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Queries to calculate the Sum of Array elements in the range [L, R] having indices as multiple of K
  • Difficulty Level : Expert
  • Last Updated : 24 Aug, 2020

Given an array arr[] consisting of N integers, and a matrix Q[][] consisting of queries of the form (L, R, K), the task for each query is to calculate the sum of array elements from the range [L, R] which are present at indices(0- based indexing) which are multiples of K and 

Examples:

Input: arr[]={1, 2, 3, 4, 5, 6}, Q[][]={{2, 5, 2}, {0, 5, 1}}
Output: 
8
21
Explanation: 
Query1: Indexes (2, 4) are multiple of K(= 2) from the range [2, 5]. Therefore, required Sum = 3+5 = 8.
Query2: Since all indices are a multiple of K(= 1), therefore, the required sum from the range [0, 5] = 1 + 2 + 3 + 4 + 5 + 6 = 21

Input: arr[]={4, 3, 5, 1, 9}, Q[][]={{1, 4, 1}, {3, 4, 3}}
Output:
18
1

Approach: The problem can be solved using Prefix Sum Array and Range sum query technique. Follow the steps below to solve the problem:



  1. Initialize a matrix of size prefixSum[][] such that prefixSum[i][j] stores the sum of elements present in indices which are a multiple of i up to jth index.
  2. Traverse the array and precompute the prefix sums.
  3. Traverse each query, print the result of prefixSum[K][R] – prefixSum[K][L – 1].

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above appoach
#include <bits/stdc++.h>
using namespace std;
  
// Structure of a Query
struct Node {
    int L;
    int R;
    int K;
};
  
// Function to calculate the sum of array
// elements at indices from range [L, R]
// which are multiples of K for each query
int kMultipleSum(int arr[], Node Query[],
                 int N, int Q)
{
    // Stores Prefix Sum
    int prefixSum[N + 1][N];
  
    // prefixSum[i][j] : Stores the sum from
    // indices [0, j] which are multiples of i
    for (int i = 1; i <= N; i++) {
        prefixSum[i][0] = arr[0];
        for (int j = 0; j < N; j++) {
  
            // If index j is a multiple of i
            if (j % i == 0) {
  
                // Compute prefix sum
                prefixSum[i][j]
                    = arr[j] + prefixSum[i][j - 1];
            }
  
            // Otherwise
            else {
                prefixSum[i][j]
                    = prefixSum[i][j - 1];
            }
        }
    }
  
    // Traverse each query
    for (int i = 0; i < Q; i++) {
  
        // Sum of all indices upto R which
        // are a multiple of K
        int last
            = prefixSum[Query[i].K][Query[i].R];
        int first;
  
        // Sum of all indices upto L - 1 which
        // are a multiple of K
        if (Query[i].L == 0) {
            first
                = prefixSum[Query[i].K][Query[i].L];
        }
        else {
            first
                = prefixSum[Query[i].K][Query[i].L - 1];
        }
  
        // Calculate the difference
        cout << last - first << endl;
    }
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int Q = 2;
    Node Query[Q];
    Query[0].L = 2, Query[0].R = 5, Query[0].K = 2;
    Query[1].L = 3, Query[1].R = 5, Query[1].K = 5;
    kMultipleSum(arr, Query, N, Q);
}

Java




// Java program to implement
// the above appoach
import java.util.*;
  
class GFG{
  
// Structure of a Query
static class Node
{
    int L;
    int R;
    int K;
};
  
// Function to calculate the sum of array
// elements at indices from range [L, R]
// which are multiples of K for each query
static void kMultipleSum(int arr[], Node Query[], 
                         int N, int Q)
{
      
    // Stores Prefix Sum
    int prefixSum[][] = new int[N + 1][N];
  
    // prefixSum[i][j] : Stores the sum from
    // indices [0, j] which are multiples of i
    for(int i = 1; i <= N; i++)
    {
        prefixSum[i][0] = arr[0];
        for(int j = 0; j < N; j++)
        {
              
            // If index j is a multiple of i
            if (j % i == 0)
            {
                  
                // Compute prefix sum
                if (j != 0)
                    prefixSum[i][j] = arr[j] + 
                                prefixSum[i][j - 1];
            }
  
            // Otherwise
            else
            {
                prefixSum[i][j] = prefixSum[i][j - 1];
            }
        }
    }
  
    // Traverse each query
    for(int i = 0; i < Q; i++)
    {
          
        // Sum of all indices upto R which
        // are a multiple of K
        int last = prefixSum[Query[i].K][Query[i].R];
        int first;
  
        // Sum of all indices upto L - 1 which
        // are a multiple of K
        if (Query[i].L == 0)
        {
            first = prefixSum[Query[i].K][Query[i].L];
        }
        else 
        {
            first = prefixSum[Query[i].K][Query[i].L - 1];
        }
  
        // Calculate the difference
        System.out.print(last - first + "\n");
    }
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6 };
    int N = arr.length;
    int Q = 2;
      
    Node Query[] = new Node[Q];
    for(int i = 0; i < Q; i++)
        Query[i] = new Node();
          
    Query[0].L = 2;
    Query[0].R = 5;
    Query[0].K = 2;
    Query[1].L = 3;
    Query[1].R = 5;
    Query[1].K = 5;
      
    kMultipleSum(arr, Query, N, Q);
}
}
  
// This code is contributed by 29AjayKumar

C#




// C# program to implement
// the above appoach
using System;
  
class GFG{
  
// Structure of a Query
class Node
{
    public int L;
    public int R;
    public int K;
};
  
// Function to calculate the sum of array
// elements at indices from range [L, R]
// which are multiples of K for each query
static void kMultipleSum(int []arr, Node []Query, 
                         int N, int Q)
{
      
    // Stores Prefix Sum
    int [,]prefixSum = new int[N + 1, N];
      
    // prefixSum[i,j] : Stores the sum from
    // indices [0, j] which are multiples of i
    for(int i = 1; i <= N; i++)
    {
        prefixSum[i, 0] = arr[0];
        for(int j = 0; j < N; j++)
        {
              
            // If index j is a multiple of i
            if (j % i == 0)
            {
                  
                // Compute prefix sum
                if (j != 0)
                    prefixSum[i, j] = arr[j] + 
                                prefixSum[i, j - 1];
            }
  
            // Otherwise
            else
            {
                prefixSum[i, j] = prefixSum[i, j - 1];
            }
        }
    }
  
    // Traverse each query
    for(int i = 0; i < Q; i++)
    {
          
        // Sum of all indices upto R which
        // are a multiple of K
        int last = prefixSum[Query[i].K,Query[i].R];
        int first;
  
        // Sum of all indices upto L - 1 which
        // are a multiple of K
        if (Query[i].L == 0)
        {
            first = prefixSum[Query[i].K,Query[i].L];
        }
        else
        {
            first = prefixSum[Query[i].K,Query[i].L - 1];
        }
  
        // Calculate the difference
        Console.Write(last - first + "\n");
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5, 6 };
    int N = arr.Length;
    int Q = 2;
      
    Node []Query = new Node[Q];
    for(int i = 0; i < Q; i++)
        Query[i] = new Node();
          
    Query[0].L = 2;
    Query[0].R = 5;
    Query[0].K = 2;
    Query[1].L = 3;
    Query[1].R = 5;
    Query[1].K = 5;
      
    kMultipleSum(arr, Query, N, Q);
}
}
  
// This code is contributed by 29AjayKumar
Output: 
8
6

Time Complexity: O(N2 + O(Q)), Computing the prefix sum array requires O(N2) computational complexity and each query requires O(1) computational complexity. 
Auxiliary Space: O(N2
 

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