Count subarrays consisting of only 0’s and only 1’s in a binary array

Given a binary array consisting of only zeroes and ones. The task is to find:

  • The number of subarrays which has only 1 in it.
  • The number of subarrays which has only 0 in it.

Examples:

Input: arr[] = {0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1}
Output:
The number of subarrays consisting of 0 only: 7
The number of subarrays consisting of 1 only: 7

Input: arr[] = {1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1}
Output:
The number of subarrays consisting of 0 only: 5
The number of subarrays consisting of 1 only: 15



Approach: To count 1’s, the idea is to start traversing the array using a counter. If the current element is 1, increment the counter otherwise add counter*(counter+1)/2 to the number of subarrays and reinitialize counter to 0. Similarly, find the number of subarrays with only 0’s in it.

Below is the implementation of the above approach:

C++

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// C++ program to count the number of subarrays
// that having only 0's and only 1's
#include <bits/stdc++.h>
using namespace std;
  
// Function to count number of subarrays
void countSubarraysof1and0(int a[], int n)
{
    int count1 = 0, count0 = 0;
  
    int number1 = 0, number0 = 0;
  
    // Iterate in the array to find count
    // of subarrays with only 1 in it
    for (int i = 0; i < n; i++) {
        // check if array element
        // is 1 or not
        if (a[i] == 1) {
            count1 += 1;
        }
        else {
            number1 += (count1) * (count1 + 1) / 2;
            count1 = 0;
        }
    }
  
    // Iterate in the array to find count
    // of subarrays with only 0 in it
    for (int i = 0; i < n; i++) {
        // check if array element
        // is 0 or not
        if (a[i] == 0) {
            count0 += 1;
        }
        else {
            number0 += (count0) * (count0 + 1) / 2;
            count0 = 0;
        }
    }
  
    // After iteration completes,
    // check for the last set of subarrays
    if (count1)
        number1 += (count1) * (count1 + 1) / 2;
  
    if (count0)
        number0 += (count0) * (count0 + 1) / 2;
  
    cout << "Count of subarrays of 0 only: " << number0;
    cout << "\nCount of subarrays of 1 only: " << number1;
}
  
// Driver Code
int main()
{
    int a[] = { 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1 };
    int n = sizeof(a) / sizeof(a[0]);
  
    countSubarraysof1and0(a, n);
  
    return 0;
}

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Java














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// Java program to count the number of subarrays 


// that having only 0's and only 1's 


  


import java.io.*; 


  


class GFG { 


      


// Function to count number of subarrays 


static void countSubarraysof1and0(int a[], int n) 


{ 


    int count1 = 0, count0 = 0; 


  


    int number1 = 0, number0 = 0; 


  


    // Iterate in the array to find count 


    // of subarrays with only 1 in it 


    for (int i = 0; i < n; i++) { 


        // check if array element 


        // is 1 or not 


        if (a[i] == 1) { 


            count1 += 1; 


        } 


        else { 


            number1 += (count1) * (count1 + 1) / 2; 


            count1 = 0; 


        } 


    } 


  


    // Iterate in the array to find count 


    // of subarrays with only 0 in it 


    for (int i = 0; i < n; i++) { 


        // check if array element 


        // is 0 or not 


        if (a[i] == 0) { 


            count0 += 1; 


        } 


        else { 


            number0 += (count0) * (count0 + 1) / 2; 


            count0 = 0; 


        } 


    } 


  


    // After iteration completes, 


    // check for the last set of subarrays 


    if (count1>0) 


        number1 += (count1) * (count1 + 1) / 2; 


  


    if (count0>0) 


        number0 += (count0) * (count0 + 1) / 2; 


  


    System.out.println("Count of subarrays of 0 only: " + number0); 


    System.out.println( "\nCount of subarrays of 1 only: " + number1); 


} 


  


// Driver Code 


  


  


    public static void main (String[] args) { 


        int a[] = { 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1 }; 


    int n = a.length; 


  


    countSubarraysof1and0(a, n);; 


    } 


} 


// This code is contributed by inder_verma.. 



















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Python3





# Python 3 program to count the number of

# subarrays that having only 0’s and only 1’s


# Function to count number of subarrays

def countSubarraysof1and0(a, n):

	count1 = 0

	count0 = 0


	number1 = 0

	number0 = 0


	# Iterate in the array to find count

	# of subarrays with only 1 in it

	for i in range(0, n, 1):


		# check if array element is 1 or not

		if (a[i] == 1):

			count1 += 1

		else:

			number1 += ((count1) *

			            (count1 + 1) / 2)

			count1 = 0


	# Iterate in the array to find count

	# of subarrays with only 0 in it

	for i in range(0, n, 1):


		# check if array element

		# is 0 or not

		if (a[i] == 0):

			count0 += 1

		else:

			number0 += (count0) * (count0 + 1) / 2

			count0 = 0


	# After iteration completes,

	# check for the last set of subarrays

	if (count1):

		number1 += (count1) * (count1 + 1) / 2


	if (count0):

		number0 += (count0) * (count0 + 1) / 2


	print(“Count of subarrays of 0 only:”,

	                         int(number0))

	print(“Count of subarrays of 1 only:”,

	                         int(number1))


# Driver Code

if __name__ == ‘__main__’:

	a = [1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1]

	n = len(a)


	countSubarraysof1and0(a, n)


# This code is contributed by

# Surendra_Gangwar



C#














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// C# program to count the number of subarrays 


// that having only 0's and only 1's 


  


using System; 


  


class GFG { 


      


// Function to count number of subarrays 


static void countSubarraysof1and0(int []a, int n) 


{ 


    int count1 = 0, count0 = 0; 


  


    int number1 = 0, number0 = 0; 


  


    // Iterate in the array to find count 


    // of subarrays with only 1 in it 


    for (int i = 0; i < n; i++) { 


        // check if array element 


        // is 1 or not 


        if (a[i] == 1) { 


            count1 += 1; 


        } 


        else { 


            number1 += (count1) * (count1 + 1) / 2; 


            count1 = 0; 


        } 


    } 


  


    // Iterate in the array to find count 


    // of subarrays with only 0 in it 


    for (int i = 0; i < n; i++) { 


        // check if array element 


        // is 0 or not 


        if (a[i] == 0) { 


            count0 += 1; 


        } 


        else { 


            number0 += (count0) * (count0 + 1) / 2; 


            count0 = 0; 


        } 


    } 


  


    // After iteration completes, 


    // check for the last set of subarrays 


    if (count1>0) 


        number1 += (count1) * (count1 + 1) / 2; 


  


    if (count0>0) 


        number0 += (count0) * (count0 + 1) / 2; 


  


    Console.WriteLine("Count of subarrays of 0 only: " + number0); 


    Console.WriteLine( "\nCount of subarrays of 1 only: " + number1); 


} 


  


// Driver Code 


  


  


    public static void Main () { 


        int []a = { 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1 }; 


    int n = a.Length; 


  


    countSubarraysof1and0(a, n);; 


    } 


} 


// This code is contributed by inder_verma.. 



















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PHP














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<?php 


// PHP program to count the number  


// of subarrays that having only  


// 0's and only 1's 


  


// Function to count number of subarrays 


function countSubarraysof1and0($a, $n) 


{ 


    $count1 = 0; $count0 = 0; 


  


    $number1 = 0; $number0 = 0; 


  


    // Iterate in the array to find count 


    // of subarrays with only 1 in it 


    for ($i = 0; $i <$n; $i++)  


    { 


        // check if array element 


        // is 1 or not 


        if ($a[$i] == 1)  


        { 


            $count1 += 1; 


        } 


        else 


        { 


            $number1 += ($count1) *  


                        ($count1 + 1) / 2; 


            $count1 = 0; 


        } 


    } 


  


    // Iterate in the array to find count 


    // of subarrays with only 0 in it 


    for ($i = 0; $i <$n; $i++)  


    { 


        // check if array element 


        // is 0 or not 


        if ($a[$i] == 0)  


        { 


            $count0 += 1; 


        } 


        else 


        { 


            $number0 += ($count0) *  


                        ($count0 + 1) / 2; 


            $count0 = 0; 


        } 


    } 


  


    // After iteration completes, 


    // check for the last set of subarrays 


    if ($count1) 


        $number1 += ($count1) *  


                    ($count1 + 1) / 2; 


  


    if ($count0) 


        $number0 += ($count0) *  


                    ($count0 + 1) / 2; 


  


    echo "Count of subarrays of 0 only: " , $number0; 


    echo "\nCount of subarrays of 1 only: " , $number1; 


} 


  


// Driver Code 


$a = array( 1, 1, 0, 0, 1, 0,  


            1, 0, 1, 1, 1, 1 ); 


$n = count($a); 


  


countSubarraysof1and0($a, $n); 


  


// This code is contributed by inder_verma 


?> 



















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Output:


Count of subarrays of 0 only: 5
Count of subarrays of 1 only: 15





Time Complexity: O(N)

Auxiliary Space: O(1)



          
          
          
          

            

    

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