# Calculate Sum of ratio of special characters to length of substrings of the given string

Given a string str and an array of special characters, specialArray[], the task is to find the sum of the ratio of the count of special characters to the length of the substring for all possible substring of the given string.

Ratio of count of special characters in a substring to the length of subtrings of the given string is given by

Sum of the ratios calculated above is given by

Examples:

Input: str = “abcdabc”, specialArray[] = {‘a’, ‘b’, ‘c’, ‘d’}
Output: 28.00000
Explanation:
Length of string = 7
Count of all possible substrings = (7 * (8 + 1)) / 2 = 28
Since, all the characters of the string are included in sprecialArray[], ratio of count of special characters to the length of substring for every substring will always be 1.
Hence, the sum of ratio = Number of substrings * 1 = 28.
Input: str = “abcd”, specialArray[] = {‘b’, ‘c’}
Output: 5.83333

Approach:
Follow the steps below to solve the problem:

• For every possible length of substrings from 1 to N, find the count of special characters in every substring of length x and add the ratio of count and x to the answer.
• To find the count of special characters in each substring in constant time, create a prefix sum array of the count of special characters using the relation:

prefix[i] = prefix[i – 1] + special(s[i]);

• Calculate the count of special characters in a substring within the indices [i, j] is given by the relation:

prefix[j – 1] – prefix[i – 1]
Therefore, the ratio of the count of special characters to the length of substring,
f(i, j) = (prefix[j – 1] – prefix[i – 1])/(j – i + 1)

Below is the implementation of the above approach:

## C++

 // C++ Program to impleemnt // the above approach #include  using namespace std;   const int N = 1e5 + 5;   // Stores frequency of special // characters in the array vector<int> prefix(N, 0);   // Stores prefix sum vector<int> sum(N, 0);   // Function to check whether a character // is special or not bool isSpecial(char c,                vector<char>& special) {     for (auto& i : special)           // If current character         // is special         if (i == c)             return true;       // Otherwise     return false; }   // Function to find sum of ratio of // count of special characters and // length of substrings double countRatio(string& s,                   vector<char>& special) {       int n = s.length();     for (int i = 0; i < n; i++) {           // Calculate the prefix sum of         // special nodes         prefix[i] = int(isSpecial(s[i],                                   special));         if (i > 0)             prefix[i] += prefix[i - 1];     }       for (int i = 0; i < n; i++) {           // Generate prefix sum array         sum[i] = prefix[i];         if (i > 0)             sum[i] += sum[i - 1];     }       double ans = 0;     for (int i = 1; i <= n; i++) {           // Calculate ratio for substring         int count = sum[n - 1]                     - (i > 1 ? sum[i - 2] : 0);         count             -= (i < n ? sum[n - i - 1] : 0);           ans += double(count) / double(i);     }       return ans; }   // Driver Code; int main() {     string s = "abcd";     vector<char> special = { 'b', 'c' };       double ans = countRatio(s, special);       cout << fixed << setprecision(6)          << ans << endl;       return 0; }

## Java

 // Java Program to impleemnt // the above approach import java.util.*; class GFG{   static int N = 1000000 + 5;   // Stores frequency of special // characters in the array static int []prefix = new int[N];     // Stores prefix sum static int []sum = new int[N];   // Function to check  // whether a character // is special or not static int isSpecial(char c,                      char[] special) {   for (char i : special)       // If current character     // is special     if (i == c)       return 1;     // Otherwise   return 0; }   // Function to find sum of ratio of // count of special characters and // length of subStrings static  double countRatio(char []s,                           char[] special) {   int n = s.length;   for (int i = 0; i < n; i++)    {     // Calculate the prefix sum of     // special nodes     prefix[i] = (isSpecial(s[i],                  special));     if (i > 0)       prefix[i] += prefix[i - 1];   }     for (int i = 0; i < n; i++)    {     // Generate prefix sum array     sum[i] = prefix[i];     if (i > 0)       sum[i] += sum[i - 1];   }     double ans = 0;   for (int i = 1; i <= n; i++)    {     // Calculate ratio for subString     int count = sum[n - 1] - (i > 1 ?                  sum[i - 2] : 0);     count -= (i < n ?                sum[n - i - 1] : 0);     ans += (double)count / (double)i;   }     return ans; }   // Driver Code; public static void main(String[] args) {   String s = "abcd";   char[] special = {'b', 'c'};   double ans = countRatio(s.toCharArray(),                            special);   System.out.format("%.6f",ans); } }   // This code is contributed by gauravrajput1

## Python3

 # Python3 program to implement # the above approach N = 100005   # Stores frequency of special # characters in the array prefix = [0] * N   # Stores prefix sum sum = [0] * N   # Function to check whether a character # is special or not def isSpecial(c, special):       for i in special:           # If current character         # is special         if (i == c):             return True       # Otherwise     return False   # Function to find sum of ratio of # count of special characters and # length of substrings def countRatio(s, special):       n = len(s)     for i in range(n):           # Calculate the prefix sum of         # special nodes         prefix[i] = int(isSpecial(s[i],                                   special))         if (i > 0):             prefix[i] += prefix[i - 1]       for i in range(n):           # Generate prefix sum array         sum[i] = prefix[i]         if (i > 0):             sum[i] += sum[i - 1]       ans = 0     for i in range(1, n + 1):           # Calculate ratio for substring         if i > 1:             count = sum[n - 1]- sum[i - 2]         else:             count = sum[n - 1]         if i < n:             count -= sum[n - i - 1]           ans += count / i           return ans   # Driver Code if __name__ == "__main__":           s = "abcd"     special = [ 'b', 'c' ]       ans = countRatio(s, special)       print('%.6f' % ans)   # This code is contributed by chitranayal

Output:

5.833333



Time Complexity: O(N)
Space Complexity: O(N)

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Improved By : chitranayal, GauravRajput1