Move all Zeros

Software Development
#1

Given an array of random numbers, Push all the zero’s of a given array to the end of the array. For example, if the given arrays is {1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0}, it should be changed to {1, 9, 8, 4, 2, 7, 6, 0, 0, 0, 0}. The order of all other elements should be same. Expected time complexity is O(n) and extra space is O(1).
Example:

Input : arr[] = {1, 2, 0, 4, 3, 0, 5, 0};
Output : arr[] = {1, 2, 4, 3, 5, 0, 0};

Input : arr[] = {1, 2, 0, 0, 0, 3, 6};
Output : arr[] = {1, 2, 3, 6, 0, 0, 0};

There can be many ways to solve this problem. Following is a simple and interesting way to solve this problem.
Traverse the given array ‘arr’ from left to right. While traversing, maintain count of non-zero elements in array. Let the count be ‘count’. For every non-zero element arr[i], put the element at ‘arr[count]’ and increment ‘count’. After complete traversal, all non-zero elements have already been shifted to front end and ‘count’ is set as index of first 0. Now all we need to do is that run a loop which makes all elements zero from ‘count’ till end of the array.
Below is the implementation of the above approach.

/* Java program to push zeroes to back of array */

import java.io.*;

class PushZero

{

// Function which pushes all zeros to end of an array.

static void pushZerosToEnd( int arr[], int n)

{

int count = 0 ; // Count of non-zero elements

// Traverse the array. If element encountered is

// non-zero, then replace the element at index 'count'

// with this element

for ( int i = 0 ; i < n; i++)

if (arr[i] != 0 )

arr[count++] = arr[i]; // here count is

// incremented

// Now all non-zero elements have been shifted to

// front and 'count' is set as index of first 0.

// Make all elements 0 from count to end.

while (count < n)

arr[count++] = 0 ;

}

/*Driver function to check for above functions*/

public static void main (String[] args)

{

int arr[] = { 1 , 9 , 8 , 4 , 0 , 0 , 2 , 7 , 0 , 6 , 0 , 9 };

int n = arr.length;

pushZerosToEnd(arr, n);

System.out.println( "Array after pushing zeros to the back: " );

for ( int i= 0 ; i<n; i++)

System.out.print(arr[i]+ " " );

}

}

Output:

Array after pushing all zeros to end of array : 1 9 8 4 2 7 6 9 0 0 0 0

Time Complexity: O(n) where n is number of elements in input array.
Auxiliary Space: O(1)