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Count of Substrings with at least K pairwise Distinct Characters having same Frequency
• Difficulty Level : Expert
• Last Updated : 06 May, 2021

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:
Explanation:
The substrings in having 2 pairwise distinct characters with same frequency are {“ab”, “ba”, “as”, “sa”, “bas”}.
Input: S = “abhay”, K = 3
Output:
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 ``using` `namespace` `std;` `// Function to find the substring with K``// pairwise distinct characters and``// with same frequency``int` `no_of_substring(string s, ``int` `N)``{``    ``// Stores the occurrence of each``    ``// character in the substring``    ``int` `fre;` `    ``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 Code``int` `main()``{``    ``string s = ``"abhay"``;``    ``int` `N = 3;` `    ``// Function call``    ``cout << no_of_substring(s, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Function to find the subString with K``// pairwise distinct characters and``// with same frequency``static` `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 Code``public` `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 frequency``def` `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 Code``s ``=` `"abhay"``N ``=` `3` `# Function call``print``(no_of_substring(s, N))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find the subString with K``// pairwise distinct characters and``// with same frequency``static` `int` `no_of_subString(String s, ``int` `N)``{``    ` `    ``// Stores the occurrence of each``    ``// character in the subString``    ``int` `[]fre = ``new` `int``;` `    ``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``;``        ` `        ``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 Code``public` `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

 ``
Output:
`4`

Time Complexity: O(N2)
Auxiliary Space: O(1)

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