Count all substrings having character K Last Updated : 25 Aug, 2021 Comments Improve Suggest changes Like Article Like Report Given a string str and a character K, the task is to find the count of all the substrings of str that contain the character K.Examples: Input: str = "geeks", K = 'g' Output: 5 "g", "ge", "gee", "geek" and "geeks" are the valid substrings.Input: str = "geeksforgeeks", K = 'k' Output: 56 Naive approach A simple approach will be to find all the substrings having character K of the given string and return the count;Efficient approach: For every index i in the string, find the first index j such that i ? j and str[j] = K. Now, the substrings str[i...j], str[i...j + 1], str[i...j + 2], ..., str[i...n - 1] will all contain the character K at least once. The approach seems to be O(n2) at first but the index j will not be calculated again for every index i, j will be a valid index for all the values of i less than j.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 index of the // next occurrence of character ch in str // starting from the given index int nextOccurrence(string str, int n, int start, char ch) { for (int i = start; i < n; i++) { // Return the index of the first // occurrence of ch if (str[i] == ch) return i; } // No occurrence found return -1; } // Function to return the count of all // the substrings of str which contain // the character ch at least one int countSubStr(string str, int n, char ch) { // To store the count of valid substrings int cnt = 0; // Index of the first occurrence of ch in str int j = nextOccurrence(str, n, 0, ch); for (int i = 0; i < n; i++) { while (j != -1 && j < i) { j = nextOccurrence(str, n, j + 1, ch); } // No occurrence of ch after index i in str if (j == -1) break; // Substrings starting at index i // and ending at indices j, j+1, ..., n-1 // are all valid substring cnt += (n - j); } return cnt; } // Driver code int main() { string str = "geeksforgeeks"; int n = str.length(); char ch = 'k'; cout << countSubStr(str, n, ch); return 0; } Java // Java implementation of the approach import java.util.*; class GFG { // Function to return the index of the // next occurrence of character ch in str // starting from the given index static int nextOccurrence(String str, int n, int start, char ch) { for (int i = start; i < n; i++) { // Return the index of the first // occurrence of ch if (str.charAt(i) == ch) return i; } // No occurrence found return -1; } // Function to return the count of all // the substrings of str which contain // the character ch at least one static int countSubStr(String str, int n, char ch) { // To store the count of valid substrings int cnt = 0; // Index of the first occurrence of ch in str int j = nextOccurrence(str, n, 0, ch); for (int i = 0; i < n; i++) { while (j != -1 && j < i) { j = nextOccurrence(str, n, j + 1, ch); } // No occurrence of ch after index i in str if (j == -1) break; // Substrings starting at index i // and ending at indices j, j+1, ..., n-1 // are all valid substring cnt += (n - j); } return cnt; } // Driver code static public void main ( String []arg) { String str = "geeksforgeeks"; int n = str.length(); char ch = 'k'; System.out.println(countSubStr(str, n, ch)); } } // This code is contributed by PrinciRaj1992 Python3 # Python3 implementation of the approach # Function to return the index of the # next occurrence of character ch in strr # starting from the given index def nextOccurrence(strr, n, start, ch): for i in range(start, n): # Return the index of the first # occurrence of ch if (strr[i] == ch): return i # No occurrence found return -1 # Function to return the count of all # the substrings of strr which contain # the character ch at least one def countSubStr(strr, n, ch): # To store the count of valid substrings cnt = 0 # Index of the first occurrence of ch in strr j = nextOccurrence(strr, n, 0, ch) for i in range(n): while (j != -1 and j < i): j = nextOccurrence(strr, n, j + 1, ch) # No occurrence of ch after index i in strr if (j == -1): break # Substrings starting at index i # and ending at indices j, j+1, ..., n-1 # are all valid substring cnt += (n - j) return cnt # Driver code strr = "geeksforgeeks" n = len(strr) ch = 'k' print(countSubStr(strr, n, ch)) # This code is contributed by Mohit Kumar C# // C# implementation of the approach using System; class GFG { // Function to return the index of the // next occurrence of character ch in str // starting from the given index static int nextOccurrence(String str, int n, int start, char ch) { for (int i = start; i < n; i++) { // Return the index of the first // occurrence of ch if (str[i] == ch) return i; } // No occurrence found return -1; } // Function to return the count of all // the substrings of str which contain // the character ch at least one static int countSubStr(String str, int n, char ch) { // To store the count of valid substrings int cnt = 0; // Index of the first occurrence of ch in str int j = nextOccurrence(str, n, 0, ch); for (int i = 0; i < n; i++) { while (j != -1 && j < i) { j = nextOccurrence(str, n, j + 1, ch); } // No occurrence of ch after index i in str if (j == -1) break; // Substrings starting at index i // and ending at indices j, j+1, ..., n-1 // are all valid substring cnt += (n - j); } return cnt; } // Driver code static public void Main () { String str = "geeksforgeeks"; int n = str.Length; char ch = 'k'; Console.WriteLine(countSubStr(str, n, ch)); } } // This code is contributed by AnkitRai01 JavaScript <script> // JavaScript implementation of the approach // Function to return the index of the // next occurrence of character ch in str // starting from the given index function nextOccurrence(str, n, start, ch) { for (var i = start; i < n; i++) { // Return the index of the first // occurrence of ch if (str[i] === ch) return i; } // No occurrence found return -1; } // Function to return the count of all // the substrings of str which contain // the character ch at least one function countSubStr(str, n, ch) { // To store the count of valid substrings var cnt = 0; // Index of the first occurrence of ch in str var j = nextOccurrence(str, n, 0, ch); for (var i = 0; i < n; i++) { while (j !== -1 && j < i) { j = nextOccurrence(str, n, j + 1, ch); } // No occurrence of ch after index i in str if (j === -1) break; // Substrings starting at index i // and ending at indices j, j+1, ..., n-1 // are all valid substring cnt += n - j; } return cnt; } // Driver code var str = "geeksforgeeks"; var n = str.length; var ch = "k"; document.write(countSubStr(str, n, ch)); </script> Output: 56 Comment More infoAdvertise with us Next Article Analysis of Algorithms P PrashantYadav6 Follow Improve Article Tags : DSA substring Similar Reads Basics & PrerequisitesLogic Building ProblemsLogic building is about creating clear, step-by-step methods to solve problems using simple rules and principles. 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