Sum of elements in given range from Array formed by infinitely concatenating given array
Last Updated :
03 Aug, 2021
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Given an array arr[](1-based indexing) consisting of N positive integers and two positive integers L and R, the task is to find the sum of array elements over the range [L, R] if the given array arr[] is concatenating to itself infinite times.
Examples:
Input: arr[] = {1, 2, 3}, L = 2, R = 8
Output: 14
Explanation:
The array, arr[] after concatenation is {1, 2, 3, 1, 2, 3, 1, 2, ...} and the sum of elements from index 2 to 8 is 2 + 3 + 1 + 2 + 3 + 1 + 2 = 14.Input: arr[] = {5, 2, 6, 9}, L = 10, R = 13
Output: 22
Naive Approach: The simplest approach to solve the given problem is to iterate over the range [L, R] using the variable i and add the value of arr[i % N] to the sum for each index. After completing the iteration, print the value of the sum as the resultant sum.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of elements
// in a given range of an infinite array
void rangeSum(int arr[], int N, int L, int R)
{
// Stores the sum of array elements
// from L to R
int sum = 0;
// Traverse from L to R
for (int i = L - 1; i < R; i++) {
sum += arr[i % N];
}
// Print the resultant sum
cout << sum;
}
// Driver Code
int main()
{
int arr[] = { 5, 2, 6, 9 };
int L = 10, R = 13;
int N = sizeof(arr) / sizeof(arr[0]);
rangeSum(arr, N, L, R);
return 0;
}
// C++ program for the above approachusing namespace std;// Function to find the sum of elements// in a given range of an infinite arrayvoid rangeSum(int arr[], int N, int L, int R){ // Stores the sum of array elements // from L to R int sum = 0; // Traverse from L to R for (int i = L - 1; i < R; i++) { sum += arr[i % N]; } // Print the resultant sum cout << sum;}// Driver Codeint main(){ int arr[] = { 5, 2, 6, 9 }; int L = 10, R = 13; int N = sizeof(arr) / sizeof(arr[0]); rangeSum(arr, N, L, R); return 0;}// Java program for the above approach
import java.io.*;
class GFG
{
// Function to find the sum of elements
// in a given range of an infinite array
static void rangeSum(int arr[], int N, int L, int R)
{
// Stores the sum of array elements
// from L to R
int sum = 0;
// Traverse from L to R
for (int i = L - 1; i < R; i++) {
sum += arr[i % N];
}
// Print the resultant sum
System.out.println(sum);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 2, 6, 9 };
int L = 10, R = 13;
int N = arr.length;
rangeSum(arr, N, L, R);
}
}
// This code is contributed by Potta Lokesh
# Python 3 program for the above approach
# Function to find the sum of elements
# in a given range of an infinite array
def rangeSum(arr, N, L, R):
# Stores the sum of array elements
# from L to R
sum = 0
# Traverse from L to R
for i in range(L - 1,R,1):
sum += arr[i % N]
# Print the resultant sum
print(sum)
# Driver Code
if __name__ == '__main__':
arr = [5, 2, 6, 9 ]
L = 10
R = 13
N = len(arr)
rangeSum(arr, N, L, R)
# This code is contributed by divyeshrabadiya07
// C# program for the above approach
using System;
class GFG {
// Function to find the sum of elements
// in a given range of an infinite array
static void rangeSum(int[] arr, int N, int L, int R)
{
// Stores the sum of array elements
// from L to R
int sum = 0;
// Traverse from L to R
for (int i = L - 1; i < R; i++) {
sum += arr[i % N];
}
// Print the resultant sum
Console.Write(sum);
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 5, 2, 6, 9 };
int L = 10, R = 13;
int N = arr.Length;
rangeSum(arr, N, L, R);
}
}
// This code is contributed by ukasp.
<script>
// Javascript program for the above approach
// Function to find the sum of elements
// in a given range of an infinite array
function rangeSum(arr, N, L, R)
{
// Stores the sum of array elements
// from L to R
let sum = 0;
// Traverse from L to R
for(let i = L - 1; i < R; i++)
{
sum += arr[i % N];
}
// Print the resultant sum
document.write(sum);
}
// Driver Code
let arr = [ 5, 2, 6, 9 ];
let L = 10, R = 13;
let N = arr.length
rangeSum(arr, N, L, R);
// This code is contributed by _saurabh_jaiswal
</script>
Output:
22
Time Complexity: O(R - L)
Auxiliary Space: O(1)
Efficient Approach: The above approach can also be optimized by using the Prefix Sum. Follow the steps below to solve the problem:
- Initialize an array, say prefix[] of size (N + 1) with all elements as 0s.
- Traverse the array, arr[] using the variable i and update prefix[i] to sum of prefix[i - 1] and arr[i - 1].
- Now, the sum of elements over the range [L, R] is given by:
the sum of elements in the range [1, R] - sum of elements in the range [1, L - 1].
- Initialize a variable, say leftSum as ((L - 1)/N)*prefix[N] + prefix[(L - 1)%N] to store the sum of elements in the range [1, L-1].
- Similarly, initialize another variable rightSum as (R/N)*prefix[N] + prefix[R%N] to store the sum of elements in the range [1, R].
- After completing the above steps, print the value of (rightSum - leftSum) as the resultant sum of elements over the given range [L, R].
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the sum of elements
// in a given range of an infinite array
void rangeSum(int arr[], int N, int L,
int R)
{
// Stores the prefix sum
int prefix[N + 1];
prefix[0] = 0;
// Calculate the prefix sum
for (int i = 1; i <= N; i++) {
prefix[i] = prefix[i - 1]
+ arr[i - 1];
}
// Stores the sum of elements
// from 1 to L-1
int leftsum
= ((L - 1) / N) * prefix[N]
+ prefix[(L - 1) % N];
// Stores the sum of elements
// from 1 to R
int rightsum = (R / N) * prefix[N]
+ prefix[R % N];
// Print the resultant sum
cout << rightsum - leftsum;
}
// Driver Code
int main()
{
int arr[] = { 5, 2, 6, 9 };
int L = 10, R = 13;
int N = sizeof(arr) / sizeof(arr[0]);
rangeSum(arr, N, L, R);
return 0;
}
// C++ program for the above approachusing namespace std;// Function to find the sum of elements// in a given range of an infinite arrayvoid rangeSum(int arr[], int N, int L, int R){ // Stores the prefix sum int prefix[N + 1]; prefix[0] = 0; // Calculate the prefix sum for (int i = 1; i <= N; i++) { prefix[i] = prefix[i - 1] + arr[i - 1]; } // Stores the sum of elements // from 1 to L-1 int leftsum = ((L - 1) / N) * prefix[N] + prefix[(L - 1) % N]; // Stores the sum of elements // from 1 to R int rightsum = (R / N) * prefix[N] + prefix[R % N]; // Print the resultant sum cout << rightsum - leftsum;}// Driver Codeint main(){ int arr[] = { 5, 2, 6, 9 }; int L = 10, R = 13; int N = sizeof(arr) / sizeof(arr[0]); rangeSum(arr, N, L, R); return 0;}// Java program for the above approach
import java.io.*;
class GFG{
// Function to find the sum of elements
// in a given range of an infinite array
static void rangeSum(int arr[], int N, int L, int R)
{
// Stores the prefix sum
int prefix[] = new int[N+1];
prefix[0] = 0;
// Calculate the prefix sum
for (int i = 1; i <= N; i++) {
prefix[i] = prefix[i - 1]
+ arr[i - 1];
}
// Stores the sum of elements
// from 1 to L-1
int leftsum
= ((L - 1) / N) * prefix[N]
+ prefix[(L - 1) % N];
// Stores the sum of elements
// from 1 to R
int rightsum = (R / N) * prefix[N]
+ prefix[R % N];
// Print the resultant sum
System.out.print( rightsum - leftsum);
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 5, 2, 6, 9 };
int L = 10, R = 13;
int N = arr.length;
rangeSum(arr, N, L, R);
}
}
// This code is contributed by shivanisinghss2110
# Python 3 program for the above approach
# Function to find the sum of elements
# in a given range of an infinite array
def rangeSum(arr, N, L, R):
# Stores the prefix sum
prefix = [0 for i in range(N + 1)]
prefix[0] = 0
# Calculate the prefix sum
for i in range(1,N+1,1):
prefix[i] = prefix[i - 1] + arr[i - 1]
# Stores the sum of elements
# from 1 to L-1
leftsum = ((L - 1) // N) * prefix[N] + prefix[(L - 1) % N]
# Stores the sum of elements
# from 1 to R
rightsum = (R // N) * prefix[N] + prefix[R % N]
# Print the resultant sum
print(rightsum - leftsum)
# Driver Code
if __name__ == '__main__':
arr = [5, 2, 6, 9]
L = 10
R = 13
N = len(arr)
rangeSum(arr, N, L, R)
# This code is contributed by SURENDRA_GANGWAR.
// C# program for the above approach
using System;
class GFG{
// Function to find the sum of elements
// in a given range of an infinite array
static void rangeSum(int []arr, int N, int L, int R)
{
// Stores the prefix sum
int []prefix = new int[N+1];
prefix[0] = 0;
// Calculate the prefix sum
for (int i = 1; i <= N; i++) {
prefix[i] = prefix[i - 1]
+ arr[i - 1];
}
// Stores the sum of elements
// from 1 to L-1
int leftsum
= ((L - 1) / N) * prefix[N]
+ prefix[(L - 1) % N];
// Stores the sum of elements
// from 1 to R
int rightsum = (R / N) * prefix[N]
+ prefix[R % N];
// Print the resultant sum
Console.Write( rightsum - leftsum);
}
// Driver Code
public static void Main (String[] args)
{
int []arr = { 5, 2, 6, 9 };
int L = 10, R = 13;
int N = arr.Length;
rangeSum(arr, N, L, R);
}
}
// This code is contributed by shivanisinghss2110
<script>
// JavaScript program for the above approach
// Function to find the sum of elements
// in a given range of an infinite array
function rangeSum(arr, N, L, R) {
// Stores the prefix sum
let prefix = new Array(N + 1);
prefix[0] = 0;
// Calculate the prefix sum
for (let i = 1; i <= N; i++) {
prefix[i] = prefix[i - 1] + arr[i - 1];
}
// Stores the sum of elements
// from 1 to L-1
let leftsum = ((L - 1) / N) * prefix[N] + prefix[(L - 1) % N];
// Stores the sum of elements
// from 1 to R
let rightsum = (R / N) * prefix[N] + prefix[R % N];
// Print the resultant sum
document.write(rightsum - leftsum);
}
// Driver Code
let arr = [5, 2, 6, 9];
let L = 10,
R = 13;
let N = arr.length;
rangeSum(arr, N, L, R);
</script>
Output:
22
Time Complexity: O(N)
Auxiliary Space: O(N)
