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]

**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:**

Array after segregation is 0 0 1 1 1 1

**Time complexity:** O(n)

// Thanks san4net for suggesting this method.

Please write comments if you find any of the above algorithms/code incorrect, or a better ways to solve the same problem.