Given the string str, a character c, and an integer k > 0. The task is to find the number of sub-strings that contain the character c exactly k times.
Examples:
Input: str = “abada”, c = ‘a’, K = 2
Output: 4
All possible sub-strings are “aba”, “abad”, “bada” and “ada”.Input: str = “55555”, c = ‘5’, k = 4
Output: 2
Naive approach: A simple solution is to generate all the sub-strings and check whether the count of a given character is exactly k times.
The time complexity of this approach is O(n2).
Space complexity of this approach is O(1).
Efficient approach: An efficient solution is to use the sliding window technique. Find the sub-string that exactly contains the given character k times, starting with character c. Count the number of characters on either side of the sub-string. Multiply the counts to get the number of possible sub-strings. The time complexity of this approach is O(n).
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 count of required sub-stringsint countSubString(string s, char c, int k){ // Left and right counters for characters on // both sides of sub-string window int leftCount = 0, rightCount = 0; // Left and right pointer on both // sides of sub-string 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 sub-strings while (left < len && (right - 1) < len) { // Counting the characters on left side // of the sub-string window while (s[left] != c && left < len) { left++; leftCount++; } // Counting the characters on right side // of the sub-string window while (right < len && s[right] != c) { if (s[right] == c) freq++; right++; rightCount++; } // Add the possible sub-strings // on both sides to result result = result + (leftCount + 1) * (rightCount + 1); // Setting the frequency for next // sub-string window freq = k - 1; // Reset the left and right counters leftCount = 0; rightCount = 0; left++; right++; } return result;}// Driver codeint main(){ string s = "abada"; char c = 'a'; int k = 2; cout << countSubString(s, c, k) << "\n"; return 0;} |
Java
import java.io.*; // Java implementation of the approachclass GFG { // Function to return the count of required sub-strings static int countSubString(char[] s, char c, int k) { // Left and right counters for characters on // both sides of sub-string window int leftCount = 0, rightCount = 0; // Left and right pointer on both // sides of sub-string 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 sub-strings while (left < len && (right - 1) < len) { // Counting the characters on left side // of the sub-string window while (s[left] != c && left < len) { left++; leftCount++; } // Counting the characters on right side // of the sub-string window while (right < len && s[right] != c) { if (s[right] == c) freq++; right++; rightCount++; } // Add the possible sub-strings // on both sides to result result = result + (leftCount + 1) * (rightCount + 1); // Setting the frequency for next // sub-string window freq = k - 1; // Reset the left and right counters leftCount = 0; rightCount = 0; left++; right++; } return result; } // Driver code public static void main(String[] args) { String s = "abada"; char c = 'a'; int k = 2; System.out.println( countSubString(s.toCharArray(), c, k)); }}// This code has been contributed by 29AjayKumar |
Python3
# Python 3 implementation of the approach# Function to return the count# of required sub-stringsdef countSubString(s, c, k): # Left and right counters for characters # on both sides of sub-string window leftCount = 0 rightCount = 0 # Left and right pointer on both # sides of sub-string window left = 0 right = 0 # Initialize the frequency freq = 0 # Result and length of string result = 0 len1 = len(s) # Initialize the left pointer while (s[left] != c and left < len1): left += 1 leftCount += 1 # Initialize the right pointer right = left + 1 while (freq != (k - 1) and (right - 1) < len1): if (s[right] == c): freq += 1 right += 1 # Traverse all the window sub-strings while (left < len1 and (right - 1) < len1): # Counting the characters on left side # of the sub-string window while (s[left] != c and left < len1): left += 1 leftCount += 1 # Counting the characters on right side # of the sub-string window while (right < len1 and s[right] != c): if (s[right] == c): freq += 1 right += 1 rightCount += 1 # Add the possible sub-strings # on both sides to result result = (result + (leftCount + 1) * (rightCount + 1)) # Setting the frequency for next # sub-string window freq = k - 1 # Reset the left and right counters leftCount = 0 rightCount = 0 left += 1 right += 1 return result# Driver codeif __name__ == '__main__': s = "abada" c = 'a' k = 2 print(countSubString(s, c, k))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG { // Function to return the count of required sub-strings static int countSubString(char[] s, char c, int k) { // Left and right counters for characters on // both sides of sub-string window int leftCount = 0, rightCount = 0; // Left and right pointer on both // sides of sub-string 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 sub-strings while (left < len && (right - 1) < len) { // Counting the characters on left side // of the sub-string window while (s[left] != c && left < len) { left++; leftCount++; } // Counting the characters on right side // of the sub-string window while (right < len && s[right] != c) { if (s[right] == c) freq++; right++; rightCount++; } // Add the possible sub-strings // on both sides to result result = result + (leftCount + 1) * (rightCount + 1); // Setting the frequency for next // sub-string window freq = k - 1; // Reset the left and right counters leftCount = 0; rightCount = 0; left++; right++; } return result; } // Driver code public static void Main(String[] args) { String s = "abada"; char c = 'a'; int k = 2; Console.WriteLine( countSubString(s.ToCharArray(), c, k)); }}// This code contributed by Rajput-Ji |
PHP
<?php// PHP implementation of the approach // Function to return the count of required sub-strings function countSubString($s, $c, $k) { // Left and right counters for characters on // both sides of sub-string window $leftCount = 0; $rightCount = 0; // Left and right pointer on both // sides of sub-string window $left = 0; $right = 0; // Initialize the frequency $freq = 0; // Result and length of string $result = 0; $len = strlen($s); // 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++; $right++; } // Traverse all the window sub-strings while ($left < $len && ($right - 1) < $len) { // Counting the characters on left side // of the sub-string window while ($s[$left] != $c && $left < $len) { $left++; $leftCount++; } // Counting the characters on right side // of the sub-string window while ($right < $len && $s[$right] != $c) { if ($s[$right] == $c) $freq++; $right++; $rightCount++; } // Add the possible sub-strings // on both sides to result $result = $result + ($leftCount + 1) * ($rightCount + 1); // Setting the frequency for next // sub-string window $freq = $k - 1; // Reset the left and right counters $leftCount = 0; $rightCount = 0; $left++; $right++; } return $result; } // Driver code $s = "abada"; $c = 'a'; $k = 2; echo countSubString($s, $c, $k), "\n"; // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript implementation of the approach// Function to return the count of required sub-stringsfunction countSubString(s, c, k){ // Left and right counters for characters on // both sides of sub-string window var leftCount = 0, rightCount = 0; // Left and right pointer on both // sides of sub-string 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 sub-strings while (left < len && (right - 1) < len) { // Counting the characters on left side // of the sub-string window while (s[left] != c && left < len) { left++; leftCount++; } // Counting the characters on right side // of the sub-string window while (right < len && s[right] != c) { if (s[right] == c) freq++; right++; rightCount++; } // Add the possible sub-strings // on both sides to result result = result + (leftCount + 1) * (rightCount + 1); // Setting the frequency for next // sub-string window freq = k - 1; // Reset the left and right counters leftCount = 0; rightCount = 0; left++; right++; } return result;}// Driver codevar s = "abada";var c = 'a';var k = 2;document.write( countSubString(s, c, k) + "<br>");</script> |
Output:
4
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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