LeetCode in Kotlin

977. Squares of a Sorted Array

Easy

Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.

Example 1:

Input: nums = [-4,-1,0,3,10]

Output: [0,1,9,16,100]

Explanation: After squaring, the array becomes [16,1,0,9,100]. After sorting, it becomes [0,1,9,16,100].

Example 2:

Input: nums = [-7,-3,2,3,11]

Output: [4,9,9,49,121]

Constraints:

• 1 <= nums.length <= 104
• -104 <= nums[i] <= 104
• nums is sorted in non-decreasing order.

Follow up: Squaring each element and sorting the new array is very trivial, could you find an O(n) solution using a different approach?

Solution

import kotlin.math.abs

class Solution {
    fun sortedSquares(nums: IntArray): IntArray {
        var l = 0
        var r = nums.size - 1
        val res = IntArray(nums.size)
        // Iterate res from back to front. put the bigger of abs(l) * abs(l) and abs(r) * abs(r) at
        // res[i] and increment respectively
        for (i in nums.indices.reversed()) {
            // If the negative is larger, put it at the end and increment left ptr to next lower
            // negative
            if (abs(nums[l]) > nums[r]) {
                res[i] = nums[l] * nums[l]
                l++
            } else {
                res[i] = nums[r] * nums[r]
                r--
            }
        }
        return res
    }
}

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