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

• Difficulty Level : Easy
• Last Updated : 14 Jun, 2022

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.

Naive approach: The idea is to sort the array of elements, this will make sure that all the negative elements will come before all the positive elements.
Below is the implementation of the above approach:

## C++

 #include #include#includeusing namespace std;void move(vector& arr){  sort(arr.begin(),arr.end());}int main() {     vector arr = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };      move(arr);    for (int e : arr)       cout<

## Java

 // Java program to move all negative numbers to the// beginning and all positive numbers to the end with// constant extra spaceimport java.util.*;public class Gfg {        public static void move(int[] arr)    {        Arrays.sort(arr);    }     // Driver code    public static void main(String[] args)    {        int[] arr = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };        move(arr);        for (int e : arr)            System.out.print(e + " ");    }}// This article is contributed by aadityapburujwale

## Python3

 # Python code for the same approachdef move(arr):  arr.sort() # driver codearr = [ -1, 2, -3, 4, 5, 6, -7, 8, 9 ]move(arr)for e in arr:    print(e , end = " ") # This code is contributed by shinjanpatra

## C#

 // C# program to move all negative numbers to the// beginning and all positive numbers to the end with// constant extra spaceusing System;using System.Collections;public class Gfg {        public static void move(int[] arr)    {        Array.Sort(arr);    }     // Driver code    public static void Main()    {        int[] arr = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };        move(arr);        foreach (int e in arr)            Console.Write(e + " ");    }}// This code is contributed by Saurabh Jaiswal

## Javascript



Output

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

Time Complexity: O(n*log(n)), Where n is the length of the given array.
Auxiliary Space: O(n)

Efficient 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 arrayvoid printArray(int arr[], int n){    for (int i = 0; i < n; i++)        printf("%d ", arr[i]);} // Driver codeint main(){    int arr[] = { -1, 2, -3, 4, 5, 6, -7, 8, 9 };    int n = sizeof(arr) / sizeof(arr[0]);    rearrange(arr, n);    printArray(arr, n);    return 0;}

## Java

 // Java program to put all negative// numbers before positive numbersimport 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 codearr = [-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 numbersusing 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.



## Javascript



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 and 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 â‰¤ 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 sidevoid 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 arrayvoid 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// approachimport java.io.*; class GFG{ // Function to shift all the// negative elements on left sidestatic 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 arraystatic 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 codepublic 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 arraydef 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    else if 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    else if arr[left]>0 and arr[right]>0:      right-=1    else:      left+=1      right-=1       # Function to print the arraydef display(arr):  for i in range(len(arr)):    print(arr[i], end=" ")  print() # Driver Codeif __name__ == "__main__":  arr=[-12, 11, -13, -5, \       6, -7, 5, -3, 11]  n=len(arr)  shiftall(arr,0,n-1)  display(arr) # Sumit Singh

## C#

 // C# program of the above// approachusing System.IO;using System;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)        {            Console.Write(arr[i] + " ");                     }        Console.WriteLine();    }         // Drive code                      static void Main()    {        int[] arr = {-12, 11, -13, -5, 6, -7, 5, -3, 11};        int arr_size = arr.Length;        shiftall(arr, 0, arr_size - 1);        display(arr, arr_size - 1);    }} // This code is contributed by avanitrachhadiya2155

## Javascript



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.

This article is contributed by Apoorva. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeek’s main page and help other Geeks.

Approach 3:
Here, we will use the famous Dutch National Flag Algorithm for two “colors”. The first color will be for all negative integers and the second color will be for all positive integers. We will divide the array into three partitions with the help of two pointers, low and high.

1. ar[1…low-1] negative integers
2. ar[low…high] unknown
3. ar[high+1…N] positive integers

Now, we explore the array with the help of low pointer, shrinking the unknown partition, and moving elements to their correct partition in the process. We do this until we have explored all the elements, and size of the unknown partition shrinks to zero.

Below is the implementation of the above approach:

## C++

 #include using namespace std; // Swap Function.void swap(int &a,int &b){  int temp =a;  a=b;  b=temp;}   // Using Dutch National Flag Algorithm.void reArrange(int arr[],int n){      int low =0,high = n-1;      while(low0){          high--;      }else{        swap(arr[low],arr[high]);      }    }}void displayArray(int arr[],int n){  for(int i=0;i

## Java

 // Java program to move all negative numbers to the// beginning and all positive numbers to the end with// constant extra space public class Gfg {     // a utility function to swap two elements of an array    public static void swap(int[] ar, int i, int j)    {        int t = ar[i];        ar[i] = ar[j];        ar[j] = t;    }     // function to shilf all negative integers to the left    // and all positive integers to the right    // using Dutch National Flag Algorithm    public static void move(int[] ar)    {        int low = 0;        int high = ar.length - 1;        while (low <= high) {            if (ar[low] <= 0)                low++;            else                swap(ar, low, high--);        }    }     // Driver code    public static void main(String[] args)    {        int[] ar = { 1, 2,  -4, -5, 2, -7, 3,                     2, -6, -8, -9, 3, 2,  1 };        move(ar);        for (int e : ar)            System.out.print(e + " ");    }} // This code is contributed by Vedant Harshit

## Python3

 # Python code for the approach # Using Dutch National Flag Algorithm.def reArrange(arr, n):    low,high = 0, n - 1    while(low 0):            high -= 1        else:            arr[low],arr[high] = arr[high],arr[low] def displayArray(arr, n):     for i in range(n):        print(arr[i],end = " ")       print('') # driver code# Dataarr = [1, 2,  -4, -5, 2, -7, 3, 2, -6, -8, -9, 3, 2,  1]n = len(arr)reArrange(arr,n)displayArray(arr,n) # This code is contributed by shinjanpatra

## C#

 // Include namespace systemusing System; // C# program to move all negative numbers to the// beginning and all positive numbers to the end with// constant extra spacepublic class Gfg{  // a utility function to swap two elements of an array  public static void swap(int[] ar, int i, int j)  {    var t = ar[i];    ar[i] = ar[j];    ar[j] = t;  }  // function to shilf all negative integers to the left  // and all positive integers to the right  // using Dutch National Flag Algorithm  public static void move(int[] ar)  {    var low = 0;    var high = ar.Length - 1;    while (low <= high)    {      if (ar[low] <= 0)      {        low++;      }      else      {        Gfg.swap(ar, low, high--);      }    }  }   // Driver code  public static void Main(String[] args)  {    int[] ar = {1, 2, -4, -5, 2, -7, 3, 2, -6, -8, -9, 3, 2, 1};    Gfg.move(ar);    foreach (int e in ar)    {            Console.Write(e.ToString() + " ");    }  }} // This code is contributed by mukulsomukesh

## Javascript



Output

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

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

The order of elements does not matter here. Explanation and code contributed by Vedant Harshit

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