Queries to calculate the Sum of Array elements in the range [L, R] having indices as multiple of K
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 = 21Input: 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:
- 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.
- Traverse the array and precompute the prefix sums.
- 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); } |
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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 |
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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 |
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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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