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

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Algorithm:

```moveZerosToEnd(arr, n)
Initialize count = 0
for i = 0 to n-1
if (arr[i] != 0) then
swap(arr[count++], arr[i])
```

## 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 ` `    ``// swap the element at index 'count' with the ` `    ``// element at index 'i' ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(arr[i] != 0) ` `            ``swap(arr[count++], arr[i]); ` `} ` ` `  `// function to print the array elements ` `void` `printArray(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``cout << arr[i] << ``" "``; ` `} ` ` `  `// Driver program to test above ` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 1, 9, 8, 4, 0, 0, 2, ` `                         ``7, 0, 6, 0, 9 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << ``"Original array: "``; ` `    ``printArray(arr, n); ` ` `  `    ``moveZerosToEnd(arr, n); ` ` `  `    ``cout << ``"\nModified array: "``; ` `    ``printArray(arr, n); ` ` `  `    ``return` `0; ` `} `

## 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``; ` `    ``int` `temp; ` ` `  `    ``// Traverse the array. If arr[i] is  ` `    ``// non-zero, then swap the element at  ` `    ``// index 'count' with the element at  ` `    ``// index 'i' ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `    ``if` `((arr[i] != ``0``)) { ` `        ``temp = arr[count]; ` `        ``arr[count] = arr[i]; ` `        ``arr[i] = temp; ` `        ``count = count + ``1``; ` `    ``} ` `    ``} ` `} ` ` `  `// function to print the array elements ` `static` `void` `printArray(``int` `arr[], ``int` `n) { ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``System.out.print(arr[i] + ``" "``); ` `} ` ` `  `// Driver program to test above ` `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); ` `} ` `} ` ` `  `// This code is contributed by Nikita Tiwari. `

## 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 ` `    ``# swap the element at index 'count' with the ` `    ``# element at index 'i' ` `    ``for` `i ``in` `range``(``0``, n): ` `        ``if` `(arr[i] !``=` `0``): ` `            ``arr[count], arr[i] ``=` `arr[i], arr[count] ` `            ``count``+``=``1` ` `  ` `  `# 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 ` `# Smitha Dinesh Semwal `

## 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; ` `        ``int` `temp; ` ` `  `        ``// Traverse the array. If arr[i] is ` `        ``// non-zero, then swap the element at ` `        ``// index 'count' with the element at ` `        ``// index 'i' ` `        ``for` `(``int` `i = 0; i < n; i++) ` `        ``{ ` `            ``if` `((arr[i] != 0))  ` `            ``{ ` `                ``temp = arr[count]; ` `                ``arr[count] = arr[i]; ` `                ``arr[i] = temp; ` `                ``count = count + 1; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// function to print the array elements ` `    ``static` `void` `printArray(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``Console.Write(arr[i] + ``" "``); ` `    ``} ` ` `  `    ``// Driver program to test above ` `    ``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); ` `    ``} ` `} ` ` `  `// This code is contributed by Sam007 `

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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