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

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

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

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