Java Program to Segregate 0s on Left Side & 1s on Right Side of the Array

Last Updated : 18 Aug, 2021

You are given an array of 0s and 1s in random order. Segregate 0s on the left side and 1s on the right side of the array. The basic goal is to traverse array elements and sort in segregating 0s and 1s.

Illustration:

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

Approaches:

Using segregation by counting
Using sorting of an array
Using pointers

Below all three approaches all discussed in detail:

Approach 1:

Count the number of 0s.
Traversing over the whole array for looking out indices where zeros are present
Maintaining a count and incrementing as 0 appears
Print all zeros to the front
The remaining number of 1s will be 1- (total number of 0s)
Print the remaining elements

Below is the implementation to segregate 0s and 1s using the above algorithm:

Java

// Java code to Segregate 0s and 1s in an array
 
// Importing generic libraries
import java.util.*;
// Importing Array libraries
import java.util.Arrays;
 
public class GFG {
 
    // Function to segregate 0s and 1s
    static void segregate0and1(int arr[], int n)
    {
        // Counts the no of zeros in array
        int count = 0;
 
        // Iteration over array
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
 
                // Incrementing the count
                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 ");
 
        // Iteration over array
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
 
    // Main driver function
    public static void main(String[] args)
    {
        // Array taken for consideration
        int arr[] = new int[] { 0, 1, 0, 1, 1, 1 };
 
        // Using inbuilt function to store array size
        int n = arr.length;
 
        // Calling function that segregates array
        segregate0and1(arr, n);
 
        // Printing the above segregated array
        print(arr, n);
    }
}

Output:

Array after segregation is 0 0 1 1 1 1

Time Complexity: O(n)

Approach 2: Using sort() function

sort() Is a method is a java.util.Arrays class method.

Syntax:

public static void sort(int[] arr, int from_Index, int to_Index)

Parameters:

arr        - the array to be sorted
from_Index - the index of the first element, inclusive, to be sorted
to_Index   - the index of the last element, exclusive, to be sorted

Return Type:

This method doesn't return any value

Java

// Java code to Segregate 0s and 1s in an array
 
// Importing generic libraries
import java.util.*;
 
public class GFG {
 
    // Function to print segregated array
    // Taking arguments- array and array size
    static void print(int arr[], int n)
    {
        System.out.print("Array after segregation is ");
 
        // Iteration over array
        for (int i = 0; i < n; ++i)
 
            // Printing array elements
            System.out.print(arr[i] + " ");
    }
 
    // Main driver function
    public static void main(String[] args)
    {
        // Array taken for consideration
        int arr[] = new int[] { 0, 1, 0, 1, 1, 1 };
 
        // Using length inbuilt function to
        int n = arr.length;
 
        // Using sort inbuilt function
        Arrays.sort(arr);
 
        // Printing elements after executing sorting
        print(arr, n);
    }
}

Output:

Array after segregation is 0 0 1 1 1 1

Time Complexity O(nlongn)

Approach 3: Maintain the left pointer and swap with the position of the left when zero found in the array and increment the left pointer.

Java

// Java code to Segregate 0s and 1s in an array
 
// Importing generic libraries
import java.util.*;
import java.io.*;
 
class GFG {
 
    // Print function outside main to print elements
    static void print(int a[])
    {
 
        System.out.print("Array after segregation is: ");
 
        // Iteration over array using array
        // class inbuilt function .length()
        for (int i = 0; i < a.length; ++i) {
 
            // Printing elements in array
            System.out.print(a[i] + " ");
        }
    }
 
    // Main driver method
    public static void main(String[] args)
    {
        // Random array taken for consideration
        int a[] = { 1, 1, 0, 0, 0, 0, 1, 1 };
 
        // Maintaining left pointer
        int left = 0;
 
        // Iteration over array using length function
        for (int i = 0; i < a.length; ++i) {
 
            // If zeros are present
            if (a[i] == 0) {
 
                // Swap the elements using
                // temporary variable
                int temp = a[left];
                a[left] = a[i];
                a[i] = temp;
 
                // Pre incrementing left pointer
                ++left;
            }
        }
 
        // Calling above function to
        // print updated array
        print(a);
    }
}

Output:

Array after segregation is: 0 0 0 0 1 1 1 1

Time Complexity O(n)

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