Count of substrings that start and end with 1 in given Binary String

Given a binary string, count the number of substrings that start and end with 1. 

Examples:

Input: “00100101”
Output: 3
Explanation: three substrings are “1001”, “100101” and “101”

Input: “1001”
Output: 1
Explanation: one substring “1001”

Count of substrings that start and end with 1 in given Binary String using Nested Loop:

A Simple Solution is to run two loops. Outer loops pick every 1 as a starting point and the inner loop searches for ending 1 and increments count whenever it finds 1.

Below is the implementation of above approach:

C++

// A simple C++ program to count number of
// substrings starting and ending with 1
#include <iostream>

using namespace std;

int countSubStr(char str[])
{
    int res = 0; // Initialize result

    // Pick a starting point
    for (int i = 0; str[i] != '\0'; i++) {
        if (str[i] == '1') {
            // Search for all possible ending point
            for (int j = i + 1; str[j] != '\0'; j++)
                if (str[j] == '1')
                    res++;
        }
    }
    return res;
}

// Driver program to test above function
int main()
{
    char str[] = "00100101";
    cout << countSubStr(str);
    return 0;
}

Java

// A simple Java program to count number of
// substrings starting and ending with 1

class CountSubString {
    int countSubStr(char str[], int n)
    {
        int res = 0; // Initialize result

        // Pick a starting point
        for (int i = 0; i < n; i++) {
            if (str[i] == '1') {
                // Search for all possible ending point
                for (int j = i + 1; j < n; j++) {
                    if (str[j] == '1')
                        res++;
                }
            }
        }
        return res;
    }

    // Driver program to test the above function
    public static void main(String[] args)
    {
        CountSubString count = new CountSubString();
        String string = "00100101";
        char str[] = string.toCharArray();
        int n = str.length;
        System.out.println(count.countSubStr(str, n));
    }
}

Python3

# A simple Python 3 program to count number of
# substrings starting and ending with 1


def countSubStr(st, n):

    # Initialize result
    res = 0

   # Pick a starting point
    for i in range(0, n):
        if (st[i] == '1'):

            # Search for all possible ending point
            for j in range(i+1, n):
                if (st[j] == '1'):
                    res = res + 1

    return res


# Driver program to test above function
st = "00100101"
list(st)
n = len(st)
print(countSubStr(st, n), end="")


# This code is contributed
# by Nikita Tiwari.

C#

// A simple C# program to count number of
// substrings starting and ending with 1
using System;

class GFG {
    public virtual int countSubStr(char[] str, int n)
    {
        int res = 0; // Initialize result

        // Pick a starting point
        for (int i = 0; i < n; i++) {
            if (str[i] == '1') {
                // Search for all possible
                // ending point
                for (int j = i + 1; j < n; j++) {
                    if (str[j] == '1') {
                        res++;
                    }
                }
            }
        }
        return res;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        GFG count = new GFG();
        string s = "00100101";
        char[] str = s.ToCharArray();
        int n = str.Length;
        Console.WriteLine(count.countSubStr(str, n));
    }
}

// This code is contributed by Shrikant13

PHP

<?php
// A simple PHP program to count number of
// substrings starting and ending with 1

function countSubStr($str)
{
    $res = 0; // Initialize result

    // Pick a starting point
    for ($i = 0; $i < strlen($str); $i++)
    {
            if ($str[$i] == '1')
            {
                // Search for all possible
                // ending point
                for ($j = $i + 1;
                     $j < strlen($str); $j++)
                if ($str[$j] == '1')
                    $res++;
            }
    }
    return $res;
}

// Driver Code
$str = "00100101";
echo countSubStr($str);

// This code is contributed by ita_c
?>

Javascript

<script>

// A simple javascript program to count number of
// substrings starting and ending with 1

    function countSubStr(str,n)
    {
        let res = 0;  // Initialize result
        // Pick a starting point
        for (let i = 0; i<n; i++)
        {
            if (str[i] == '1')
            {
                // Search for all possible ending point
                for (let j = i + 1; j< n; j++)
                {
                    if (str[j] == '1')
                        res++;
                }
            }
        }
        return res;
    }

    // Driver program to test the above function
    let string = "00100101";
    let n=string.length;
    document.write(countSubStr(string,n));


    // This code is contributed by rag2127

</script>

Output

3

Time Complexity: O(N²), 
Auxiliary Space: O(1)

Count of substrings that start and end with 1 in a given Binary String using Subarray count:

We know that if count of 1’s is m, then there will be m * (m – 1) / 2 possible subarrays.

Follow the steps to solve the problem:

• Count the number of 1’s. Let the count of 1’s be m. 
• Return m(m-1)/2 

Below is the implementation of above approach:

C++

// A O(n) C++ program to count number of
// substrings starting and ending with 1
#include <iostream>

using namespace std;

int countSubStr(char str[])
{
    int m = 0; // Count of 1's in input string

    // Traverse input string and count of 1's in it
    for (int i = 0; str[i] != '\0'; i++) {
        if (str[i] == '1')
            m++;
    }

    // Return count of possible pairs among m 1's
    return m * (m - 1) / 2;
}

// Driver program to test above function
int main()
{
    char str[] = "00100101";
    cout << countSubStr(str);
    return 0;
}

Java

// A O(n) Java program to count number of substrings
// starting and ending with 1

class CountSubString {
    int countSubStr(char str[], int n)
    {
        int m = 0; // Count of 1's in input string

        // Traverse input string and count of 1's in it
        for (int i = 0; i < n; i++) {
            if (str[i] == '1')
                m++;
        }

        // Return count of possible pairs among m 1's
        return m * (m - 1) / 2;
    }

    // Driver program to test the above function
    public static void main(String[] args)
    {
        CountSubString count = new CountSubString();
        String string = "00100101";
        char str[] = string.toCharArray();
        int n = str.length;
        System.out.println(count.countSubStr(str, n));
    }
}

Python3

# A Python3 program to count number of
# substrings starting and ending with 1


def countSubStr(st, n):

    # Count of 1's in input string
    m = 0

    # Traverse input string and
    # count of 1's in it
    for i in range(0, n):
        if (st[i] == '1'):
            m = m + 1

    # Return count of possible
    # pairs among m 1's
    return m * (m - 1) // 2


# Driver program to test above function
st = "00100101"
list(st)
n = len(st)
print(countSubStr(st, n), end="")


# This code is contributed
# by Nikita Tiwari.

C#

// A O(n) C# program to count
// number of substrings starting
// and ending with 1
using System;

class GFG {
    int countSubStr(char[] str, int n)
    {
        int m = 0; // Count of 1's in
                   // input string

        // Traverse input string and
        // count of 1's in it
        for (int i = 0; i < n; i++) {
            if (str[i] == '1')
                m++;
        }

        // Return count of possible
        // pairs among m 1's
        return m * (m - 1) / 2;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        GFG count = new GFG();
        String strings = "00100101";
        char[] str = strings.ToCharArray();
        int n = str.Length;
        Console.Write(count.countSubStr(str, n));
    }
}

// This code is contributed by princiraj

PHP

<?php
// A simple PHP program to count number of
// substrings starting and ending with 1

function countSubStr($str)
{
    $m = 0; // Initialize result

    // Pick a starting point
    for ($i = 0; $i < strlen($str); $i++)
    {
        if ($str[$i] == '1')
        {
            $m++;
        }
    }

    // Return count of possible
    // pairs among m 1's
    return $m * ($m - 1) / 2;
}

// Driver Code
$str = "00100101";
echo countSubStr($str);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>
// A O(n) javascript program to count number of substrings
//starting and ending with 1

    function countSubStr(str,n)
    {
        let m = 0; // Count of 1's in input string

        // Traverse input string and count of 1's in it
        for (let i = 0; i < n; i++)
        {
            if (str[i] == '1')
                m++;
        }

        // Return count of possible pairs among m 1's
        return m * Math.floor((m - 1) / 2);
    }

    // Driver program to test the above function
    let str = "00100101";
    let n = str.length;
    document.write(countSubStr(str, n));

    //  This code is contributed by avanitrachhadiya2155
</script>

Output

3

Time Complexity: O(N), where n is the length of the string.
Auxiliary Space: O(1).

Count of substrings that start and end with 1 in given Binary String using Recursion:

This approach is the same as the above approach but here to calculate the count of 1s we use recursion.

Follow the steps to solve the problem:

• Count the number of 1’s using recursion. Let the count of 1’s be m. 
• Return m(m-1)/2 

Below is the implementation of above approach:

C++

// A O(n) C++ program to count number of
// substrings starting and ending with 1
#include <bits/stdc++.h>
using namespace std;

int helper(int n, char str[], int i)
{
    // if 'i' is on the last index
    if (i == n - 1)
        return (str[i] == '1') ? 1 : 0;

    // if current char is 1
    // add 1 to the answer
    if (str[i] == '1')
        return 1 + helper(n, str, i + 1);

    // if it is zero
    else
        return helper(n, str, i + 1);
}

int countSubStr(char str[])
{
    int n = strlen(str);
    // counting the number of 1's in the string
    int count = helper(n, str, 0);

    // return the number of combinations
    return (count * (count - 1)) / 2;
}

// Driver program to test above function
int main()
{
    char str[] = "00100101";
    cout << countSubStr(str);
    return 0;
}

// this code is contributed by rajdeep999

Java

/*package whatever //do not write package name here */
import java.io.*;

class GFG {
  static int helper(int n, char str[], int i)
  {
    // if 'i' is on the last index
    if (i == n - 1)
      return (str[i] == '1') ? 1 : 0;

    // if current char is 1
    // add 1 to the answer
    if (str[i] == '1')
      return 1 + helper(n, str, i + 1);

    // if it is zero
    else
      return helper(n, str, i + 1);
  }

  static int countSubStr(char str[])
  {
    int n = str.length;
    // counting the number of 1's in the string
    int count = helper(n, str, 0);

    // return the number of combinations
    return (count * (count - 1)) / 2;
  }

  public static void main (String[] args) {
    char str[] = "00100101".toCharArray();
    System.out.println(countSubStr(str));
  }
}

// This code is contributed by aadityaburujwale.

Python3

class GFG :
    @staticmethod
    def  helper( n,  str,  i) :

        # if 'i' is on the last index
        if (i == n - 1) :
            return 1 if (str[i] == '1') else 0

        # if current char is 1
        # add 1 to the answer
        if (str[i] == '1') :
            return 1 + GFG.helper(n, str, i + 1)
        else :
            return GFG.helper(n, str, i + 1)
    @staticmethod
    def  countSubStr( str) :
        n = len(str)

        # counting the number of 1's in the string
        count = GFG.helper(n, str, 0)

        # return the number of combinations
        return int((count * (count - 1)) / 2)
    @staticmethod
    def main( args) :
        str = list("00100101")
        print(GFG.countSubStr(str))


if __name__=="__main__":
    GFG.main([])

    # This code is contributed by aadityaburujwale.

C#

// Include namespace system
using System;


public class GFG
{
    public static int helper(int n, char[] str, int i)
    {
        // if 'i' is on the last index
        if (i == n - 1)
        {
            return (str[i] == '1') ? 1 : 0;
        }

        // if current char is 1
        // add 1 to the answer
        if (str[i] == '1')
        {
            return 1 + GFG.helper(n, str, i + 1);
        }
        else
        {
            return GFG.helper(n, str, i + 1);
        }
    }
    public static int countSubStr(char[] str)
    {
        var n = str.Length;

        // counting the number of 1's in the string
        var count = GFG.helper(n, str, 0);

        // return the number of combinations
        return (int)((count * (count - 1)) / 2);
    }
    public static void Main(String[] args)
    {
        char[] str = "00100101".ToCharArray();
        Console.WriteLine(GFG.countSubStr(str));
    }
}

// This code is contributed by aadityaburujwale.

Javascript

// A O(n) JS program to count number of
// substrings starting and ending with 1
function helper(n, str, i)
{

    // if 'i' is on the last index
    if (i == n - 1) {
        return (str[i] == '1') ? 1 : 0;
    }

    // if current char is 1
    // add 1 to the answer
    if (str[i] == '1') {
        return 1 + helper(n, str, i + 1);
    }

    // if it is zero
    else {
        return helper(n, str, i + 1);
    }
}

function countSubStr(str) {
    let n = str.length;

    // counting the number of 1's in the string
    let count = helper(n, str, 0);

    // return the number of combinations
    return (count * (count - 1)) / 2;
}

// Driver program to test above function
console.log(countSubStr("00100101"));

// This code is contributed by akashish_

Output

3

Time Complexity: O(N), Traversing over the string of size N
Auxiliary Space: O(N), for recursion call stack