Segregate 0s and 1s in an array

You are given an array of 0s and 1s in random order. Segregate 0s on left side and 1s on right side of the array. Traverse array only once.

Input array   =  [0, 1, 0, 1, 0, 0, 1, 1, 1, 0] 
Output array =  [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Method 1 (Count 0s or 1s)
Thanks to Naveen for suggesting this method.
1) Count the number of 0s. Let count be C.
2) Once we have count, we can put C 0s at the beginning and 1s at the remaining n – C positions in array.

Time Complexity : O(n)

C++

// C++ code to Segregate 0s and 1s in an array
#include <bits/stdc++.h>
using namespace std;

// Function to segregate 0s and 1s
void segregate0and1(int arr[], int n)
{
    int count = 0; // Counts the no of zeros in arr

    for (int i = 0; i < n; i++) {
        if (arr[i] == 0)
            count++;
    }

    // Loop fills the arr with 0 until count
    for (int i = 0; i < count; i++)
        arr[i] = 0;

    // Loop fills remaining arr space with 1
    for (int i = count; i < n; i++)
        arr[i] = 1;
}

// Function to print segregated array
void print(int arr[], int n)
{
    cout << "Array after segregation is ";

    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Driver function
int main()
{
    int arr[] = { 0, 1, 0, 1, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    
    segregate0and1(arr, n);
    print(arr, n);
    
    return 0;
}

// This code is contributed by Sahil_Bansall

Java

// Java code to Segregate 0s and 1s in an array
class GFG {
    
    // function to segregate 0s and 1s
    static void segregate0and1(int arr[], int n)
    {
        int count = 0; // counts the no of zeros in arr
    
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
                count++;
        }

        // loop fills the arr with 0 until count
        for (int i = 0; i < count; i++)
            arr[i] = 0;

        // loop fills remaining arr space with 1
        for (int i = count; i < n; i++)
            arr[i] = 1;
    }
    
    // function to print segregated array
    static void print(int arr[], int n)
    {
        System.out.print("Array after segregation is ");
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");    
    }
    
    // driver function
    public static void main(String[] args)
    {
        int arr[] = new int[]{ 0, 1, 0, 1, 1, 1 };
        int n = arr.length;

        segregate0and1(arr, n);
        print(arr, n);
        
    }
}

// This code is contributed by Kamal Rawal

Python3

# Python 3 code to Segregate
# 0s and 1s in an array

# Function to segregate 0s and 1s
def segregate0and1(arr, n) :
    
    # Counts the no of zeros in arr
    count = 0 

    for i in range(0, n) :
        if (arr[i] == 0) :
            count = count + 1

    # Loop fills the arr with 0 until count
    for i in range(0, count) :
        arr[i] = 0

    # Loop fills remaining arr space with 1
    for i in range(count, n) :
        arr[i] = 1
        

# Function to print segregated array
def print_arr(arr , n) :
    print( "Array after segregation is ",end = "")

    for i in range(0, n) :
        print(arr[i] , end = " ")
        

# Driver function
arr = [ 0, 1, 0, 1, 1, 1 ]
n = len(arr)
    
segregate0and1(arr, n)
print_arr(arr, n)


        
# This code is contributed by Nikita Tiwari.

C#

// C# code to Segregate 0s and 1s in an array
using System;

class GFG {
    
    // function to segregate 0s and 1s
    static void segregate0and1(int []arr, int n)
    {   
        // counts the no of zeros in arr
        int count = 0; 
    
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
                count++;
        }

        // loop fills the arr with 0 until count
        for (int i = 0; i < count; i++)
            arr[i] = 0;

        // loop fills remaining arr space with 1
        for (int i = count; i < n; i++)
            arr[i] = 1;
    }
    
    // function to print segregated array
    static void print(int []arr, int n)
    {
        Console.WriteLine("Array after segregation is ");
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " "); 
    }
    
    // driver function
    public static void Main()
    {
        int []arr = new int[]{ 0, 1, 0, 1, 1, 1 };
        int n = arr.Length;

        segregate0and1(arr, n);
        print(arr, n);
        
    }
}

//This code is contributed by vt_m.

Output :

Array after segregation is 0 0 1 1 1 1

The method 1 traverses the array two times. Method 2 does the same in a single pass.

Method 2 (Use two indexes to traverse)
Maintain two indexes. Initialize first index left as 0 and second index right as n-1.

Do following while left < right
a) Keep incrementing index left while there are 0s at it
b) Keep decrementing index right while there are 1s at it
c) If left < right then exchange arr[left] and arr[right]

Implementation:

C/C++

// C program to sort a binary array in one pass
#include<stdio.h>

/*Function to put all 0s on left and all 1s on right*/
void segregate0and1(int arr[], int size)
{
    /* Initialize left and right indexes */
    int left = 0, right = size-1;

    while (left < right)
    {
        /* Increment left index while we see 0 at left */
        while (arr[left] == 0 && left < right)
            left++;

        /* Decrement right index while we see 1 at right */
        while (arr[right] == 1 && left < right)
            right--;

        /* If left is smaller than right then there is a 1 at left
          and a 0 at right.  Exchange arr[left] and arr[right]*/
        if (left < right)
        {
            arr[left] = 0;
            arr[right] = 1;
            left++;
            right--;
        }
    }
}

/* driver program to test */
int main()
{
    int arr[] = {0, 1, 0, 1, 1, 1};
    int i, arr_size = sizeof(arr)/sizeof(arr[0]);

    segregate0and1(arr, arr_size);

    printf("Array after segregation ");
    for (i = 0; i < 6; i++)
        printf("%d ", arr[i]);

    getchar();
    return 0;
}

Java

class Segregate 
{
    /*Function to put all 0s on left and all 1s on right*/
    void segregate0and1(int arr[], int size) 
    {
        /* Initialize left and right indexes */
        int left = 0, right = size - 1;

        while (left < right) 
        {
            /* Increment left index while we see 0 at left */
            while (arr[left] == 0 && left < right)
               left++;

            /* Decrement right index while we see 1 at right */
            while (arr[right] == 1 && left < right)
                right--;

            /* If left is smaller than right then there is a 1 at left
               and a 0 at right.  Exchange arr[left] and arr[right]*/
            if (left < right) 
            {
                arr[left] = 0;
                arr[right] = 1;
                left++;
                right--;
            }
        }
    }
    
    /* Driver Program to test above functions */
    public static void main(String[] args) 
    {
        Segregate seg = new Segregate();
        int arr[] = new int[]{0, 1, 0, 1, 1, 1};
        int i, arr_size = arr.length;

        seg.segregate0and1(arr, arr_size);

        System.out.print("Array after segregation is ");
        for (i = 0; i < 6; i++)
            System.out.print(arr[i] + " ");
    }
}

Python

# Python program to sort a binary array in one pass

# Function to put all 0s on left and all 1s on right
def segregate0and1(arr, size):
    # Initialize left and right indexes
    left, right = 0, size-1
    
    while left < right:
        # Increment left index while we see 0 at left
        while arr[left] == 0 and left < right:
            left += 1

        # Decrement right index while we see 1 at right
        while arr[right] == 1 and left < right:
            right -= 1

        # If left is smaller than right then there is a 1 at left
        # and a 0 at right. Exchange arr[left] and arr[right]
        if left < right:
            arr[left] = 0
            arr[right] = 1
            left += 1
            right -= 1

    return arr

# driver program to test
arr = [0, 1, 0, 1, 1, 1]
arr_size = len(arr)
print("Array after segregation")
print(segregate0and1(arr, arr_size))

# This code is contributed by Pratik Chhajer

C#

// C# program to sort a binary array in one pass
using System;

class Segregate 
{
    /*Function to put all 0s on
      left and all 1s on right*/
    void segregate0and1(int []arr, int size) 
    {
        /* Initialize left and right indexes */
        int left = 0, right = size - 1;

        while (left < right) 
        {
            /* Increment left index while
               we see 0 at left */
            while (arr[left] == 0 && left < right)
            left++;

            /* Decrement right index while 
               we see 1 at right */
            while (arr[right] == 1 && left < right)
                right--;

            /* If left is smaller than right then
               there is a 1 at left and a 0 at right. 
               Exchange arr[left] and arr[right]*/
            if (left < right) 
            {
                arr[left] = 0;
                arr[right] = 1;
                left++;
                right--;
            }
        }
    }
    
    /* Driver Program to test above functions */
    public static void Main() 
    {
        Segregate seg = new Segregate();
        int []arr = new int[]{0, 1, 0, 1, 1, 1};
        int i, arr_size = arr.Length;

        seg.segregate0and1(arr, arr_size);

        Console.WriteLine("Array after segregation is ");
        for (i = 0; i < 6; i++)
            Console.Write(arr[i] + " ");
    }
}

//This code is contributed by vt_m.

Output:

Array after segregation is 0 0 1 1 1 1

Time Complexity: O(n)

Another approach :
1. Take two pointer type0(for element 0) starting from beginning (index = 0) and type1(for element 1) starting from end (index = array.length-1).
Initialize type0 = 0 and type1 = array.length-1
2. It is intended to Put 1 to the right side of the array. Once it is done, then 0 will definitely towards left side of array.

C++

// C++ program to sort a 
// binary array in one pass
#include <bits/stdc++.h>
using namespace std;

/*Function to put all 0s on 
left and all 1s on right*/
void segregate0and1(int arr[], 
                    int size)
{
    int type0 = 0;
    int type1 = size - 1;
    
    while(type0 < type1)
    {
        if(arr[type0] == 1)
        {
            swap(arr[type0], 
                 arr[type1]);
            type1--;
        }
        else
        type0++;
    }
}

// Driver Code
int main()
{
    int arr[] = {0, 1, 0, 1, 1, 1};
    int i, arr_size = sizeof(arr) / 
                      sizeof(arr[0]);

    segregate0and1(arr, arr_size);

    cout << "Array after segregation is ";
    for (i = 0; i < arr_size; i++)
        cout << arr[i] << " ";

    return 0;
}

Java

// Java code to segregate 0 and 1
import java.util.*;

class GFG{
/**
Method for segregation 0 and 1 given input array
*/
static void segregate0and1(int arr[]) {
        int type0 = 0;
        int type1 = arr.length - 1;
        
        while (type0 < type1) {
            if (arr[type0] == 1) {
                // swap
                arr[type1] = arr[type1]+ arr[type0];
                arr[type0] = arr[type1]-arr[type0];
                arr[type1] = arr[type1]-arr[type0];
                type1--;
            } else {
                type0++;
            }
        }

    }
    
// Driver program
public static void main(String[] args) {     
        
        int[] array = {1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 1};
        
        segregate0and1(array);
        
        for(int a : array){
            System.out.print(a+" ");
        }
    }
}

Output: