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Count of substrings which contains a given character K times

  • Difficulty Level : Medium
  • Last Updated : 06 Oct, 2021

Given a string consisting of numerical alphabets, a character C and an integer K, the task is to find the number of possible substrings which contains the character C exactly Ktimes. 

Examples: 

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Input : str = "212", c = '2', k = 1 
Output : 4
Possible substrings are {"2", "21", "12", "2"}
that contains 2 exactly 1 time.

Input  : str = "55555", c = '5', k = 4
Output : 2
Possible substrings are {"5555", "5555"} 
that contains 5 exactly 4 times

Naive Approach: A simple solution is to generate all the substrings of string and count the number of substrings in which a given character occurs exactly k times.



The time complexity of this solution is O(N2) where N is the length of the string.

Efficient Approach: An efficient solution is to use sliding window technique. Find the substring that contains character C exactly K times starting with character C. Count the number of characters on either side of the substring. Multiply the counts to get the number of possible substrings with given substring 

Below is the implementation of the above approach:  

C++




// C++ program to count the number of substrings
// which contains the character C exactly K times
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of substrings
// which contains the character C exactly K times
int countSubString(string s, char c, int k)
{
    // left and right counters for characters on
    // both sides of substring window
    int leftCount = 0, rightCount = 0;
 
    // left and right pointer on both sides
    // of substring window
    int left = 0, right = 0;
 
    // initialize the frequency
    int freq = 0;
 
    // result and length of string
    int result = 0, len = s.length();
 
    // initialize the left pointer
    while (s[left] != c && left < len) {
        left++;
        leftCount++;
    }
 
    // initialize the right pointer
    right = left + 1;
    while (freq != (k - 1) && (right - 1) < len) {
        if (s[right] == c)
            freq++;
        right++;
    }
 
    // traverse all the window substrings
    while (left < len && (right - 1) < len) {
 
        // counting the characters on leftSide
        // of substring window
        while (s[left] != c && left < len) {
            left++;
            leftCount++;
        }
 
        // counting the characters on rightSide of
        // substring window
        while (right < len && s[right] != c) {
            if (s[right] == c)
                freq++;
            right++;
            rightCount++;
        }
 
        // Add the possible substrings on both
        // sides to result
        result = result + (leftCount + 1) * (rightCount + 1);
 
        // Setting the frequency for next
        // substring window
        freq = k - 1;
 
        // reset the left, right counters
        leftCount = 0;
        rightCount = 0;
 
        left++;
        right++;
    }
 
    return result;
}
 
// Driver code
int main()
{
    string s = "3123231";
    char c = '3';
    int k = 2;
 
    cout << countSubString(s, c, k);
 
    return 0;
}

Java




// Java program to count the number of substrings
// which contains the character C exactly K times
class GFG
{
 
    // Function to count the number of substrings
    // which contains the character C exactly K times
    static int countSubString(String s, char c, int k)
    {
        // left and right counters for characters on
        // both sides of substring window
        int leftCount = 0, rightCount = 0;
 
        // left and right pointer on both sides
        // of substring window
        int left = 0, right = 0;
 
        // initialize the frequency
        int freq = 0;
 
        // result and length of string
        int result = 0, len = s.length();
 
        // initialize the left pointer
        while (s.charAt(left) != c && left < len)
        {
            left++;
            leftCount++;
        }
 
        // initialize the right pointer
        right = left + 1;
        while (freq != (k - 1) && (right - 1) < len)
        {
            if (s.charAt(right) == c)
            {
                freq++;
            }
            right++;
        }
 
        // traverse all the window substrings
        while (left < len && (right - 1) < len)
        {
 
            // counting the characters on leftSide
            // of substring window
            while (s.charAt(left) != c && left < len)
            {
                left++;
                leftCount++;
            }
 
            // counting the characters on rightSide of
            // substring window
            while (right < len && s.charAt(right) != c)
            {
                if (s.charAt(right) == c)
                {
                    freq++;
                }
                right++;
                rightCount++;
            }
 
            // Add the possible substrings on both
            // sides to result
            result = result + (leftCount + 1) * (rightCount + 1);
 
            // Setting the frequency for next
            // substring window
            freq = k - 1;
 
            // reset the left, right counters
            leftCount = 0;
            rightCount = 0;
 
            left++;
            right++;
        }
 
        return result;
    }
 
    // Driver code
    public static void main(String args[])
    {
        String s = "3123231";
        char c = '3';
        int k = 2;
 
        System.out.println(countSubString(s, c, k));
    }
}
 
/* This code is contributed by PrinciRaj1992 */

Python3




# Python3 program to count the number of substrings
# which contains the character C exactly K times
 
# Function to count the number of substrings
# which contains the character C exactly K times
def countSubString(s, c, k):
 
    # left and right counters for characters
    # on both sides of subwindow
    leftCount = 0
    rightCount = 0
 
    # left and right pointer on both sides
    # of subwindow
    left = 0
    right = 0
 
    # Initialize the frequency
    freq = 0
 
    # result and Length of string
    result = 0
    Len = len(s)
 
    # initialize the left pointer
    while (s[left] != c and left < Len):
        left += 1
        leftCount += 1
 
    # initialize the right pointer
    right = left + 1
    while (freq != (k - 1) and (right - 1) < Len):
        if (s[right] == c):
            freq += 1
        right += 1
 
    # traverse all the window substrings
    while (left < Len and (right - 1) < Len):
 
        # counting the characters on leftSide
        # of subwindow
        while (s[left] != c and left < Len):
            left += 1
            leftCount += 1
             
        # counting the characters on rightSide of
        # subwindow
        while (right < Len and s[right] != c):
            if (s[right] == c):
                freq += 1
            right += 1
            rightCount += 1
 
        # Add the possible substrings on both
        # sides to result
        result = (result + (leftCount + 1) *
                           (rightCount + 1))
 
        # Setting the frequency for next
        # subwindow
        freq = k - 1
 
        # reset the left, right counters
        leftCount = 0
        rightCount = 0
 
        left += 1
        right += 1
 
    return result
 
# Driver code
s = "3123231"
c = '3'
k = 2
 
print(countSubString(s, c, k))
 
# This code is contributed by Mohit Kumar

C#




// C# program to count the number of substrings
// which contains the character C exactly K times
using System;
 
class GFG
{
         
    // Function to count the number of substrings
    // which contains the character C exactly K times
    static int countSubString(string s, char c, int k)
    {
        // left and right counters for characters on
        // both sides of substring window
        int leftCount = 0, rightCount = 0;
     
        // left and right pointer on both sides
        // of substring window
        int left = 0, right = 0;
     
        // initialize the frequency
        int freq = 0;
     
        // result and length of string
        int result = 0, len = s.Length;
     
        // initialize the left pointer
        while (s[left] != c && left < len)
        {
            left++;
            leftCount++;
        }
     
        // initialize the right pointer
        right = left + 1;
        while (freq != (k - 1) && (right - 1) < len)
        {
            if (s[right] == c)
                freq++;
            right++;
        }
     
        // traverse all the window substrings
        while (left < len && (right - 1) < len)
        {
     
            // counting the characters on leftSide
            // of substring window
            while (s[left] != c && left < len)
            {
                left++;
                leftCount++;
            }
     
            // counting the characters on rightSide of
            // substring window
            while (right < len && s[right] != c)
            {
                if (s[right] == c)
                    freq++;
                right++;
                rightCount++;
            }
     
            // Add the possible substrings on both
            // sides to result
            result = result + (leftCount + 1) * (rightCount + 1);
     
            // Setting the frequency for next
            // substring window
            freq = k - 1;
     
            // reset the left, right counters
            leftCount = 0;
            rightCount = 0;
     
            left++;
            right++;
        }
     
        return result;
    }
     
    // Driver code
    static public void Main ()
    {
        string s = "3123231";
        char c = '3';
        int k = 2;
     
        Console.WriteLine(countSubString(s, c, k));
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
 
// JavaScript program to count the number of substrings
// which contains the character C exactly K times
 
// Function to count the number of substrings
// which contains the character C exactly K times
function countSubString(s, c, k)
{
     
    // left and right counters for characters
    // on both sides of substring window
    var leftCount = 0,
    rightCount = 0;
     
    // left and right pointer on both
    // sides of substring window
    var left = 0,
    right = 0;
     
    // Initialize the frequency
    var freq = 0;
     
    // Result and length of string
    var result = 0,
    len = s.length;
     
    // Initialize the left pointer
    while (s[left] !== c && left < len)
    {
        left++;
        leftCount++;
    }
     
    // Initialize the right pointer
    right = left + 1;
     
    while (freq !== k - 1 && right - 1 < len)
    {
        if (s[right] === c) freq++;
            right++;
    }
     
    // Traverse all the window substrings
    while (left < len && right - 1 < len)
    {
         
        // Counting the characters on leftSide
        // of substring window
        while (s[left] !== c && left < len)
        {
            left++;
            leftCount++;
        }
         
        // Counting the characters on rightSide of
        // substring window
        while (right < len && s[right] !== c)
        {
            if (s[right] === c)
                freq++;
                 
            right++;
            rightCount++;
        }
         
        // Add the possible substrings on both
        // sides to result
        result = result + (leftCount + 1) *
                         (rightCount + 1);
         
        // Setting the frequency for next
        // substring window
        freq = k - 1;
         
        // Reset the left, right counters
        leftCount = 0;
        rightCount = 0;
         
        left++;
        right++;
    }
    return result;
}
 
// Driver code
var s = "3123231";
var c = "3";
var k = 2;
 
document.write(countSubString(s, c, k));
 
// This code is contributed by rdtank
 
</script>
Output: 
8

 

Time Complexity: O(N)
 




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