# Move all zeroes to end of array | Set-2 (Using single traversal)

• Difficulty Level : Easy
• Last Updated : 23 Jun, 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

 `// C++ implementation to move all zeroes at` `// the end of array` `#include ` `using` `namespace` `std;`   `// function to move all zeroes at` `// the end of array` `void` `moveZerosToEnd(``int` `arr[], ``int` `n)` `{` `    ``// Count of non-zero elements` `    ``int` `count = 0;`   `    ``// Traverse the array. If arr[i] is non-zero, then` `    ``// update the value of arr at index count to arr[i]` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``if` `(arr[i] != 0)` `            ``arr[count++] = arr[i];` `   `  `    ``// Update all elements at index >=count with value 0` `    ``for` `(``int` `i = count; i

## Java

 `// Java implementation to move ` `// all zeroes at the end of array` `import` `java.io.*;`   `class` `GFG {`   `// function to move all zeroes at` `// the end of array` `static` `void` `moveZerosToEnd(``int` `arr[], ``int` `n) {` `    `  `    ``// Count of non-zero elements` `    ``int` `count = ``0``;`   `    ``// Traverse the array. If arr[i] is non-zero, then` `    ``// update the value of arr at index count to arr[i]` `    ``for` `(``int` `i = ``0``; i < n; i++)` `        ``if` `(arr[i] != ``0``)` `            ``arr[count++] = arr[i];` `   `  `    ``// Update all elements at index >=count with value 0` `    ``for` `(``int` `i = count; i

## Python3

 `# Python implementation to move all zeroes at` `# the end of array`   `# function to move all zeroes at` `# the end of array` `def` `moveZerosToEnd (arr, n):`   `    ``# Count of non-zero elements` `    ``count ``=` `0``;`     `    ``# Traverse the array. If arr[i] is non-zero, then` `    ``# update the value of arr at index count to arr[i]` `    ``for` `i ``in` `range``(``0``, n):` `        ``if` `(arr[i] !``=` `0``):` `            ``arr[count] ``=` `arr[i]` `            ``count``+``=``1` `   `  `    ``# Update all elements at index >=count with value 0` `    ``for` `i ``in` `range``(count, n):` `        ``arr[i] ``=` `0`       `# function to print the array elements` `def` `printArray(arr, n):`   `    ``for` `i ``in` `range``(``0``, n):` `        ``print``(arr[i],end``=``" "``)`     `# Driver program to test above` `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)`   `# This code is contributed by` `# Ashutosh Singh`

## C#

 `// C# implementation to move` `// all zeroes at the end of array` `using` `System;`   `class` `GFG {`   `    ``// function to move all zeroes at` `    ``// the end of array` `    ``static` `void` `moveZerosToEnd(``int``[] arr, ``int` `n)` `    ``{` `        ``// Count of non-zero elements` `        ``int` `count = 0;`   `      ``// Traverse the array. If arr[i] is non-zero, then` `      ``// update the value of arr at index count to arr[i]` `      ``for` `(``int` `i = 0; i < n; i++)` `        ``if` `(arr[i] != 0)` `            ``arr[count++] = arr[i];` `   `  `      ``// Update all elements at index >=count with value 0` `      ``for` `(``int` `i = count; i

## Javascript

 ``

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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