Given a string S and an integer K, the task is to find the number of substrings which consists of at least K pairwise distinct characters having same frequency.
Examples:
Input: S = “abasa”, K = 2
Output: 5
Explanation:
The substrings in having 2 pairwise distinct characters with same frequency are {“ab”, “ba”, “as”, “sa”, “bas”}.
Input: S = “abhay”, K = 3
Output: 4
Explanation:
The substrings having 3 pairwise distinct characters with same frequency are {“abh”, “bha”, “hay”, “bhay”}.
Naive Approach: The simplest approach to solve this problem is to generate all possible substrings of the given string and check if both the conditions are satisfied. If found to be true, increase count. Finally, print count.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, follow the steps below to solve the problem:
- Check if the frequencies of each character is same. If found to be true, simply generate all the substrings to check if each character satisfies the condition of at least N pairwise distinct characters.
- Precompute the frequencies of characters to check the conditions for each substring.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the substring with K// pairwise distinct characters and// with same frequencyint no_of_substring(string s, int N){ // Stores the occurrence of each // character in the substring int fre[26]; int str_len; // Length of the string str_len = (int)s.length(); int count = 0; // Iterate over the string for (int i = 0; i < str_len; i++) { // Set all values at each index to zero memset(fre, 0, sizeof(fre)); int max_index = 0; // Stores the count of // unique characters int dist = 0; // Moving the substring ending at j for (int j = i; j < str_len; j++) { // Calculate the index of // character in frequency array int x = s[j] - 'a'; if (fre[x] == 0) dist++; // Increment the frequency fre[x]++; // Update the maximum index max_index = max(max_index, fre[x]); // Check for both the conditions if (dist >= N && ((max_index * dist) == (j - i + 1))) count++; } } // Return the answer return count;}// Driver Codeint main(){ string s = "abhay"; int N = 3; // Function call cout << no_of_substring(s, N); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the subString with K// pairwise distinct characters and// with same frequencystatic int no_of_subString(String s, int N){ // Stores the occurrence of each // character in the subString int fre[] = new int[26]; int str_len; // Length of the String str_len = (int)s.length(); int count = 0; // Iterate over the String for(int i = 0; i < str_len; i++) { // Set all values at each index to zero Arrays.fill(fre, 0); int max_index = 0; // Stores the count of // unique characters int dist = 0; // Moving the subString ending at j for(int j = i; j < str_len; j++) { // Calculate the index of // character in frequency array int x = s.charAt(j) - 'a'; if (fre[x] == 0) dist++; // Increment the frequency fre[x]++; // Update the maximum index max_index = Math.max(max_index, fre[x]); // Check for both the conditions if (dist >= N && ((max_index * dist) == (j - i + 1))) count++; } } // Return the answer return count;}// Driver Codepublic static void main(String[] args){ String s = "abhay"; int N = 3; // Function call System.out.print(no_of_subString(s, N));}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program to implement# the above approach# Function to find the substring with K# pairwise distinct characters and# with same frequencydef no_of_substring(s, N): # Length of the string str_len = len(s) count = 0 # Iterate over the string for i in range(str_len): # Stores the occurrence of each # character in the substring # Set all values at each index to zero fre = [0] * 26 max_index = 0 # Stores the count of # unique characters dist = 0 # Moving the substring ending at j for j in range(i, str_len): # Calculate the index of # character in frequency array x = ord(s[j]) - ord('a') if (fre[x] == 0): dist += 1 # Increment the frequency fre[x] += 1 # Update the maximum index max_index = max(max_index, fre[x]) # Check for both the conditions if(dist >= N and ((max_index * dist) == (j - i + 1))): count += 1 # Return the answer return count# Driver Codes = "abhay"N = 3# Function callprint(no_of_substring(s, N))# This code is contributed by Shivam Singh |
C#
// C# program for the above approachusing System;class GFG{// Function to find the subString with K// pairwise distinct characters and// with same frequencystatic int no_of_subString(String s, int N){ // Stores the occurrence of each // character in the subString int []fre = new int[26]; int str_len; // Length of the String str_len = (int)s.Length; int count = 0; // Iterate over the String for(int i = 0; i < str_len; i++) { // Set all values at each index to zero fre = new int[26]; int max_index = 0; // Stores the count of // unique characters int dist = 0; // Moving the subString ending at j for(int j = i; j < str_len; j++) { // Calculate the index of // character in frequency array int x = s[j] - 'a'; if (fre[x] == 0) dist++; // Increment the frequency fre[x]++; // Update the maximum index max_index = Math.Max(max_index, fre[x]); // Check for both the conditions if (dist >= N && ((max_index * dist) == (j - i + 1))) count++; } } // Return the answer return count;}// Driver Codepublic static void Main(String[] args){ String s = "abhay"; int N = 3; // Function call Console.Write(no_of_subString(s, N));}}// This code is contributed by Amit Katiyar |
Javascript
<script>// javascript program for the above approach// Function to find the subString with K// pairwise distinct characters and// with same frequencyfunction no_of_subString(s , N){ // Stores the occurrence of each // character in the subString var fre = Array.from({length: 26}, (_, i) => 0); var str_len; // Length of the String str_len = parseInt(s.length); var count = 0; // Iterate over the String for(i = 0; i < str_len; i++) { // Set all values at each index to zero fre = Array(26).fill(0); var max_index = 0; // Stores the count of // unique characters var dist = 0; // Moving the subString ending at j for(j = i; j < str_len; j++) { // Calculate the index of // character in frequency array var x = s.charAt(j).charCodeAt(0) - 'a'.charCodeAt(0); if (fre[x] == 0) dist++; // Increment the frequency fre[x]++; // Update the maximum index max_index = Math.max(max_index, fre[x]); // Check for both the conditions if (dist >= N && ((max_index * dist) == (j - i + 1))) count++; } } // Return the answer return count;}// Driver Codevar s = "abhay";var N = 3;// Function calldocument.write(no_of_subString(s, N));// This code contributed by shikhasingrajput </script> |
Output:
4
Time Complexity: O(N2)
Auxiliary Space: O(1)
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