The Challenge #2

Understanding how arrays work & their overall time complexity

Question

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Note that you must do this in-place without making a copy of the array.

Example 1:

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]

Example 2:

Input: nums = [0]
Output: [0]

Code

import kotlin.test.*
/**
* Without changing anything in the main make all test cases pass
*/
fun main() {
    
    val list = listOf(
        Pair(intArrayOf(0,1,0,3,12), intArrayOf(1,3,12,0,0)),
        Pair(intArrayOf(0), intArrayOf(0)),
        Pair(intArrayOf(0,1,0), intArrayOf(1,0,0)),
        Pair(intArrayOf(88,4,2,6), intArrayOf(88,4,2,6)),
        Pair(intArrayOf(9,3,4,0,0,4,6), intArrayOf(9,3,4,4,6,0,0)),
    )
    
    list.forEach{ (arr, expected) -> 
        println("-".repeat(30))
        println("Expected -> [${expected}]")
        println("Actual   -> [${moveZeroes(arr)}]")
        assertContentEquals(
            expected,
            moveZeroes(arr),
        )
    }
    
    println("Everything Passed!")
}

fun moveZeroes(nums: IntArray): IntArray {
    // add your solution here
}

Solution

Toggle me!

You thought you could get the easy way out? 😉 😜 😉 😜 😉 😜 Come back later, for the solution