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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

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:



3

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:

3


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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