Maximum circular subarray sum of size K

Given an array arr of size N and an integer K, the task is to find the maximum sum subarray of size k among all contiguous sub-array (considering circular subarray also).

Examples:

Input: arr = {18, 4, 3, 4, 5, 6, 7, 8, 2, 10}, k = 3 
Output:
max circular sum = 32 
start index = 9 
end index = 1 
Explanation:
Maximum Sum = 10 + 18 + 4 = 32

Input: arr = {8, 2, 5, 9}, k = 4 
Output:
max circular sum = 24 
start index = 0 
end index = 3  

Approach:

• Iterate the loop till (n + k) times and
• Take (i % n) to handle the case when the array index becomes greater than n.

Below is the implementation of above approach:  

C++

// C++ program to find maximum circular
// subarray sum of size k

#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// maximum sum
void maxCircularSum(int arr[], int n, int k)
{
    // k must be greater
    if (n < k) {
        cout << "Invalid";
        return;
    }

    int sum = 0, start = 0, end = k - 1;

    // calculate the sum of first k elements.
    for (int i = 0; i < k; i++) {
        sum += arr[i];
    }

    int ans = sum;

    for (int i = k; i < n + k; i++) {

        // add current element to sum
        // and subtract the first element
        // of the previous window.
        sum += arr[i % n] - arr[(i - k) % n];

        if (sum > ans) {
            ans = sum;
            start = (i - k + 1) % n;
            end = i % n;
        }
    }

    cout << "max circular sum = "
         << ans << endl;
    cout << "start index = " << start
         << "\nend index = " << end << endl;
}

// Driver Code
int main()
{
    int arr[] = { 18, 4, 3, 4, 5, 6, 7, 8, 2, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;

    maxCircularSum(arr, n, k);
    return 0;
}

Java

// Java program to find maximum circular
// subarray sum of size k

import java.util.*;

class GFG
{

    // Function to calculate
    // maximum sum
    static void maxCircularSum(int[] arr, int n, int k)
    {

        // k must be greater
        if (n < k)
        {
            System.out.println("Invalid");
            return;
        }

        int sum = 0, start = 0, end = k - 1;

        // calculate the sum of first k elements.
        for (int i = 0; i < k; i++)
            sum += arr[i];

        int ans = sum;

        for (int i = k; i < n + k; i++)
        {

            // add current element to sum
            // and subtract the first element
            // of the previous window.
            sum += arr[i % n] - arr[(i - k) % n];

            if (sum > ans)
            {
                ans = sum;
                start = (i - k + 1) % n;
                end = i % n;
            }
        }

        System.out.println("max circular sum = " + ans);
        System.out.println("start index = " + start + "\nend index = " + end);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 18, 4, 3, 4, 5, 6, 7, 8, 2, 10 };
        int n = arr.length;
        int k = 3;

        maxCircularSum(arr, n, k);
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 program to find maximum circular
# subarray sum of size k

# Function to calculate
# maximum sum
def maxCircularSum(arr, n, k) :

    # k must be greater
    if (n < k) :
        print("Invalid");
        return;

    sum = 0; start = 0; end = k - 1;

    # calculate the sum of first k elements.
    for i in range(k) :
        sum += arr[i];

    ans = sum;

    for i in range(k, n + k) :

        # add current element to sum
        # and subtract the first element
        # of the previous window.
        sum += arr[i % n] - arr[(i - k) % n];

        if (sum > ans) :
            ans = sum;
            start = (i - k + 1) % n;
            end = i % n;

    print("max circular sum = ",ans);
    print("start index = ", start,
          "\nend index = ", end);

# Driver Code
if __name__ == "__main__" :

    arr = [ 18, 4, 3, 4, 5, 6, 7, 8, 2, 10 ];
    n = len(arr);
    k = 3;

    maxCircularSum(arr, n, k);

# This code is contributed by AnkitRai01

C#

// C# program to find maximum circular
// subarray sum of size k
using System;

class GFG
{

    // Function to calculate
    // maximum sum
    static void maxCircularSum(int[] arr,
                               int n, int k)
    {

        // k must be greater
        if (n < k)
        {
            Console.WriteLine("Invalid");
            return;
        }

        int sum = 0, start = 0, end = k - 1;

        // calculate the sum of first k elements.
        for (int i = 0; i < k; i++)
            sum += arr[i];

        int ans = sum;

        for (int i = k; i < n + k; i++)
        {

            // add current element to sum
            // and subtract the first element
            // of the previous window.
            sum += arr[i % n] - arr[(i - k) % n];

            if (sum > ans)
            {
                ans = sum;
                start = (i - k + 1) % n;
                end = i % n;
            }
        }

        Console.WriteLine("max circular sum = " + ans);
        Console.WriteLine("start index = " + start +
                          "\nend index = " + end);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = { 18, 4, 3, 4, 5,
                      6, 7, 8, 2, 10 };
        int n = arr.Length;
        int k = 3;

        maxCircularSum(arr, n, k);
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to find maximum circular
// subarray sum of size k

// Function to calculate
// maximum sum
function maxCircularSum(arr, n, k)
{

    // k must be greater
    if (n < k)
    {
        document.write("Invalid");
        return;
    }

    let sum = 0, start = 0, end = k - 1;

    // Calculate the sum of first k elements.
    for(let i = 0; i < k; i++)
    {
        sum += arr[i];
    }

    let ans = sum;

    for(let i = k; i < n + k; i++)
    {

        // Add current element to sum
        // and subtract the first element
        // of the previous window.
        sum += arr[i % n] - arr[(i - k) % n];

        if (sum > ans)
        {
            ans = sum;
            start = (i - k + 1) % n;
            end = i % n;
        }
    }

    document.write("max circular sum = " +
                   ans + "<br>");
    document.write("start index = " + start +
                   "<br>end index = " + end +
                   "<br>");
}

// Driver Code
let arr = [ 18, 4, 3, 4, 5,
            6, 7, 8, 2, 10 ];
let n = arr.length
let k = 3;

maxCircularSum(arr, n, k);

// This code is contributed by _saurabh_jaiswal

</script>

Output:

max circular sum = 32
start index = 9
end index = 1

Time Complexity:
Auxiliary Space: O(1), no extra space is required, so it is a constant.