.

Example 2:

Input: nums = [-1,4,-1]

Output: [-1,-1,4]

Explanation:

• For nums[0] that is equal to -1, If we move 1 step to left, we reach nums[2]. So result[0] should be -1.
• For nums[1] that is equal to 4, If we move 4 steps to right, we reach nums[2]. So result[1] should be -1.
• For nums[2] that is equal to -1, If we move 1 step to left, we reach nums[1]. So result[2] should be 4.

Constraints:

• 1 <= nums.length <= 100
• -100 <= nums[i] <= 100

Solution

import kotlin.math.abs

class Solution {
    fun constructTransformedArray(nums: IntArray): IntArray {
        val n = nums.size
        val res = IntArray(n)
        for (i in 0..<n) {
            if (nums[i] > 0) {
                res[i] = nums[(i + nums[i]) % n]
            } else if (nums[i] < 0) {
                val r: Int = abs(nums[i]) / n
                res[i] = nums[abs(i + nums[i] + r * n + n) % n]
            }
        }
        return res
    }
}

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