Move all zeroes to end of array | Set-2 (Using single traversal)
Last Updated :
02 Aug, 2022
Given an array of n numbers. The problem is to move all the 0’s to the end of the array while maintaining the order of the other elements. Only single traversal of the array is required.
Examples:
Input : arr[] = {1, 2, 0, 0, 0, 3, 6}
Output : 1 2 3 6 0 0 0
Input: arr[] = {0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9}
Output: 1 9 8 4 2 7 6 9 0 0 0 0 0
Algorithm:
moveZerosToEnd(arr, n)
Initialize count = 0
for i = 0 to n-1
if (arr[i] != 0) then
arr[count++]=arr[i]
for i = count to n-1
arr[i] = 0
Flowchart
CPP
#include <iostream>
using namespace std;
void moveZerosToEnd( int arr[], int n)
{
int count = 0;
for ( int i = 0; i < n; i++)
if (arr[i] != 0)
arr[count++] = arr[i];
for ( int i = count; i < n; i++)
arr[i] = 0;
}
void printArray( int arr[], int n)
{
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
}
int main()
{
int arr[] = { 0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Original array: " ;
printArray(arr, n);
moveZerosToEnd(arr, n);
cout << "\nModified array: " ;
printArray(arr, n);
return 0;
}
|
C
#include <stdio.h>
void moveZerosToEnd( int arr[], int n)
{
int count = 0;
for ( int i = 0; i < n; i++)
if (arr[i] != 0)
arr[count++] = arr[i];
for ( int i = count; i < n; i++)
arr[i] = 0;
}
void printArray( int arr[], int n)
{
for ( int i = 0; i < n; i++)
printf ( "%d " , arr[i]);
}
int main()
{
int arr[] = { 0, 1, 9, 8, 4, 0, 0, 2, 7, 0, 6, 0, 9 };
int n = sizeof (arr) / sizeof (arr[0]);
printf ( "Original array: " );
printArray(arr, n);
moveZerosToEnd(arr, n);
printf ( "\nModified array: " );
printArray(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void moveZerosToEnd( int arr[], int n) {
int count = 0 ;
for ( int i = 0 ; i < n; i++)
if (arr[i] != 0 )
arr[count++] = arr[i];
for ( int i = count; i<n;i++)
arr[i]= 0 ;
}
static void printArray( int arr[], int n) {
for ( int i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
}
public static void main(String args[]) {
int arr[] = { 0 , 1 , 9 , 8 , 4 , 0 , 0 , 2 ,
7 , 0 , 6 , 0 , 9 };
int n = arr.length;
System.out.print( "Original array: " );
printArray(arr, n);
moveZerosToEnd(arr, n);
System.out.print( "\nModified array: " );
printArray(arr, n);
}
}
|
Python3
def moveZerosToEnd (arr, n):
count = 0 ;
for i in range ( 0 , n):
if (arr[i] ! = 0 ):
arr[count] = arr[i]
count + = 1
for i in range (count, n):
arr[i] = 0
def printArray(arr, n):
for i in range ( 0 , n):
print (arr[i],end = " " )
arr = [ 0 , 1 , 9 , 8 , 4 , 0 , 0 , 2 ,
7 , 0 , 6 , 0 , 9 ]
n = len (arr)
print ( "Original array:" , end = " " )
printArray(arr, n)
moveZerosToEnd(arr, n)
print ( "\nModified array: " , end = " " )
printArray(arr, n)
|
C#
using System;
class GFG {
static void moveZerosToEnd( int [] arr, int n)
{
int count = 0;
for ( int i = 0; i < n; i++)
if (arr[i] != 0)
arr[count++] = arr[i];
for ( int i = count; i<n;i++)
arr[i]=0;
}
static void printArray( int [] arr, int n)
{
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
}
public static void Main()
{
int [] arr = { 0, 1, 9, 8, 4, 0, 0, 2,
7, 0, 6, 0, 9 };
int n = arr.Length;
Console.Write( "Original array: " );
printArray(arr, n);
moveZerosToEnd(arr, n);
Console.Write( "\nModified array: " );
printArray(arr, n);
}
}
|
Javascript
<script>
function moveZerosToEnd(arr, n)
{
let count = 0;
for (let i = 0; i < n; i++)
if (arr[i] != 0)
{
arr[count] = arr[i];
count = count + 1;
}
for (let i = count; i < n; i++)
arr[i] = 0
}
function printArray(arr, n)
{
for (let i = 0; i < n; i++)
document.write(arr[i] + " " );
}
let arr = [ 0, 1, 9, 8, 4, 0, 0, 2,
7, 0, 6, 0, 9 ];
let n = arr.length;
document.write( "Original array: " );
printArray(arr, n);
moveZerosToEnd(arr, n);
document.write( "<br>" + "Modified array: " );
printArray(arr, n);
</script>
|
Output
Original array: 0 1 9 8 4 0 0 2 7 0 6 0 9
Modified array: 1 9 8 4 2 7 6 9 0 0 0 0 0
Time Complexity: O(n).
Auxiliary Space: O(1).
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