K个连续删除后的最大总和

给定一个大小为N的数组arr[]和一个整数K ，任务是从数组中删除K个连续元素，使得剩余元素的总和最大。这里我们需要打印数组的剩余元素。

例子：

Input: arr[] = {-1, 1, 2, -3, 2, 2}, K = 3 
Output: -1 2 2 
Delete 1, 2, -3 and the sum of the remaining 
elements will be 3 which is maximum possible.
Input: arr[] = {1, 2, -3, 4, 5}, K = 1 
Output: 1 2 4 5  

方法：计算k个连续元素的总和，去掉总和最小的元素。打印数组的其余元素。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to print the array after removing
// k consecutive elements such that the sum
// of the remaining elements is maximized
void maxSumArr(int arr[], int n, int k)
{
    int cur = 0, index = 0;
 
    // Find the sum of first k elements
    for (int i = 0; i < k; i++)
        cur += arr[i];
 
    // To store the minimum sum of k
    // consecutive elements of the array
    int min = cur;
    for (int i = 0; i < n - k; i++) {
 
        // Calculating sum of next k elements
        cur = cur - arr[i] + arr[i + k];
 
        // Update the minimum sum so far and the
        // index of the first element
        if (cur < min) {
            cur = min;
            index = i + 1;
        }
    }
 
    // Printing result
    for (int i = 0; i < index; i++)
        cout << arr[i] << " ";
    for (int i = index + k; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { -1, 1, 2, -3, 2, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 3;
 
    maxSumArr(arr, n, k);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to print the array after removing
    // k consecutive elements such that the sum
    // of the remaining elements is maximized
    static void maxSumArr(int arr[], int n, int k)
    {
        int cur = 0, index = 0;
 
        // Find the sum of first k elements
        for (int i = 0; i < k; i++)
            cur += arr[i];
 
        // To store the minimum sum of k
        // consecutive elements of the array
        int min = cur;
        for (int i = 0; i < n - k; i++) {
 
            // Calculating sum of next k elements
            cur = cur - arr[i] + arr[i + k];
 
            // Update the minimum sum so far and the
            // index of the first element
            if (cur < min) {
                cur = min;
                index = i + 1;
            }
        }
 
        // Printing result
        for (int i = 0; i < index; i++)
            System.out.print(arr[i] + " ");
        for (int i = index + k; i < n; i++)
            System.out.print(arr[i] + " ");
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { -1, 1, 2, -3, 2, 2 };
        int n = arr.length;
        int k = 3;
 
        maxSumArr(arr, n, k);
    }
}

Python

# Python3 implementation of the approach
 
# Function to print the array after removing
# k consecutive elements such that the sum
# of the remaining elements is maximized
def maxSumArr(arr,  n, k):
    cur = 0
    index = 0
 
    # Find the sum of first k elements
    for i in range(k):
        cur += arr[i]
 
    # To store the minimum sum of k
    # consecutive elements of the array
    min = cur;
    for i in range(n-k):
 
        # Calculating sum of next k elements
        cur = cur-arr[i]+arr[i + k]
         
        # Update the minimum sum so far and the
        # index of the first element
        if(cur

---

C#

// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Function to print the array after removing
    // k consecutive elements such that the sum
    // of the remaining elements is maximized
    static void maxSumArr(int []arr, int n, int k)
    {
        int cur = 0, index = 0;
 
        // Find the sum of first k elements
        for (int i = 0; i < k; i++)
            cur = cur + arr[i];
 
        // To store the minimum sum of k
        // consecutive elements of the array
        int min = cur;
        for (int i = 0; i < n - k; i++)
        {
 
            // Calculating sum of next k elements
            cur = (cur - arr[i]) + (arr[i + k]);
 
            // Update the minimum sum so far and the
            // index of the first element
            if (cur < min)
            {
                cur = min;
                index = i + 1;
            }
        }
 
        // Printing result
        for (int i = 0; i < index; i++)
            Console.Write(arr[i] + " ");
        for (int i = index + k; i < n; i++)
            Console.Write(arr[i] + " ");
    }
 
    // Driver code
    static public void Main ()
    {
        int []arr = { -1, 1, 2, -3, 2, 2 };
        int n = arr.Length;
        int k = 3;
 
        maxSumArr(arr, n, k);
    }
}
 
// This code is contributed by ajit..

Javascript

    // Javascript implementation of the above approach
     
    // Function to print the array after removing
    // k consecutive elements such that the sum
    // of the remaining elements is maximized
    function maxSumArr(arr, n, k)
    {
        let cur = 0, index = 0;
   
        // Find the sum of first k elements
        for (let i = 0; i < k; i++)
            cur = cur + arr[i];
   
        // To store the minimum sum of k
        // consecutive elements of the array
        let min = cur;
        for (let i = 0; i < n - k; i++)
        {
   
            // Calculating sum of next k elements
            cur = (cur - arr[i]) + (arr[i + k]);
   
            // Update the minimum sum so far and the
            // index of the first element
            if (cur < min)
            {
                cur = min;
                index = i + 1;
            }
        }
   
        // Printing result
        for (let i = 0; i < index; i++)
            document.write(arr[i] + " ");
        for (let i = index + k; i < n; i++)
            document.write(arr[i] + " ");
    }
     
    let arr = [ -1, 1, 2, -3, 2, 2 ];
    let n = arr.length;
    let k = 3;
 
    maxSumArr(arr, n, k);
 

输出：

-1 2 2

时间复杂度： O(n)