Number of substrings with count of each character as k
Given a string and an integer k, find number of substrings in which all the different characters occurs exactly k times.
Examples:
Input : s = "aabbcc"
k = 2
Output : 6
The substrings are aa, bb, cc,
aabb, bbcc and aabbcc.
Input : s = "aabccc"
k = 2
Output : 3
There are three substrings aa,
cc and cc
The idea is to traverse through all substrings. We fix a starting point, traverse through all substrings starring with the picked point, we keep incrementing frequencies of all characters. If all frequencies become k, we increment result. If count of any frequency becomes more than k, we break and change starting point.
C++
// C++ program to count number of substrings // with counts of distinct characters as k. #include <bits/stdc++.h> using namespace std; const int MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. bool check(int freq[], int k) { for (int i = 0; i < MAX_CHAR; i++) if (freq[i] && freq[i] != k) return false; return true; } // Returns count of substrings where frequency // of every present character is k int substrings(string s, int k) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; s[i]; i++) { // Initialize all frequencies as 0 // for this starting point int freq[MAX_CHAR] = { 0 }; // One by one pick ending points for (int j = i; s[j]; j++) { // Increment frequency of current char int index = s[j] - 'a'; freq[index]++; // If frequency becomes more than // k, we can't have more substrings // starting with i if (freq[index] > k) break; // If frequency becomes k, then check // other frequencies as well. else if (freq[index] == k && check(freq, k) == true) res++; } } return res; } // Driver code int main() { string s = "aabbcc"; int k = 2; cout << substrings(s, k) << endl; s = "aabbc"; k = 2; cout << substrings(s, k) << endl; } |
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Java
// Java program to count number of substrings // with counts of distinct characters as k. class GFG { static int MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. static boolean check(int freq[], int k) { for (int i = 0; i < MAX_CHAR; i++) if (freq[i] !=0 && freq[i] != k) return false; return true; } // Returns count of substrings where frequency // of every present character is k static int substrings(String s, int k) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; i< s.length(); i++) { // Initialize all frequencies as 0 // for this starting point int freq[] = new int[MAX_CHAR]; // One by one pick ending points for (int j = i; j<s.length(); j++) { // Increment frequency of current char int index = s.charAt(j) - 'a'; freq[index]++; // If frequency becomes more than // k, we can't have more substrings // starting with i if (freq[index] > k) break; // If frequency becomes k, then check // other frequencies as well. else if (freq[index] == k && check(freq, k) == true) res++; } } return res; } // Driver code public static void main(String[] args) { String s = "aabbcc"; int k = 2; System.out.println(substrings(s, k)); s = "aabbc"; k = 2; System.out.println(substrings(s, k)); } } // This code has been contributed by 29AjayKumar |
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Python 3
# Python3 program to count number of substrings # with counts of distinct characters as k. MAX_CHAR = 26 # Returns true if all values # in freq[] are either 0 or k. def check(freq, k): for i in range(0, MAX_CHAR): if(freq[i] and freq[i] != k): return False return True # Returns count of substrings where # frequency of every present character is k def substrings(s, k): res = 0 # Initialize result # Pick a starting point for i in range(0, len(s)): # Initialize all frequencies as 0 # for this starting point freq = [0] * MAX_CHAR # One by one pick ending points for j in range(i, len(s)): # Increment frequency of current char index = ord(s[j]) - ord('a') freq[index] += 1 # If frequency becomes more than # k, we can't have more substrings # starting with i if(freq[index] > k): break # If frequency becomes k, then check # other frequencies as well elif(freq[index] == k and check(freq, k) == True): res += 1 return res # Driver Code if __name__ == "__main__": s = "aabbcc" k = 2 print(substrings(s, k)) s = "aabbc"; k = 2; print(substrings(s, k)) # This code is contributed # by Sairahul Jella |
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C#
// C# program to count number of substrings // with counts of distinct characters as k. using System; class GFG { static int MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. static bool check(int []freq, int k) { for (int i = 0; i < MAX_CHAR; i++) if (freq[i] != 0 && freq[i] != k) return false; return true; } // Returns count of substrings where frequency // of every present character is k static int substrings(String s, int k) { int res = 0; // Initialize result // Pick a starting point for (int i = 0; i < s.Length; i++) { // Initialize all frequencies as 0 // for this starting point int []freq = new int[MAX_CHAR]; // One by one pick ending points for (int j = i; j < s.Length; j++) { // Increment frequency of current char int index = s[j] - 'a'; freq[index]++; // If frequency becomes more than // k, we can't have more substrings // starting with i if (freq[index] > k) break; // If frequency becomes k, then check // other frequencies as well. else if (freq[index] == k && check(freq, k) == true) res++; } } return res; } // Driver code public static void Main(String[] args) { String s = "aabbcc"; int k = 2; Console.WriteLine(substrings(s, k)); s = "aabbc"; k = 2; Console.WriteLine(substrings(s, k)); } } /* This code contributed by PrinciRaj1992 */ |
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PHP
<?php // PHP program to count number of substrings // with counts of distinct characters as k. $MAX_CHAR = 26; // Returns true if all values // in freq[] are either 0 or k. function check(&$freq, $k) { global $MAX_CHAR; for ($i = 0; $i < $MAX_CHAR; $i++) if ($freq[$i] && $freq[$i] != $k) return false; return true; } // Returns count of substrings where frequency // of every present character is k function substrings($s, $k) { global $MAX_CHAR; $res = 0; // Initialize result // Pick a starting point for ($i = 0; $i < strlen($s); $i++) { // Initialize all frequencies as 0 // for this starting point $freq = array_fill(0, $MAX_CHAR,NULL); // One by one pick ending points for ($j = $i; $j < strlen($s); $j++) { // Increment frequency of current char $index = ord($s[$j]) - ord('a'); $freq[$index]++; // If frequency becomes more than // k, we can't have more substrings // starting with i if ($freq[$index] > $k) break; // If frequency becomes k, then check // other frequencies as well. else if ($freq[$index] == $k && check($freq, $k) == true) $res++; } } return $res; } // Driver code $s = "aabbcc"; $k = 2; echo substrings($s, $k)."\n"; $s = "aabbc"; $k = 2; echo substrings($s, $k)."\n"; // This code is contributed by Ita_c. ?> |
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Output:
6 3
Time Complexity : O(n2) where n is length of input string.
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