# Move all negative numbers to beginning and positive to end with constant extra space

An array contains both positive and negative numbers in random order. Rearrange the array elements so that all negative numbers appear before all positive numbers.
Examples :

```Input: -12, 11, -13, -5, 6, -7, 5, -3, -6
Output: -12 -13 -5 -7 -3 -6 11 6 5

```

Note: Order of elements is not important here.

Approach 1:
The idea is to simply apply the partition process of quicksort

## C++

 `// A C++ program to put all negative` `// numbers before positive numbers` `#include ` `using` `namespace` `std;`   `void` `rearrange(``int` `arr[], ``int` `n)` `{` `    ``int` `j = 0;` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(arr[i] < 0) {` `            ``if` `(i != j)` `                ``swap(arr[i], arr[j]);` `            ``j++;` `        ``}` `    ``}` `}`   `// A utility function to print an array` `void` `printArray(``int` `arr[], ``int` `n)` `{` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``printf``(``"%d "``, arr[i]);` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``rearrange(arr, n);` `    ``printArray(arr, n);` `    ``return` `0;` `}`

## Java

 `// Java program to put all negative` `// numbers before positive numbers` `import` `java.io.*;`   `class` `GFG {`   `    ``static` `void` `rearrange(``int` `arr[], ``int` `n)` `    ``{` `        ``int` `j = ``0``, temp;` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(arr[i] < ``0``) {` `                ``if` `(i != j) {` `                    ``temp = arr[i];` `                    ``arr[i] = arr[j];` `                    ``arr[j] = temp;` `                ``}` `                ``j++;` `            ``}` `        ``}` `    ``}`   `    ``// A utility function to print an array` `    ``static` `void` `printArray(``int` `arr[], ``int` `n)` `    ``{` `        ``for` `(``int` `i = ``0``; i < n; i++)` `            ``System.out.print(arr[i] + ``" "``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``int` `arr[] = { -``1``, ``2``, -``3``, ``4``, ``5``, ``6``, -``7``, ``8``, ``9` `};` `        ``int` `n = arr.length;`   `        ``rearrange(arr, n);` `        ``printArray(arr, n);` `    ``}` `}`   `// This code is contributed by Nikita Tiwari.`

## Python3

 `# A Python 3 program to put` `# all negative numbers before` `# positive numbers`   `def` `rearrange(arr, n ) :`   `    ``# Please refer partition() in ` `    ``# below post` `    ``# https://www.geeksforgeeks.org / quick-sort / j = 0` `    ``j ``=` `0` `    ``for` `i ``in` `range``(``0``, n) :` `        ``if` `(arr[i] < ``0``) :` `            ``temp ``=` `arr[i]` `            ``arr[i] ``=` `arr[j]` `            ``arr[j]``=` `temp` `            ``j ``=` `j ``+` `1` `    ``print``(arr)`   `# Driver code` `arr ``=` `[``-``1``, ``2``, ``-``3``, ``4``, ``5``, ``6``, ``-``7``, ``8``, ``9``]` `n ``=` `len``(arr)` `rearrange(arr, n)`     `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to put all negative` `// numbers before positive numbers` `using` `System;`   `class` `GFG {` `    ``static` `void` `rearrange(``int``[] arr, ``int` `n)` `    ``{`   `        ``int` `j = 0, temp;` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``if` `(arr[i] < 0) {` `                ``temp = arr[i];` `                ``arr[i] = arr[j];` `                ``arr[j] = temp;` `                ``j++;` `            ``}` `        ``}` `    ``}`   `    ``// A utility function to print an array` `    ``static` `void` `printArray(``int``[] arr, ``int` `n)` `    ``{` `        ``for` `(``int` `i = 0; i < n; i++)` `            ``Console.Write(arr[i] + ``" "``);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };` `        ``int` `n = arr.Length;`   `        ``rearrange(arr, n);` `        ``printArray(arr, n);` `    ``}` `}`   `// This code is contributed by nitin mittal.`

## PHP

 ``

Output

```-1 -3 -7 4 5 6 2 8 9

```

Time complexity: O(N)
Auxiliary Space: O(1)

Two Pointer Approach: The idea is to solve this problem with constant space and linear time is by using a two-pointer or two-variable approach where we simply take two variables like left and right which hold the 0 and N-1 indexes. Just need to check that :

1. Check If the left and right pointer elements are negative then simply increment the left pointer.
2. Otherwise, if the left element is positive and the right element is negative then simply swap the elements, and Simultaneously increment or decrement the left and right pointers.
3. Else if the left element is positive and the right element is also positive then simply decrement the right pointer.
4. Repeat the above 3 steps until the left pointer ≤ right pointer.

Below is the implementation of the above approach:

## C++

 `// C++ program of the above ` `// approach`   `#include ` `using` `namespace` `std;`   `// Function to shift all the` `// negative elements on left side` `void` `shiftall(``int` `arr[], ``int` `left, ` `              ``int` `right)` `{` `  `  `  ``// Loop to iterate over the ` `  ``// array from left to the right` `  ``while` `(left<=right)` `  ``{` `    ``// Condition to check if the left` `    ``// and the right elements are ` `    ``// negative` `    ``if` `(arr[left] < 0 && arr[right] < 0)` `      ``left+=1;` `    `  `    ``// Condition to check if the left ` `    ``// pointer element is positive and ` `    ``// the right pointer element is negative` `    ``else` `if` `(arr[left]>0 && arr[right]<0)` `    ``{` `      ``int` `temp=arr[left];` `      ``arr[left]=arr[right];` `      ``arr[right]=temp;` `      ``left+=1;` `      ``right-=1;` `    ``}` `    `  `    ``// Condition to check if both the ` `    ``// elements are positive` `    ``else` `if` `(arr[left]>0 && arr[right] >0)` `      ``right-=1;` `    ``else``{` `      ``left += 1;` `      ``right -= 1;` `    ``}` `  ``}` `}`   `// Function to print the array` `void` `display(``int` `arr[], ``int` `right){` `  `  `  ``// Loop to iterate over the element` `  ``// of the given array` `  ``for` `(``int` `i=0;i<=right;++i){` `    ``cout<

## Java

 `// Java program of the above` `// approach` `import` `java.io.*;`   `class` `GFG{`   `// Function to shift all the` `// negative elements on left side` `static` `void` `shiftall(``int``[] arr, ``int` `left,` `                     ``int` `right)` `{` `    `  `    ``// Loop to iterate over the` `    ``// array from left to the right` `    ``while` `(left <= right) ` `    ``{` `        `  `        ``// Condition to check if the left` `        ``// and the right elements are` `        ``// negative` `        ``if` `(arr[left] < ``0` `&& arr[right] < ``0``)` `            ``left++;`   `        ``// Condition to check if the left` `        ``// pointer element is positive and` `        ``// the right pointer element is negative` `        ``else` `if` `(arr[left] > ``0` `&& arr[right] < ``0``)` `        ``{` `            ``int` `temp = arr[left];` `            ``arr[left] = arr[right];` `            ``arr[right] = temp;` `            ``left++;` `            ``right--;` `        ``}`   `        ``// Condition to check if both the` `        ``// elements are positive` `        ``else` `if` `(arr[left] > ``0` `&& arr[right] > ``0``)` `            ``right--;` `        ``else` `        ``{` `            ``left++;` `            ``right--;` `        ``}` `    ``}` `}`   `// Function to print the array` `static` `void` `display(``int``[] arr, ``int` `right)` `{` `    `  `    ``// Loop to iterate over the element` `    ``// of the given array` `    ``for``(``int` `i = ``0``; i <= right; ++i)` `        ``System.out.print(arr[i] + ``" "``);` `        `  `    ``System.out.println();` `}`   `// Drive code` `public` `static` `void` `main(String[] args)` `{` `    ``int``[] arr = { -``12``, ``11``, -``13``, -``5``, ` `                   ``6``, -``7``, ``5``, -``3``, ``11` `};` `                   `  `    ``int` `arr_size = arr.length;`   `    ``// Function Call` `    ``shiftall(arr, ``0``, arr_size - ``1``);` `    ``display(arr, arr_size - ``1``);` `}` `}`   `// This code is contributed by dhruvgoyal267`

## Python3

 `# Python3 program of the ` `# above approach`   `# Function to shift all the ` `# the negative elements to` `# the left of the array` `def` `shiftall(arr,left,right):` `  `  `  ``# Loop to iterate while the ` `  ``# left pointer is less than` `  ``# the right pointer` `  ``while` `left<``=``right:` `    `  `    ``# Condition to check if the left` `    ``# and right pointer negative` `    ``if` `arr[left] < ``0` `and` `arr[right] < ``0``:` `      ``left``+``=``1` `      `  `    ``# Condition to check if the left ` `    ``# pointer element is positive and ` `    ``# the right pointer element is` `    ``# negative` `    ``elif` `arr[left]>``0` `and` `arr[right]<``0``:` `      ``arr[left], arr[right] ``=` `\` `              ``arr[right],arr[left]` `      ``left``+``=``1` `      ``right``-``=``1` `      `  `    ``# Condition to check if the left` `    ``# pointer is positive and right ` `    ``# pointer as well` `    ``elif` `arr[left]>``0` `and` `arr[right]>``0``:` `      ``right``-``=``1` `    ``else``:` `      ``left``+``=``1` `      ``right``-``=``1` `      `  `# Function to print the array` `def` `display(arr):` `  ``for` `i ``in` `range``(``len``(arr)):` `    ``print``(arr[i], end``=``" "``)` `  ``print``()`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `  ``arr``=``[``-``12``, ``11``, ``-``13``, ``-``5``, \` `       ``6``, ``-``7``, ``5``, ``-``3``, ``11``]` `  ``n``=``len``(arr)` `  ``shiftall(arr,``0``,n``-``1``)` `  ``display(arr)`   `# Sumit Singh`

Output

```-12 -3 -13 -5 -7 6 5 11 11

```

This is an in-place rearranging algorithm for arranging the positive and negative numbers where the order of elements is not maintained.

Time Complexity: O(N)
Auxiliary Space: O(1)

The problem becomes difficult if we need to maintain the order of elements. Please refer to Rearrange positive and negative numbers with constant extra space for details.

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