# Count of all sub-strings with sum of weights at most K

Given a string S consisting of small English letters and a string W consisting of weight of all characters of English alphabet where for all i, . We have to find the total numbers of a unique substring with sum of weights at most K.

Examples:

Input : P = “ababab”, Q = “12345678912345678912345678”, K=5
Output : 7
Explanation :
The unique substrings are: “a”, “ab”, “b”, “bc”, “c”, “d”, “e”
Hence, the count is 7.

Input : P = “acbacbacaa”, Q = “12300045600078900012345000”, K=2
Output : 3
Explanation :The unique substrings are: “a”, “b”, “aa”
Hence, the count is 3.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:
To solve the above problem, the main idea is to simply iterate through all the substrings and maintain a sum of the weight of all characters encountered so far. If the sum of characters is not greater than K, then insert it in a hashmap otherwise discard it and move forward with another substring. Finally, the result will be the size of the hashmap because it stores all the substring which have weight less than or equal to K.

Below is the implementation of the above approach:

## C++

 // C++ implementation to Count all  // sub-strings with sum of weights at most K     #include  using namespace std;     // Function to count all substrings  int distinctSubstring(string& P, string& Q,                        int K, int N)  {         // Hashmap to store substrings      unordered_set S;         // iterate over all substrings      for (int i = 0; i < N; ++i) {             // variable to maintain sum          // of all characters encountered          int sum = 0;             // variable to maintain          // substring till current position          string s;             for (int j = i; j < N; ++j) {              // get position of              // character in string W              int pos = P[j] - 'a';                 // add weight to current sum              sum += Q[pos] - '0';                 // add current character to substring              s += P[j];                 // check if sum of characters              // is <=K insert in Hashmap              if (sum <= K) {                  S.insert(s);              }              else {                  break;              }          }      }         return S.size();  }     // Driver code  int main()  {      // initialise string      string S = "abcde";         // initialise weight      string W = "12345678912345678912345678";         int K = 5;         int N = S.length();         cout << distinctSubstring(S, W, K, N);         return 0;  }

## Python3

 # Python3 implementation to Count all  # sub-strings with sum of weights at most K     # Function to count all substrings  def distinctSubstring(P, Q, K, N):         # Hashmap to store substrings      S = set()         # iterate over all substrings      for i in range(N):             # variable to maintain sum          # of all characters encountered          sum = 0            # variable to maintain          # substring till current position          s = ""             for j in range(i, N):                 # get position of              # character in string W              pos = ord(P[j]) - 97                # add weight to current sum              sum += ord(Q[pos]) - 48                # add current character to substring              s += P[j]                 # check if sum of characters              # is <=K insert in Hashmap              if (sum <= K):                  S.add(s)                 else:                  break        return len(S)     # Driver code  if __name__ == '__main__':      # initialise string      S = "abcde"        # initialise weight      W = "12345678912345678912345678"        K = 5        N = len(S)         print(distinctSubstring(S, W, K, N))     # This code is contributed by Surendra_Gangwar

Output:

7


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Improved By : SURENDRA_GANGWAR