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;
} |
// 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> |
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;
} |
// 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> |
22
Time Complexity: O(N)
Auxiliary Space: O(N)