# Coding Interview — Move Zeros (Array)

Oct 18, 2018 · 2 min read

Given an array of numbers, write a function to move all `0`'s to the end of it while maintaining the relative order of the non-zero elements.

`Input: [0,-1,0,15,7]Output: [-1,15,7,0,0]`

## Thought process:

In order to check all the elements within the input array, we need to loop through the whole array, and that’s `O(n)` for Time Complexity, which is inevitable. How about Space? Naturally we would think to create an empty array `result=[]` and loop through the Input array and do the operations to get the result we want. However that would require additional `O(n)` space. Is it possible to do it without additional space? Yes! Let’s do it `In-Place`!

## Solution:

`moveZeros(numbers) {    // Keep good habit to check corner cases at the beginning.    if (numbers === null || numbers.length < 1) return numbers;    // Use a pointer to keep track our 'zero' location    let idx = 0;    // Loop through the input to find the non-zero elements and update the input array in place!    for (let num of numbers) {        if (num !== 0) {            numbers[idx++] = num;        }    }        while (idx < numbers.length) {        numbers[idx++] = 0;    }};`

## Key Notes:

1. Save Space
2. How to track & update the input array? Use pointer/Swap technique … etc

## What’s the next question?

Stocks 101 — When to buy/sell the stock? Given an array contains the daily price for a stock. Try to find the maximum profit. (thought process and solution)

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## Victor Lin

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