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Count of sub-strings that do not consist of the given character

  • Difficulty Level : Medium
  • Last Updated : 18 May, 2021

Given a string str and a character c. The task is to find the number of sub-strings that do not consist of the character c
Examples: 
 

Input: str = “baa”, c = ‘b’ 
Output:
The sub-strings are “a”, “a” and “aa” 
Input: str = “ababaa”, C = ‘b’ 
Output:
 

 

Approach: Initially take a counter that counts the number of characters continuously with no character c. Iterate in the string and increase the counter till str[i] != c. Once str[i] == c, the number of sub-strings from the contiguous length cnt will be (cnt * (cnt + 1)) / 2. After the complete traversal of the string also add (cnt *(cnt + 1)) / 2 to the result for the group of characters appearing after the last occurrence of c.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the number
// of sub-strings that do not contain
// the given character c
int countSubstrings(string s, char c)
{
 
    // Length of the string
    int n = s.length();
 
    int cnt = 0;
    int sum = 0;
 
    // Traverse in the string
    for (int i = 0; i < n; i++) {
 
        // If current character is different
        // from the given character
        if (s[i] != c)
            cnt++;
        else {
 
            // Update the number of sub-strings
            sum += (cnt * (cnt + 1)) / 2;
 
            // Reset count to 0
            cnt = 0;
        }
    }
 
    // For the characters appearing
    // after the last occurrence of c
    sum += (cnt * (cnt + 1)) / 2;
    return sum;
}
 
// Driver code
int main()
{
    string s = "baa";
    char c = 'b';
    cout << countSubstrings(s, c);
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
     
// Function to return the number
// of sub-strings that do not contain
// the given character c
static int countSubstrings(String s, char c)
{
 
    // Length of the string
    int n = s.length();
 
    int cnt = 0;
    int sum = 0;
 
    // Traverse in the string
    for (int i = 0; i < n; i++)
    {
 
        // If current character is different
        // from the given character
        if (s.charAt(i) != c)
            cnt++;
        else
        {
 
            // Update the number of sub-strings
            sum += (cnt * (cnt + 1)) / 2;
 
            // Reset count to 0
            cnt = 0;
        }
    }
 
    // For the characters appearing
    // after the last occurrence of c
    sum += (cnt * (cnt + 1)) / 2;
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "baa";
    char c = 'b';
    System.out.println(countSubstrings(s, c));
}
}
 
// This code is contributed by Code_Mech.

Python3




# Python3 implementation of the approach
 
# Function to return the number
# of sub-strings that do not contain
# the given character c
def countSubstrings(s, c):
 
    # Length of the string
    n = len(s)
 
    cnt = 0
    Sum = 0
 
    # Traverse in the string
    for i in range(n):
 
        # If current character is different
        # from the given character
        if (s[i] != c):
            cnt += 1
        else:
 
            # Update the number of sub-strings
            Sum += (cnt * (cnt + 1)) // 2
 
            # Reset count to 0
            cnt = 0
         
    # For the characters appearing
    # after the last occurrence of c
    Sum += (cnt * (cnt + 1)) // 2
    return Sum
 
# Driver code
s = "baa"
c = 'b'
print(countSubstrings(s, c))
 
# This code is contributed
# by mohit kumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the number
// of sub-strings that do not contain
// the given character c
static int countSubstrings(string s, char c)
{
 
    // Length of the string
    int n = s.Length;
 
    int cnt = 0;
    int sum = 0;
 
    // Traverse in the string
    for (int i = 0; i < n; i++)
    {
 
        // If current character is different
        // from the given character
        if (s[i] != c)
            cnt++;
        else
        {
 
            // Update the number of sub-strings
            sum += (cnt * (cnt + 1)) / 2;
 
            // Reset count to 0
            cnt = 0;
        }
    }
 
    // For the characters appearing
    // after the last occurrence of c
    sum += (cnt * (cnt + 1)) / 2;
    return sum;
}
 
// Driver code
public static void Main()
{
    string s = "baa";
    char c = 'b';
    Console.Write(countSubstrings(s, c));
}
}
 
// This code is contributed by Akanksha Rai

PHP




<?php
// PHP implementation of the approach
 
// Function to return the number
// of sub-strings that do not contain
// the given character c
function countSubstrings($s, $c)
{
 
    // Length of the string
    $n = strlen($s);
 
    $cnt = 0;
    $sum = 0;
 
    // Traverse in the string
    for ($i = 0; $i < $n; $i++)
    {
 
        // If current character is different
        // from the given character
        if ($s[$i] != $c)
            $cnt++;
        else
        {
 
            // Update the number of sub-strings
            $sum += floor(($cnt * ($cnt + 1)) / 2);
 
            // Reset count to 0
            $cnt = 0;
        }
    }
 
    // For the characters appearing
    // after the last occurrence of c
    $sum += floor(($cnt * ($cnt + 1)) / 2);
    return $sum;
}
 
// Driver code
$s = "baa";
$c = 'b';
 
echo countSubstrings($s, $c);
 
// This code is contributed by Ryuga
?>

Javascript




<script>
// javascript implementation of the approach   
// Function to return the number
    // of sub-strings that do not contain
    // the given character c
    function countSubstrings( s,  c) {
 
        // Length of the string
        var n = s.length;
 
        var cnt = 0;
        var sum = 0;
 
        // Traverse in the string
        for (i = 0; i < n; i++) {
 
            // If current character is different
            // from the given character
            if (s.charAt(i) != c)
                cnt++;
            else {
 
                // Update the number of sub-strings
                sum += (cnt * (cnt + 1)) / 2;
 
                // Reset count to 0
                cnt = 0;
            }
        }
 
        // For the characters appearing
        // after the last occurrence of c
        sum += (cnt * (cnt + 1)) / 2;
        return sum;
    }
 
    // Driver code
     
        var s = "baa";
        var c = 'b';
        document.write(countSubstrings(s, c));
 
// This code contributed by umadevi9616
</script>
Output: 
3

 

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