Sum of the first M elements of Array formed by infinitely concatenating given array

Given an array arr[] consisting of N integers and a positive integer M, the task is to find the sum of the first M elements of the array formed by the infinite concatenation of the given array arr[].

Examples:

Input: arr[] = {1, 2, 3}, M = 5
Output: 9
Explanation:
The array formed by the infinite concatenation of the given array arr[] is of the form {1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, ... }.
The sum of the first M(= 5) elements of the array is 1 + 2 + 3 + 1 + 2 = 9.

Input: arr[] = {1}, M = 7
Output: 7

Approach: The given problem can be solved by using the Modulo Operator (%) and consider the given array as the circular array and find the sum of the first M elements accordingly. Follow the steps below to solve this problem:

• Initialize a variable, say sum as 0 to store the resultant sum of the first M elements of the new array.
• Iterate over the range [0, M - 1] using the variable i and increment the value of sum by arr[i%N].
• After completing the above steps, print the value of the sum as the result.

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 first
// M numbers formed by the infinite
// concatenation of the array A[]
int sumOfFirstM(int A[], int N, int M)
{
    // Stores the resultant sum
    int sum = 0;

    // Iterate over the range [0, M - 1]
    for (int i = 0; i < M; i++) {

        // Add the value A[i%N] to sum
        sum = sum + A[i % N];
    }

    // Return the resultant sum
    return sum;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3 };
    int M = 5;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sumOfFirstM(arr, N, M);

    return 0;
}

// Java program for the above approach

import java.io.*;
import java.lang.*;

class GFG {

  // Function to find the sum of first
  // M numbers formed by the infinite
  // concatenation of the array A[]
  public static int sumOfFirstM(int A[], int N, int M)
  {
    
    // Stores the resultant sum
    int sum = 0;

    // Iterate over the range [0, M - 1]
    for (int i = 0; i < M; i++) {

      // Add the value A[i%N] to sum
      sum = sum + A[i % N];
    }

    // Return the resultant sum
    return sum;
  }

  // Driver Code
  public static void main(String[] args) {
    int arr[] = { 1, 2, 3 };
    int M = 5;
    int N = arr.length;
    System.out.println(sumOfFirstM(arr, N, M));

  }
}

// This code is contributed by gfgking.

# Python3 program for the above approach

# Function to find the sum of first
# M numbers formed by the infinite
# concatenation of the array A[]
def sumOfFirstM(A, N, M):
    
    # Stores the resultant sum
    sum = 0

    # Iterate over the range [0, M - 1]
    for i in range(M):
        
        # Add the value A[i%N] to sum
        sum = sum + A[i % N]

    # Return the resultant sum
    return sum

# Driver Code
if __name__ == '__main__':
    
    arr = [ 1, 2, 3 ]
    M = 5
    N = len(arr)
    
    print(sumOfFirstM(arr, N, M))
    
# This code is contributed by ipg2016107

// C# program for the above approach
using System;
class GFG {

    // Function to find the sum of first
    // M numbers formed by the infinite
    // concatenation of the array A[]
    static int sumOfFirstM(int[] A, int N, int M)
    {

        // Stores the resultant sum
        int sum = 0;

        // Iterate over the range [0, M - 1]
        for (int i = 0; i < M; i++) {

            // Add the value A[i%N] to sum
            sum = sum + A[i % N];
        }

        // Return the resultant sum
        return sum;
    }

    // Driver Code
    public static void Main()
    {
        int[] arr = { 1, 2, 3 };
        int M = 5;
        int N = arr.Length;
        Console.WriteLine(sumOfFirstM(arr, N, M));
    }
}

// This code is contributed by subhammahato348.

 <script>
 
        // JavaScript program for the above approach


        // Function to find the sum of first
        // M numbers formed by the infinite
        // concatenation of the array A[]
        function sumOfFirstM(A, N, M) {
            // Stores the resultant sum
            let sum = 0;
            // Iterate over the range [0, M - 1]
            for (let i = 0; i < M; i++) {

                // Add the value A[i%N] to sum
                sum = sum + A[i % N];
            }

            // Return the resultant sum
            return sum;
        }

        // Driver Code

        let arr = [1, 2, 3];
        let M = 5;
        let N = arr.length;
        document.write(sumOfFirstM(arr, N, M));



  // This code is contributed by Potta Lokesh
  
</script>

9

Time Complexity: O(M)
Auxiliary Space: O(1)