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# Number of substrings with count of each character as k

Given a string and an integer k, find the number of substrings in which all the different characters occur 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

Naive Approach: The idea is to traverse through all substrings. We fix a starting point, traverse through all substrings starting with the picked point, we keep incrementing frequencies of all characters. If all frequencies become k, we increment the result. If the count of any frequency becomes more than k, we break and change starting point.

Below is the implementation of the above approach:

## C++

 // C++ program to count number of substrings// with counts of distinct characters as k.#include 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 kint 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 codeint main(){    string s = "aabbcc";    int k = 2;    cout << substrings(s, k) << endl;     s = "aabbc";    k = 2;    cout << substrings(s, k) << endl;}

## Java

 // Java program to count number of substrings// with counts of distinct characters as k.import java.io.*;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 kstatic 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 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 codepublic 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

## Python3

 # 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 kdef 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 =  * 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 Codeif __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

## 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 kstatic 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 codepublic 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 */

## PHP

  $k) break;  // If frequency becomes k, then check // other frequencies as well. else if ($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.?>

## Javascript

 

Output

6
3

Time Complexity: O(n*n) where n is the length of input string. Function Check() is running a loop of constant length from 0 to MAX_CHAR (ie; 26 always) so this function check() is running in O(MAX_CHAR) time so Time complexity is O(MAX_CHAR*n*n)=O(n^2).
Auxiliary Space: O(1)

Efficient Approach: On very careful observation, we can see that it is enough to check the same for substrings of length where is the number of distinct characters present in the given string.

Argument:
Consider a substring S_{i+1}S_{i+2}\dots S_{i+p} of length ‘p’. If this substring has ‘m’ distinct characters and each distinct character occurs exactly ‘K’ times, then the length of the substring, ‘p’, is given by p = K\times m. Since ‘ ‘ is always a multiple of ‘K’ and for the given string, it is enough to iterate over the substrings whose length is divisible by ‘K’ and having m, 1 \le m \le 26 distinct characters. We will use Sliding window to iterate over the substrings of fixed length.

#### Solution:

• Find the number of distinct characters present in the given string. Let it be D.
• For each i, 1\le i\le D, do the following
• Iterate over the substrings of length $i \times K$, using a sliding window.
• Check if they satisfy the condition – All distinct characters in the substring occur exactly K times.
• If they satisfy the condition, increment the count.

Below is the implementation of the above approach:

## C++

 #include #include #include #include  int min(int a, int b) { return a < b ? a : b; } using namespace std; bool have_same_frequency(map<char, int>& freq, int k){    for (auto& pair : freq) {        if (pair.second != k && pair.second != 0) {            return false;        }    }    return true;} int count_substrings(string s, int k){    int count = 0;    int distinct = (set<char>(s.begin(), s.end())).size();    for (int length = 1; length <= distinct; length++) {        int window_length = length * k;        map<char, int> freq;        int window_start = 0;        int window_end = window_start + window_length - 1;        for (int i = window_start;             i <= min(window_end, s.length() - 1); i++) {            freq[s[i]]++;        }        while (window_end < s.length()) {            if (have_same_frequency(freq, k)) {                count++;            }            freq[s[window_start]]--;            window_start++;            window_end++;            if (window_length < s.length()) {                freq[s[window_end]]++;            }        }    }    return count;} int main(){    string s = "aabbcc";    int k = 2;    cout << count_substrings(s, k) << endl;    s = "aabbc";    k = 2;    cout << count_substrings(s, k) << endl;    return 0;}

## C

 #include #include #include  int min(int a, int b) { return a < b ? a : b; } bool have_same_frequency(int freq[], int k){    for (int i = 0; i < 26; i++) {        if (freq[i] != 0 && freq[i] != k) {            return false;        }    }    return true;} int count_substrings(char* s, int n, int k){    int count = 0;    int distinct = 0;    bool have = { false };    for (int i = 0; i < n; i++) {        have[s[i] - 'a'] = true;    }    for (int i = 0; i < 26; i++) {        if (have[i]) {            distinct++;        }    }    for (int length = 1; length <= distinct; length++) {        int window_length = length * k;        int freq = { 0 };        int window_start = 0;        int window_end = window_start + window_length - 1;        for (int i = window_start;             i <= min(window_end, n - 1); i++) {            freq[s[i] - 'a']++;        }        while (window_end < n) {            if (have_same_frequency(freq, k)) {                count++;            }            freq[s[window_start] - 'a']--;            window_start++;            window_end++;            if (window_end < n) {                freq[s[window_end] - 'a']++;            }        }    }    return count;} int main(){    char* s = "aabbcc";    int k = 2;    printf("%d\n", count_substrings(s, 6, k));    s = "aabbc";    k = 2;    printf("%d\n", count_substrings(s, 5, k));    return 0;}

## Java

 import java.util.*; class GFG {     static boolean have_same_frequency(int[] freq, int k)    {        for (int i = 0; i < 26; i++) {            if (freq[i] != 0 && freq[i] != k) {                return false;            }        }        return true;    }     static int count_substrings(String s, int k)    {        int count = 0;        int distinct = 0;        boolean[] have = new boolean;        Arrays.fill(have, false);        for (int i = 0; i < s.length(); i++) {            have[((int)(s.charAt(i) - 'a'))] = true;        }        for (int i = 0; i < 26; i++) {            if (have[i]) {                distinct++;            }        }        for (int length = 1; length <= distinct; length++) {            int window_length = length * k;            int[] freq = new int;            Arrays.fill(freq, 0);            int window_start = 0;            int window_end                = window_start + window_length - 1;            for (int i = window_start;                 i <= Math.min(window_end, s.length() - 1);                 i++) {                freq[((int)(s.charAt(i) - 'a'))]++;            }            while (window_end < s.length()) {                if (have_same_frequency(freq, k)) {                    count++;                }                freq[(                    (int)(s.charAt(window_start) - 'a'))]--;                window_start++;                window_end++;                if (window_end < s.length()) {                    freq[((int)(s.charAt(window_end)                                - 'a'))]++;                }            }        }        return count;    }    public static void main(String[] args)    {        String s = "aabbcc";        int k = 2;        System.out.println(count_substrings(s, k));        s = "aabbc";        k = 2;        System.out.println(count_substrings(s, k));    }}

## Python3

 from collections import defaultdict  def have_same_frequency(freq: defaultdict, k: int):    return all([freq[i] == k or freq[i] == 0 for i in freq])  def count_substrings(s: str, k: int) -> int:    count = 0    distinct = len(set([i for i in s]))    for length in range(1, distinct + 1):        window_length = length * k        freq = defaultdict(int)        window_start = 0        window_end = window_start + window_length - 1        for i in range(window_start, min(window_end + 1, len(s))):            freq[s[i]] += 1        while window_end < len(s):            if have_same_frequency(freq, k):                count += 1            freq[s[window_start]] -= 1            window_start += 1            window_end += 1            if window_end < len(s):                freq[s[window_end]] += 1    return count  if __name__ == '__main__':    s = "aabbcc"    k = 2    print(count_substrings(s, k))    s = "aabbc"    k = 2    print(count_substrings(s, k))

## C#

 using System; class GFG{ static bool have_same_frequency(int[] freq, int k){    for(int i = 0; i < 26; i++)    {        if (freq[i] != 0 && freq[i] != k)        {            return false;        }    }    return true;} static int count_substrings(string s, int k){    int count = 0;    int distinct = 0;    bool[] have = new bool;    Array.Fill(have, false);         for(int i = 0; i < s.Length; i++)    {        have[((int)(s[i] - 'a'))] = true;    }         for(int i = 0; i < 26; i++)    {        if (have[i])        {            distinct++;        }    }         for(int length = 1; length <= distinct; length++)    {        int window_length = length * k;        int[] freq = new int;        Array.Fill(freq, 0);        int window_start = 0;        int window_end = window_start +                         window_length - 1;                                  for(int i = window_start;                i <= Math.Min(window_end, s.Length - 1);                i++)        {            freq[((int)(s[i] - 'a'))]++;        }        while (window_end < s.Length)        {            if (have_same_frequency(freq, k))            {                count++;            }            freq[((int)(s[window_start] - 'a'))]--;            window_start++;            window_end++;                         if (window_end < s.Length)            {                freq[((int)(s[window_end] - 'a'))]++;            }        }    }    return count;} // Driver codepublic static void Main(string[] args){    string s = "aabbcc";    int k = 2;    Console.WriteLine(count_substrings(s, k));         s = "aabbc";    k = 2;    Console.WriteLine(count_substrings(s, k));}} // This code is contributed by gaurav01

## Javascript

 

Output

6
3

Time Complexity: O(N * D) where D is the number of distinct characters present in the string and N is the length of the string.
Auxiliary Space: O(N)

This article is contributed by Aarti_Rathi and Rahul Chawla. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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