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

Difficulty Level : Easy
Last Updated : 06 Nov, 2020

Given a binary string, count number of substrings that start and end with 1. For example, if the input string is “00100101”, then there are three substrings “1001”, “100101” and “101”.

Source: Amazon Interview Experience | Set 162

Difficulty Level: Rookie

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

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

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 C++ 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 
?> 

Output:

Time Complexity of the above solution is O(n²). We can find count in O(n) using a single traversal of input string. Following are steps.
a) Count the number of 1’s. Let the count of 1’s be m.
b) Return m(m-1)/2
The idea is to count total number of possible pairs of 1’s.

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) C++ 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 
?> 

Output:

This article is contributed by Shivam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Related Articles

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

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP