Given an array arr[] and two integers L and R. The task is to find the sum of the product of all the pairs (i, j) in the range [L, R], such that i ≤ j.
Input: arr[] = { 1, 3, 5, 8 }, L = 0, R = 2
Output: 58
Explanation: As 1*1 + 1*3 + 1*5 + 3*3 + 3*5 + 5*5 = 58Input: arr[] = { 2, 1, 4, 5, 3, 2, 1 }, L = 1, R = 5
Output: 140
Naive Approach: The brute force approach can be directly implemented by multiplying the indices using two nested loops and storing the sum in a variable.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the sum// of (arr[i]*arr[j]) for all i and j// between the index L and Rint sum_of_products(int arr[], int N, int L, int R){ int sum = 0; for (int i = L; i <= R; i++) { for (int j = i; j <= R; j++) { sum += arr[i] * arr[j]; } } return sum;}// Driver codeint main(){ int arr[] = { 1, 3, 5, 8 }; int N = sizeof(arr) / sizeof(arr[0]); int L = 0; int R = 2; cout << sum_of_products(arr, N, L, R); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;public class GFG { // Function to return the sum // of (arr[i]*arr[j]) for all i and j // between the index L and R static int sum_of_products(int[] arr, int N, int L, int R) { int sum = 0; for (int i = L; i <= R; i++) { for (int j = i; j <= R; j++) { sum += arr[i] * arr[j]; } } return sum; } // Driver code public static void main(String args[]) { int[] arr = { 1, 3, 5, 8 }; int N = arr.length; int L = 0; int R = 2; System.out.println(sum_of_products(arr, N, L, R)); }}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python program for the above approach:## Function to return the sum## of (arr[i]*arr[j]) for all i and j## between the index L and Rdef sum_of_products(arr, N, L, R): sum1 = 0 for i in range(L, R + 1): for j in range(i, R + 1): sum1 += arr[i] * arr[j] return sum1## Driver codeif __name__ == "__main__": arr = [ 1, 3, 5, 8 ] N = len(arr) L = 0 R = 2 print(sum_of_products(arr, N, L, R)) # This code is contributed by entertain2022. |
C#
// C# implementation of the above approachusing System;class GFG { // Function to return the sum // of (arr[i]*arr[j]) for all i and j // between the index L and R static int sum_of_products(int[] arr, int N, int L, int R) { int sum = 0; for (int i = L; i <= R; i++) { for (int j = i; j <= R; j++) { sum += arr[i] * arr[j]; } } return sum; } // Driver code public static void Main() { int[] arr = { 1, 3, 5, 8 }; int N = arr.Length; int L = 0; int R = 2; Console.WriteLine(sum_of_products(arr, N, L, R)); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach // Function to return the sum // of (arr[i]*arr[j]) for all i and j // between the index L and R function sum_of_products(arr, N, L, R) { let sum = 0; for (let i = L; i <= R; i++) { for (let j = i; j <= R; j++) { sum += arr[i] * arr[j]; } } return sum; } // Driver code let arr = [1, 3, 5, 8]; let N = arr.length let L = 0; let R = 2; document.write(sum_of_products(arr, N, L, R)); // This code is contributed by Potta Lokesh </script> |
Output
58
Time complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: This problem can be efficiently solved by using the Prefix sum technique. In this method, store the prefix sum in pre-calculation and then iterate a single loop from L to R and multiply the corresponding prefix sum from that index to the last index.
Basically 1*1+1*3+1*5+3*3+3*5+5*5 can be written as 1*(1+3+5)+3*(3+5)+5*(5) = 1*(prefix_sum from 1 to 5)+3*(prefix_sum from 3 to 5)+5*(prefix sum from 5 to 5)
Below is the implementation of the above approach.
C++
// C++ code to implement above approach#include <bits/stdc++.h>using namespace std;// Function to return the sum of// (arr[i]*arr[j]) for all i and j// between the index L and Rint sum_of_products(int arr[], int n, int L, int R){ int sum = 0; // Pre-calculating Prefix sum int prefix_sum[n]; prefix_sum[0] = arr[0]; for (int i = 1; i < n; i++) { prefix_sum[i] = prefix_sum[i - 1] + arr[i]; } // Using prefix sum to find // summation of products for (int i = L; i <= R; i++) { // if-else for i==0 case // in prefix sum if (i != 0) sum += arr[i] * (prefix_sum[R] - prefix_sum[i - 1]); else sum += arr[i] * (prefix_sum[R]); } return sum;}// Driver codeint main(){ int arr[] = { 1, 3, 5, 8 }; int N = sizeof(arr) / sizeof(arr[0]); int L = 0; int R = 2; cout << sum_of_products(arr, N, L, R); return 0;} |
Java
// Java code to implement above approachimport java.util.*;class GFG{ // Function to return the sum of // (arr[i]*arr[j]) for all i and j // between the index L and R static int sum_of_products(int arr[], int n, int L, int R) { int sum = 0; // Pre-calculating Prefix sum int []prefix_sum = new int[n]; prefix_sum[0] = arr[0]; for (int i = 1; i < n; i++) { prefix_sum[i] = prefix_sum[i - 1] + arr[i]; } // Using prefix sum to find // summation of products for (int i = L; i <= R; i++) { // if-else for i==0 case // in prefix sum if (i != 0) sum += arr[i] * (prefix_sum[R] - prefix_sum[i - 1]); else sum += arr[i] * (prefix_sum[R]); } return sum; } // Driver code public static void main(String[] args) { int arr[] = { 1, 3, 5, 8 }; int N = arr.length; int L = 0; int R = 2; System.out.print(sum_of_products(arr, N, L, R)); }}// This code is contributed by shikhasingrajput |
Python3
# Python program for the above approach:## Function to return the sum of## (arr[i]*arr[j]) for all i and j## between the index L and Rdef sum_of_products(arr, n, L, R): sum = 0 ## Pre-calculating Prefix sum prefix_sum = [0]*n; prefix_sum[0] = arr[0]; for i in range(1, n): prefix_sum[i] = prefix_sum[i - 1] + arr[i] ## Using prefix sum to find ## summation of products for i in range(L, R + 1): ## if-else for i==0 case ## in prefix sum if (i != 0): sum += arr[i] * (prefix_sum[R] - prefix_sum[i - 1]) else: sum += arr[i] * (prefix_sum[R]) return sum## Driver codeif __name__=='__main__': arr = [1, 3, 5, 8] N = len(arr) L = 0 R = 2 print(sum_of_products(arr, N, L, R)) # This code is contributed by subhamgoyal2014. |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{ // Function to return the sum of // (arr[i]*arr[j]) for all i and j // between the index L and R static int sum_of_products(int[] arr, int n, int L, int R) { int sum = 0; // Pre-calculating Prefix sum int []prefix_sum = new int[n]; prefix_sum[0] = arr[0]; for (int i = 1; i < n; i++) { prefix_sum[i] = prefix_sum[i - 1] + arr[i]; } // Using prefix sum to find // summation of products for (int i = L; i <= R; i++) { // if-else for i==0 case // in prefix sum if (i != 0) sum += arr[i] * (prefix_sum[R] - prefix_sum[i - 1]); else sum += arr[i] * (prefix_sum[R]); } return sum; } // Driver Code public static void Main() { int[] arr = { 1, 3, 5, 8 }; int N = arr.Length; int L = 0; int R = 2; Console.Write(sum_of_products(arr, N, L, R)); }}// This code is contributed by ukasp. |
Javascript
<script>// javascript code to implement above approach // Function to return the sum of // (arr[i]*arr[j]) for all i and j // between the index L and R function sum_of_products(arr , n , L , R) { var sum = 0; // Pre-calculating Prefix sum var prefix_sum = Array(n).fill(0); prefix_sum[0] = arr[0]; for (i = 1; i < n; i++) { prefix_sum[i] = prefix_sum[i - 1] + arr[i]; } // Using prefix sum to find // summation of products for (i = L; i <= R; i++) { // if-else for i==0 case // in prefix sum if (i != 0) sum += arr[i] * (prefix_sum[R] - prefix_sum[i - 1]); else sum += arr[i] * (prefix_sum[R]); } return sum; } // Driver code var arr = [ 1, 3, 5, 8 ]; var N = arr.length; var L = 0; var R = 2; document.write(sum_of_products(arr, N, L, R));// This code is contributed by umadevi9616</script> |
Output
58
Time complexity: O(N)
Auxiliary Space: O(N)
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