Count of substrings which contains a given character K times
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 K times.
Examples:
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> |
8
Time Complexity: O(N)
Auxiliary Space: O(1) as it is using constant space for variables
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