Perform range sum queries on string as per given condition

Last Updated : 30 Jun, 2022

Given a string S with lowercase alphabets only and Q queries where each query contains a pair {L, R}. For each query {L, R}, there exists a substring S[L, R], the task is to find the value of the product of the frequency of each character in substring with their position in alphabetical order.
Note: Consider 1-based indexing.
Examples:

Input: S = "abcd", Q = { {2, 4}, {1, 3} }
Output: 9 6
Explanation:
For 1st query,
substring is S[2, 4] = "bcd". Therefore the frequency of b, c, d are 1, 1, 1 in range 2 to 4.
value = 2*(1) + 3*(1) + 4*(1) = 9.
For 2nd query,
substring is S[1, 3] = "abc". Therefore the frequency of a, b, c are 1, 1, 1 in range 1 to 3.
value = 1*(1) + 2*(1) + 3*(1) = 6.
Input: S = "geeksforgeeks", Q = { {3, 3}, {2, 6}, {1, 13} }
Output: 5 46 133

Naive Approach: The naive idea is to traverse for each ranges [L, R] in the query and keep the count of each character in an array. After traversing the range find the value of the expression 1*(occurrences of 'a') + 2*(occurrences of 'b') + 3*(occurrences of 'c') + ..+ 26*(occurrences of 'z').
Time Complexity: O(N*Q), where N is the length of the given string.
Auxiliary Space: O(1)
Efficient Approach: The idea is to use the Prefix Sum of the whole string by which we can perform each query in constant time. Below are the steps:

Create an array arr[] of length equals to the length of the string.
Traverse the given string and for each corresponding index i in the string, assign arr[i] the value of current character - 'a'.
Find the prefix sum of the array arr[]. This prefix sum array will given sum of occurrence of all characters till each index i.
Now for each query(say {L, R}) the value of arr[R - 1] - arr[L - 2] will give the value of the given expression.

Below is the implementation of the above approach:

C++

// C++ program for the above approach #include <bits/stdc++.h> using namespace std;  // Function to perform range sum queries // on string as per the given condition void Range_sum_query(string S,                      vector<pair<int, int> > Query) {     // Initialize N by string size     int N = S.length();      // Create array A[] for prefix sum     int A[N];      A[0] = S[0] - 'a' + 1;      // Iterate till N     for (int i = 1; i < N; i++) {         A[i] = S[i] - 'a' + 1;         A[i] = A[i] + A[i - 1];     }      // Traverse the queries     for (int i = 0; i < Query.size(); i++) {          if (Query[i].first == 1) {              // Check if L == 1 range             // sum will be A[R-1]             cout << A[(Query[i].second) - 1]                  << endl;         }          else {              // Condition if L > 1 range sum             // will be A[R-1] - A[L-2]             cout << A[(Query[i].second) - 1]                         - A[(Query[i].first) - 2]                  << endl;         }     } }  // Driver Code int main() {     // Given string     string S = "abcd";      vector<pair<int, int> > Query;      // Given Queries     Query.push_back(make_pair(2, 4));     Query.push_back(make_pair(1, 3));      // Function call     Range_sum_query(S, Query);     return 0; }

Java

// Java program for the above approach import java.util.*;  class GFG{      static class pair {      int first, second;      public pair(int first, int second)      {          this.first = first;          this.second = second;      }  }   // Function to perform range sum queries // on String as per the given condition static void Range_sum_query(String S,                             Vector<pair> Query) {          // Initialize N by String size     int N = S.length();      // Create array A[] for prefix sum     int []A = new int[N];      A[0] = S.charAt(0) - 'a' + 1;      // Iterate till N     for(int i = 1; i < N; i++)     {         A[i] = S.charAt(i) - 'a' + 1;         A[i] = A[i] + A[i - 1];     }      // Traverse the queries     for(int i = 0; i < Query.size(); i++)     {         if (Query.get(i).first == 1)         {                          // Check if L == 1 range             // sum will be A[R-1]             System.out.print(                 A[(Query.get(i).second) - 1] + "\n");         }          else          {                          // Condition if L > 1 range sum             // will be A[R-1] - A[L-2]             System.out.print(                 A[(Query.get(i).second) - 1] -                 A[(Query.get(i).first) - 2] + "\n");         }     } }  // Driver Code public static void main(String[] args) {          // Given String     String S = "abcd";      Vector<pair> Query = new Vector<pair>();      // Given Queries     Query.add(new pair(2, 4));     Query.add(new pair(1, 3));      // Function call     Range_sum_query(S, Query); } }  // This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach   # Function to perform range sum queries # on string as per the given condition def Range_sum_query(S, Query):      # Initialize N by string size     N = len(S)      # Create array A[] for prefix sum     A = [0] * N      A[0] = ord(S[0]) - ord('a') + 1      # Iterate till N     for i in range(1, N):         A[i] = ord(S[i]) - ord('a') + 1         A[i] = A[i] + A[i - 1]      # Traverse the queries     for i in range(len(Query)):         if(Query[i][0] == 1):              # Check if L == 1 range             # sum will be A[R-1]             print(A[Query[i][1] - 1])          else:              # Condition if L > 1 range sum             # will be A[R-1] - A[L-2]             print(A[Query[i][1] - 1] -                   A[Query[i][0] - 2])  # Driver Code  # Given string S = "abcd"  Query = []  # Given Queries Query.append([2, 4]) Query.append([1, 3])  # Function call  Range_sum_query(S, Query)  # This code is contributed by Shivam Singh

// C# program for the above approach using System; using System.Collections.Generic;   class GFG{    class pair   {     public int first, second;     public pair(int first, int second)     {       this.first = first;       this.second = second;     }   }    // Function to perform range sum queries   // on String as per the given condition   static void Range_sum_query(String S, List<pair> Query)   {      // Initialize N by String size     int N = S.Length;      // Create array []A for prefix sum     int[] A = new int[N];      A[0] = S[0] - 'a' + 1;      // Iterate till N     for (int i = 1; i < N; i++)     {       A[i] = S[i] - 'a' + 1;       A[i] = A[i] + A[i - 1];     }      // Traverse the queries     for (int i = 0; i < Query.Count; i++)      {       if (Query[i].first == 1)       {          // Check if L == 1 range         // sum will be A[R-1]         Console.Write(A[(Query[i].second) - 1] + "\n");       }        else        {          // Condition if L > 1 range sum         // will be A[R-1] - A[L-2]         Console.Write(A[(Query[i].second) - 1] -                        A[(Query[i].first) - 2] + "\n");       }     }   }    // Driver Code   public static void Main(String[] args)   {      // Given String     String S = "abcd";      List<pair> Query = new List<pair>();      // Given Queries     Query.Add(new pair(2, 4));     Query.Add(new pair(1, 3));      // Function call     Range_sum_query(S, Query);   } }  // This code is contributed by gauravrajput1

JavaScript

<script>  // Javascript program for the above approach  // Function to perform range sum queries // on string as per the given condition function Range_sum_query(S, Query) {      // Initialize N by string size     var N = S.length;      // Create array A[] for prefix sum     var A = Array(N);      A[0] = S[0].charCodeAt(0) - 'a'.charCodeAt(0) + 1;      // Iterate till N     for (var i = 1; i < N; i++) {         A[i] = S[i].charCodeAt(0) - 'a'.charCodeAt(0) + 1;         A[i] = A[i] + A[i - 1];     }      // Traverse the queries     for (var i = 0; i < Query.length; i++) {          if (Query[i][0] == 1) {              // Check if L == 1 range             // sum will be A[R-1]             document.write( A[(Query[i][1]) - 1]+ "<br>");         }          else {              // Condition if L > 1 range sum             // will be A[R-1] - A[L-2]             document.write( A[(Query[i][1]) - 1]                         - A[(Query[i][0]) - 2]+ "<br>");         }     } }  // Driver Code // Given string var S = "abcd"; var Query = [];  // Given Queries Query.push([2, 4]); Query.push([1, 3]);  // Function call Range_sum_query(S, Query);  // This code is contributed by itsok. </script>

Output:

9 6

Time Complexity: O(N), where N is the length of the given string.
Auxiliary Space: O(N)

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Perform range sum queries on string as per given condition

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