Given an array arr[] of size N, and an array Q[][] with each row representing a query of the form { X, Y }, the task for each query is to find the sum of array elements present at indices X, X + Y, X + 2 * Y + …
Examples:
Input: arr[] = { 1, 2, 7, 5, 4 }, Q[][] = { { 2, 1 }, { 3, 2 } }
Output: 16 5
Explanation:
Query1: arr[2] + arr[2 + 1] + arr[2 + 2] = 7 + 5 + 4 = 16.
Query2: arr[3] = 5.Input: arr[] = { 3, 6, 1, 8, 0 } Q[][] = { { 0, 2 } }
Output: 4
Naive Approach: The simplest approach to solve this problem is to traverse the array for each query and print the sum of arr[x] + arr[x + y] + arr[x + 2 * y] + …
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesvoid querySum(int arr[], int N, int Q[][2], int M){ // Iterate over each query for (int i = 0; i < M; i++) { int x = Q[i][0]; int y = Q[i][1]; // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array and calculate // the sum of the expression while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } cout << sum << " "; }}// Driver Codeint main(){ int arr[] = { 1, 2, 7, 5, 4 }; int Q[][2] = { { 2, 1 }, { 3, 2 } }; int N = sizeof(arr) / sizeof(arr[0]); int M = sizeof(Q) / sizeof(Q[0]); querySum(arr, N, Q, M); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesstatic void querySum(int arr[], int N, int Q[][], int M){ // Iterate over each query for(int i = 0; i < M; i++) { int x = Q[i][0]; int y = Q[i][1]; // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array and calculate // the sum of the expression while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } System.out.print(sum + " "); }}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 7, 5, 4 }; int Q[][] = { { 2, 1 }, { 3, 2 } }; int N = arr.length; int M = Q.length; querySum(arr, N, Q, M);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function to Find the sum of# arr[x]+arr[x+y]+arr[x+2*y] + ...# for all queriesdef querySum(arr, N, Q, M): # Iterate over each query for i in range(M): x = Q[i][0] y = Q[i][1] # Stores the sum of # arr[x]+arr[x+y]+arr[x+2*y] + ... sum = 0 # Traverse the array and calculate # the sum of the expression while (x < N): # Update sum sum += arr[x] # Update x x += y print(sum, end=" ")# Driver Codeif __name__ == '__main__': arr = [ 1, 2, 7, 5, 4 ]; Q = [ [ 2, 1 ], [3, 2 ] ] N = len(arr) M = len(Q) querySum(arr, N, Q, M) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{ // Function to Find the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... // for all queries static void querySum(int []arr, int N, int [,]Q, int M) { // Iterate over each query for(int i = 0; i < M; i++) { int x = Q[i, 0]; int y = Q[i, 1]; // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array and calculate // the sum of the expression while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } Console.Write(sum + " "); } } // Driver Code public static void Main(String[] args) { int []arr = { 1, 2, 7, 5, 4 }; int [,]Q = { { 2, 1 }, { 3, 2 } }; int N = arr.Length; int M = Q.GetLength(0); querySum(arr, N, Q, M); }}// This code is contributed by shikhasingrajput |
Javascript
<script>// JavaScript program for the above approach// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesfunction querySum(arr, N, Q, M){ // Iterate over each query for (let i = 0; i < M; i++) { let x = Q[i][0]; let y = Q[i][1]; // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... let sum = 0; // Traverse the array and calculate // the sum of the expression while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } document.write(sum + " "); }}// Driver Code let arr = [ 1, 2, 7, 5, 4 ]; let Q = [ [ 2, 1 ], [ 3, 2 ] ]; let N = arr.length; let M = Q.length; querySum(arr, N, Q, M);// This code is contributed by Surbhi Tyagi.</script> |
Output:
16 5
Time Complexity: O(|Q| * O(N))
Auxiliary Space: O(1)
Approach 2A: Using Dynamic programming technique and Square Root Decomposition technique
The problem can be solved by precomputing the value of the given expression for all possible values of { X, Y } using Dynamic programming technique and the Square Root Decomposition technique. The following are the recurrence relation:
if i + j < N dp[i][j] = dp[i + j][j] + arr[i] Otherwise, dp[i][j] = arr[i] dp[i][j]: Stores the sum of the given expression where X = i, Y = j
Follow the steps below to solve the problem:
- Initialize a 2D array, say dp[][], to store the sum of expression for all possible values of X and Y, where Y is less than or equal to sqrt(N).
- Fill the dp[][] array using tabulation method.
- Traverse the array Q[][]. For each query, check if the value of Q[i][1] is less than or equal to sqrt(N) or not. If found to be true, then print the value of dp[Q[i][0]][Q[i][1]].
- Otherwise, calculate the value of the expression using the above naive approach and print the calculated value.
Below is the implementation of our approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;const int sz = 20;const int sqr = int(sqrt(sz)) + 1;// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...// for all possible values of X and Y, where Y is// less than or equal to sqrt(N).void precomputeExpressionForAllVal(int arr[], int N, int dp[sz][sqr]){ // Iterate over all possible values of X for (int i = N - 1; i >= 0; i--) { // Precompute for all possible values // of an expression such that y <= sqrt(N) for (int j = 1; j <= sqrt(N); j++) { // If i + j less than N if (i + j < N) { // Update dp[i][j] dp[i][j] = arr[i] + dp[i + j][j]; } else { // Update dp[i][j] dp[i][j] = arr[i]; } } }}// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesint querySum(int arr[], int N, int Q[][2], int M){ // dp[x][y]: Stores sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int dp[sz][sqr]; precomputeExpressionForAllVal(arr, N, dp); // Traverse the query array, Q[][] for (int i = 0; i < M; i++) { int x = Q[i][0]; int y = Q[i][1]; // If y is less than or equal // to sqrt(N) if (y <= sqrt(N)) { cout << dp[x][y] << " "; continue; } // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array, arr[] while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } cout << sum << " "; }}// Driver Codeint main(){ int arr[] = { 1, 2, 7, 5, 4 }; int Q[][2] = { { 2, 1 }, { 3, 2 } }; int N = sizeof(arr) / sizeof(arr[0]); int M = sizeof(Q) / sizeof(Q[0]); querySum(arr, N, Q, M); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ static int sz = 20;static int sqr = (int)(Math.sqrt(sz)) + 1;// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...// for all possible values of X and Y, where Y is// less than or equal to Math.sqrt(N).static void precomputeExpressionForAllVal(int arr[], int N, int dp[][]){ // Iterate over all possible values of X for(int i = N - 1; i >= 0; i--) { // Precompute for all possible values // of an expression such that y <= Math.sqrt(N) for(int j = 1; j <= Math.sqrt(N); j++) { // If i + j less than N if (i + j < N) { // Update dp[i][j] dp[i][j] = arr[i] + dp[i + j][j]; } else { // Update dp[i][j] dp[i][j] = arr[i]; } } }}// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesstatic void querySum(int arr[], int N, int Q[][], int M){ // dp[x][y]: Stores sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int [][]dp = new int[sz][sqr]; precomputeExpressionForAllVal(arr, N, dp); // Traverse the query array, Q[][] for(int i = 0; i < M; i++) { int x = Q[i][0]; int y = Q[i][1]; // If y is less than or equal // to Math.sqrt(N) if (y <= Math.sqrt(N)) { System.out.print(dp[x][y] + " "); continue; } // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array, arr[] while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } System.out.print(sum + " "); }}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 7, 5, 4 }; int Q[][] = { { 2, 1 }, { 3, 2 } }; int N = arr.length; int M = Q.length; querySum(arr, N, Q, M);}}// This code is contributed by shikhasingrajput |
Python3
# python program for the above approachimport mathsz = 20sqr = int(math.sqrt(sz)) + 1# Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...# for all possible values of X and Y, where Y is# less than or equal to sqrt(N).def precomputeExpressionForAllVal(arr, N, dp): # Iterate over all possible values of X for i in range(N - 1, -1, -1) : # Precompute for all possible values # of an expression such that y <= sqrt(N) for j in range (1,int(math.sqrt(N)) + 1): # If i + j less than N if (i + j < N): # Update dp[i][j] dp[i][j] = arr[i] + dp[i + j][j] else: # Update dp[i][j] dp[i][j] = arr[i]# Function to Find the sum of# arr[x]+arr[x+y]+arr[x+2*y] + ...# for all queriesdef querySum(arr, N, Q, M): # dp[x][y]: Stores sum of # arr[x]+arr[x+y]+arr[x+2*y] + ... dp = [ [0 for x in range(sz)]for x in range(sqr)] precomputeExpressionForAllVal(arr, N, dp) # Traverse the query array, Q[][] for i in range (0,M): x = Q[i][0] y = Q[i][1] # If y is less than or equal # to sqrt(N) if (y <= math.sqrt(N)): print(dp[x][y]) continue # Stores the sum of # arr[x]+arr[x+y]+arr[x+2*y] + ... sum = 0 # Traverse the array, arr[] while (x < N): # Update sum sum += arr[x] # Update x x += y print(sum)# Driver Codearr = [ 1, 2, 7, 5, 4 ]Q = [ [ 2, 1 ], [ 3, 2]]N = len(arr)M = len(Q[0])querySum(arr, N, Q, M)# This code is contributed by amreshkumar3. |
C#
// C# program for the above approachusing System;class GFG{ static int sz = 20;static int sqr = (int)(Math.Sqrt(sz)) + 1;// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...// for all possible values of X and Y, where Y is// less than or equal to Math.Sqrt(N).static void precomputeExpressionForAllVal(int []arr, int N, int [,]dp){ // Iterate over all possible values of X for(int i = N - 1; i >= 0; i--) { // Precompute for all possible values // of an expression such that y <= Math.Sqrt(N) for(int j = 1; j <= Math.Sqrt(N); j++) { // If i + j less than N if (i + j < N) { // Update dp[i,j] dp[i, j] = arr[i] + dp[i + j, j]; } else { // Update dp[i,j] dp[i, j] = arr[i]; } } }}// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesstatic void querySum(int []arr, int N, int [,]Q, int M){ // dp[x,y]: Stores sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int [,]dp = new int[sz, sqr]; precomputeExpressionForAllVal(arr, N, dp); // Traverse the query array, Q[,] for(int i = 0; i < M; i++) { int x = Q[i, 0]; int y = Q[i, 1]; // If y is less than or equal // to Math.Sqrt(N) if (y <= Math.Sqrt(N)) { Console.Write(dp[x, y] + " "); continue; } // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array, []arr while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } Console.Write(sum + " "); }}// Driver Codepublic static void Main(String[] args){ int []arr = { 1, 2, 7, 5, 4 }; int [,]Q = { { 2, 1 }, { 3, 2 } }; int N = arr.Length; int M = Q.GetLength(0); querySum(arr, N, Q, M);}}// This code is contributed by shikhasingrajput |
Javascript
<script>// javascript program of the above approach let sz = 20; let sqr = (Math.sqrt(sz)) + 1;// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...// for all possible values of X and Y, where Y is// less than or equal to Math.sqrt(N).function precomputeExpressionForAllVal(arr, N, dp){ // Iterate over all possible values of X for(let i = N - 1; i >= 0; i--) { // Precompute for all possible values // of an expression such that y <= Math.sqrt(N) for(let j = 1; j <= Math.sqrt(N); j++) { // If i + j less than N if (i + j < N) { // Update dp[i][j] dp[i][j] = arr[i] + dp[i + j][j]; } else { // Update dp[i][j] dp[i][j] = arr[i]; } } }}// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesfunction querySum(arr, N, Q, M){ let dp = new Array(sz); // Loop to create 2D array using 1D array for (var i = 0; i < dp.length; i++) { dp[i] = new Array(2); } // Loop to create 2D array using 1D array for (var i = 0; i < dp.length; i++) { dp[i] = new Array(2); } precomputeExpressionForAllVal(arr, N, dp); // Traverse the query array, Q[][] for(let i = 0; i < M; i++) { let x = Q[i][0]; let y = Q[i][1]; // If y is less than or equal // to Math.sqrt(N) if (y <= Math.sqrt(N)) { document.write(dp[x][y] + " "); continue; } // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... let sum = 0; // Traverse the array, arr[] while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } Sdocument.write(sum + " "); }} // Driver Code // Given array let arr = [ 1, 2, 7, 5, 4 ]; let Q= [[ 2, 1 ], [ 3, 2 ]] let N = arr.length; let M = Q.length; querySum(arr, N, Q, M); // This code is contributed by chinmoy1997pal.</script> |
Output:
16 5
Time complexity: O(N * sqrt(N) + |Q| * sqrt(N))
Auxiliary Space:O(N * sqrt(N))
Approach 2B: Space optimization
In the previous approach, we can see that in the function precomputeExpressionForAllVal we use 2d dp which is not required because dp[i][j] is dependent upon arr[i] and dp[i + j][j] so we can optimize its space complexity by using 1d DP.
Implementation:
- Initialize dp to all zeros and Iterate over all possible values of X from N-1 down to 0.
- Precompute for all possible values of an expression such that y <= sqrt(N).
- If i + j is less than N, update dp[i] to add arr[i+j].
- Define the function querySum to take in the input parameters arr, N, Q, and M.
- Traverse the query array Q[][], where each query is represented by a pair of integers {x, y}.
- If y is less than or equal to sqrt(N), iterate over arr[x], arr[x+y], arr[x+2y],…, until the end of the array and print their sum.
- Otherwise, for each query, initialize sum to zero and traverse the array arr[] starting from index x and incrementing by y.
- Add each element arr[i] to sum until x is less than N.
- Print the final sum for each query.
Example:
C++
#include <bits/stdc++.h>using namespace std;// Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ...// for all possible values of X and Y, where Y is// less than or equal to sqrt(N).void precomputeExpressionForAllVal(int arr[], int N, int dp[]){ // Iterate over all possible values of X for (int i = N - 1; i >= 0; i--) { // Precompute for all possible values // of an expression such that y <= sqrt(N) for (int j = 1; j <= sqrt(N); j++) { // If i + j less than N if (i + j < N) { // Update dp[i+j] dp[i] += arr[i + j]; } } }}// Function to Find the sum of// arr[x]+arr[x+y]+arr[x+2*y] + ...// for all queriesvoid querySum(int arr[], int N, int Q[][2], int M){ // Traverse the query array, Q[][] for (int i = 0; i < M; i++) { int x = Q[i][0]; int y = Q[i][1]; // If y is less than or equal // to sqrt(N) if (y <= sqrt(N)) { int ans = 0; for (int j = x; j < N; j += y) ans += arr[j]; cout << ans << " "; continue; } // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array, arr[] while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } cout << sum << " "; }}// Driver Codeint main(){ int arr[] = { 1, 2, 7, 5, 4 }; int Q[][2] = { { 2, 1 }, { 3, 2 } }; int N = sizeof(arr) / sizeof(arr[0]); int M = sizeof(Q) / sizeof(Q[0]); // dp[x]: Stores sum of arr[x]+arr[x+y]+arr[x+2*y] + ... // for all possible values of X and Y, where Y is // less than or equal to sqrt(N). int dp[N] = { 0 }; precomputeExpressionForAllVal(arr, N, dp); querySum(arr, N, Q, M); return 0;}// this code is contributed by bhardwajji |
Java
import java.util.*;public class Main { // Function to sum of arr[x]+arr[x+y]+arr[x+2*y] + ... // for all possible values of X and Y, where Y is // less than or equal to sqrt(N). static void precomputeExpressionForAllVal(int[] arr, int N, int[] dp) { // Iterate over all possible values of X for (int i = N - 1; i >= 0; i--) { // Precompute for all possible values // of an expression such that y <= sqrt(N) for (int j = 1; j <= Math.sqrt(N); j++) { // If i + j less than N if (i + j < N) { // Update dp[i+j] dp[i] += arr[i + j]; } } } } // Function to Find the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... // for all queries static void querySum(int[] arr, int N, int[][] Q, int M) { // Traverse the query array, Q[][] for (int i = 0; i < M; i++) { int x = Q[i][0]; int y = Q[i][1]; // If y is less than or equal // to sqrt(N) if (y <= Math.sqrt(N)) { int ans = 0; for (int j = x; j < N; j += y) ans += arr[j]; System.out.print(ans + " "); continue; } // Stores the sum of // arr[x]+arr[x+y]+arr[x+2*y] + ... int sum = 0; // Traverse the array, arr[] while (x < N) { // Update sum sum += arr[x]; // Update x x += y; } System.out.print(sum + " "); } } // Driver Code public static void main(String[] args) { int[] arr = { 1, 2, 7, 5, 4 }; int[][] Q = { { 2, 1 }, { 3, 2 } }; int N = arr.length; int M = Q.length; // dp[x]: Stores sum of arr[x]+arr[x+y]+arr[x+2*y] + // ... for all possible values of X and Y, where Y // is less than or equal to sqrt(N). int[] dp = new int[N]; precomputeExpressionForAllVal(arr, N, dp); querySum(arr, N, Q, M); }} |
Python3
import mathdef precompute_expression_for_all_val(arr, N, dp): # Iterate over all possible values of X for i in range(N - 1, -1, -1): # Precompute for all possible values # of an expression such that y <= sqrt(N) for j in range(1, int(math.sqrt(N)) + 1): # If i + j less than N if i + j < N: # Update dp[i+j] dp[i] += arr[i + j]def query_sum(arr, N, Q, M): # Traverse the query array, Q[][] for i in range(M): x = Q[i][0] y = Q[i][1] # If y is less than or equal # to sqrt(N) if y <= int(math.sqrt(N)): ans = 0 for j in range(x, N, y): ans += arr[j] print(ans, end=" ") continue # Stores the sum of # arr[x]+arr[x+y]+arr[x+2*y] + ... sum = 0 # Traverse the array, arr[] while x < N: # Update sum sum += arr[x] # Update x x += y print(sum, end=" ")# Driver Codearr = [1, 2, 7, 5, 4]Q = [[2, 1], [3, 2]]N = len(arr)M = len(Q)# dp[x]: Stores sum of arr[x]+arr[x+y]+arr[x+2*y] +# ... for all possible values of X and Y, where Y# is less than or equal to sqrt(N).dp = [0] * Nprecompute_expression_for_all_val(arr, N, dp)query_sum(arr, N, Q, M) |
C#
// C# implementation of the above codeusing System;class Program { static void PrecomputeExpressionForAllVal(int[] arr, int N, int[] dp) { for (int i = N - 1; i >= 0; i--) { for (int j = 1; j <= Math.Sqrt(N); j++) { if (i + j < N) { dp[i] += arr[i + j]; } } } } static void QuerySum(int[] arr, int N, int[][] Q, int M) { for (int i = 0; i < M; i++) { int x = Q[i][0]; int y = Q[i][1]; if (y <= Math.Sqrt(N)) { int ans = 0; for (int j = x; j < N; j += y) { ans += arr[j]; } Console.Write(ans + " "); continue; } int sum = 0; while (x < N) { sum += arr[x]; x += y; } Console.Write(sum + " "); } } static void Main(string[] args) { int[] arr = { 1, 2, 7, 5, 4 }; int[][] Q = { new int[] { 2, 1 }, new int[] { 3, 2 } }; int N = arr.Length; int M = Q.Length; int[] dp = new int[N]; PrecomputeExpressionForAllVal(arr, N, dp); QuerySum(arr, N, Q, M); }}// This code is contributed by user_dtewbxkn77n |
Javascript
// JavaScript implementation of the above codefunction precomputeExpressionForAllVal(arr, N) {const dp = new Array(N).fill(0);for (let i = N - 1; i >= 0; i--) {for (let j = 1; j <= Math.sqrt(N); j++) {if (i + j < N) {dp[i] += arr[i + j];}}}return dp;}function querySum(arr, N, Q, M) {for (let i = 0; i < M; i++) {const x = Q[i][0];const y = Q[i][1];if (y <= Math.sqrt(N)) {let ans = 0;for (let j = x; j < N; j += y) {ans += arr[j];}console.log(ans + " ");continue;}let sum = 0;while (x < N) {sum += arr[x];x += y;}console.log(sum + " ");}}const arr = [1, 2, 7, 5, 4];const Q = [[2, 1],[3, 2]];const N = arr.length;const M = Q.length;const dp = precomputeExpressionForAllVal(arr, N);querySum(arr, N, Q, M); |
Output:
16 5
Time complexity: O(N * sqrt(N) + |Q| * sqrt(N))
Auxiliary Space: O(N)
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