Suma circular máxima de subarreglo de tamaño K

Dada una array arr de tamaño N y un número entero K , la tarea es encontrar la suma máxima de subarreglo de tamaño k entre todos los subarreglos contiguos (considerando también el subarreglo circular).

Ejemplos: 

Entrada: arr = {18, 4, 3, 4, 5, 6, 7, 8, 2, 10}, k = 3 
Salida: 
suma circular máxima = 32 
índice inicial = 9 
índice final = 1 
Explicación: 
Suma máxima = 10 + 18 + 4 = 32

Entrada: arr = {8, 2, 5, 9}, k = 4 
Salida: 
suma circular máxima = 24 
índice inicial = 0 
índice final = 3  

Acercarse:  

• Iterar el bucle hasta (n + k) veces y
• Tome (i % n) para manejar el caso cuando el índice de la array se vuelve mayor que n.

A continuación se muestra la implementación del enfoque anterior:  

// 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 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 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# 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

<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>

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

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