删除最多 K 个数组元素后可能的最大子数组和

给定一个大小为N的数组arr[]和一个整数K ，任务是通过从数组中删除最多K 个元素来找到最大子数组和。

例子：

Input: arr[] = { -2, 1, 3, -2, 4, -7, 20 }, K = 1 
Output: 26 
Explanation: 
Removing arr[5] from the array modifies arr[] to { -2, 1, 3, -2, 4, 20 } 
Subarray with maximum sum is { 1, 3, -2, 4, 20 }. 
Therefore, the required output is 26.

Input:arr[] = { -1, 1, -1, -1, 1, 1 }, K=2 
Output: 3 
Explanation: 
Removing arr[2] and arr[3] from the array modifies arr[] to { – 1, 1, 1, 1} 
Subarray with maximum sum is { 1, 1, 1 }. 
Therefore, the required output is 3.

方法：该问题可以使用动态规划解决。这个想法是使用 Kadane 算法的概念。请按照以下步骤解决问题：

• 遍历数组arr[]并且对于每个数组元素都需要执行以下两个操作：
  • 从子数组中删除当前数组元素。
  • 在子数组中包含当前数组元素。
• 因此，求解该问题的递推关系如下：

mxSubSum(i, j) = max(max(0, arr[i] + mxSubSum(i – 1, j)), mxSubSum(i – 1, j – 1))

i: Stores index of array element 
j: Maximum count of elements that can be removed from the subarray 
mxSubSum(i, j): Return maximum subarray sum from the subarray { arr[i], arr[N – 1] } by removing K – j array elements.

• 初始化一个二维数组，比如dp[][] ，以存储上述递推关系的重叠子问题。
• 使用记忆填充dp[][]数组。
• 从 dp[][] 数组中找到最大元素，比如res 。
• 初始化一个变量，比如Max ，以存储数组arr[] 中存在的最大元素。
• 如果所有数组元素都是负数，则更新res = max(res, Max) 。
• 最后，打印res作为所需答案。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
#define M 100
 
// Function to find the maximum subarray
// sum greater than or equal to 0 by
// removing K array elements
int mxSubSum(int i, int* arr,
             int j, int dp[][M])
{
    // Base case
    if (i == 0) {
        return dp[i][j] = max(0, arr[i]);
    }
 
    // If overlapping subproblems
    // already occurred
    if (dp[i][j] != -1) {
        return dp[i][j];
    }
 
    // Include current element in the subarray
    int X = max(0, arr[i]
                       + mxSubSum(i - 1, arr, j, dp));
 
    // If K elements already removed
    // from the subarray
    if (j == 0) {
 
        return dp[i][j] = X;
    }
 
    // Remove current element from the subarray
    int Y = mxSubSum(i - 1, arr, j - 1, dp);
 
    return dp[i][j] = max(X, Y);
}
 
// Utility function to find the maximum subarray
// sum by removing at most K array elements
int MaximumSubarraySum(int n, int* arr, int k)
{
 
    // Stores overlapping subproblems
    // of the recurrence relation
    int dp[M][M];
 
    // Initialize dp[][] to -1
    memset(dp, -1, sizeof(dp));
 
    mxSubSum(n - 1, arr, k, dp);
 
    // Stores maximum subarray sum by
    // removing at most K elements
    int res = 0;
 
    // Calculate maximum element
    // in dp[][]
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= k; j++) {
 
            // Update res
            res = max(res, dp[i][j]);
        }
    }
 
    // If all array elements are negative
    if (*max_element(arr, arr + n) < 0) {
 
        // Update res
        res = *max_element(arr, arr + n);
    }
    return res;
}
 
// Driver Code
int main()
{
    int arr[] = { -2, 1, 3, -2, 4, -7, 20 };
    int K = 1;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << MaximumSubarraySum(N, arr, K) << endl;
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
static final int M = 100;
 
// Function to find the maximum subarray
// sum greater than or equal to 0 by
// removing K array elements
static int mxSubSum(int i, int []arr,
             int j, int dp[][])
{
    // Base case
    if (i == 0) {
        return dp[i][j] = Math.max(0, arr[i]);
    }
 
    // If overlapping subproblems
    // already occurred
    if (dp[i][j] != -1) {
        return dp[i][j];
    }
 
    // Include current element in the subarray
    int X = Math.max(0, arr[i]
                       + mxSubSum(i - 1, arr, j, dp));
 
    // If K elements already removed
    // from the subarray
    if (j == 0)
    {
        return dp[i][j] = X;
    }
 
    // Remove current element from the subarray
    int Y = mxSubSum(i - 1, arr, j - 1, dp);
 
    return dp[i][j] = Math.max(X, Y);
}
 
// Utility function to find the maximum subarray
// sum by removing at most K array elements
static int MaximumSubarraySum(int n, int []arr, int k)
{
 
    // Stores overlapping subproblems
    // of the recurrence relation
    int [][]dp = new int[M][M];
 
    // Initialize dp[][] to -1
    for (int i = 0; i < M; i++)
        for (int j = 0; j < M; j++)
            dp[i][j] = -1;
 
    mxSubSum(n - 1, arr, k, dp);
 
    // Stores maximum subarray sum by
    // removing at most K elements
    int res = 0;
 
    // Calculate maximum element
    // in dp[][]
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= k; j++) {
 
            // Update res
            res = Math.max(res, dp[i][j]);
        }
    }
 
    // If all array elements are negative
    if (Arrays.stream(arr).max().getAsInt() < 0) {
 
        // Update res
        res = Arrays.stream(arr).max().getAsInt();
    }
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { -2, 1, 3, -2, 4, -7, 20 };
    int K = 1;
    int N = arr.length;
    System.out.print(MaximumSubarraySum(N, arr, K) +"\n");
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
M = 100
 
# Function to find the maximum subarray
# sum greater than or equal to 0 by
# removing K array elements
def mxSubSum(i, arr, j):
    global dp
     
    # Base case
    if (i == 0):
        dp[i][j] = max(0, arr[i])
        return dp[i][j]
 
    # If overlapping subproblems
    # already occurred
    if (dp[i][j] != -1):
        return dp[i][j]
 
    # Include current element in the subarray
    X = max(0, arr[i] + mxSubSum(i - 1, arr, j))
 
    # If K elements already removed
    # from the subarray
    if (j == 0):
 
        dp[i][j] = X
        return X
 
    # Remove current element from the subarray
    Y = mxSubSum(i - 1, arr, j - 1)
 
    dp[i][j] = max(X, Y)
 
    return dp[i][j]
 
# Utility function to find the maximum subarray
# sum by removing at most K array elements
# Utility function to find the maximum subarray
# sum by removing at most K array elements
def MaximumSubarraySum(n, arr, k):
 
    mxSubSum(n - 1, arr, k)
 
    # Stores maximum subarray sum by
    # removing at most K elements
    res = 0
 
    # Calculate maximum element
    # in dp[][]
    for i in range(n):
        for j in range(k + 1):
 
            # Update res
            res = max(res, dp[i][j])
 
    # If all array elements are negative
    if (max(arr) < 0):
 
        # Update res
        res = max(arr)
    return res
 
# Driver Code
if __name__ == '__main__':
    dp = [[-1 for i in range(100)] for i in range(100)]
    arr = [-2, 1, 3, -2, 4, -7, 20]
    K = 1
    N = len(arr)
    print(MaximumSubarraySum(N, arr, K))
 
# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
using System.Collections;
 
class GFG{
     
static int M = 100;
 
// Function to find the maximum subarray
// sum greater than or equal to 0 by
// removing K array elements
static int mxSubSum(int i, int []arr,
                    int j, int [,]dp)
{
     
    // Base case
    if (i == 0)
    {
        return dp[i, j] = Math.Max(0, arr[i]);
    }
 
    // If overlapping subproblems
    // already occurred
    if (dp[i, j] != -1)
    {
        return dp[i, j];
    }
 
    // Include current element in the subarray
    int X = Math.Max(0, arr[i] +
                    mxSubSum(i - 1, arr, j, dp));
 
    // If K elements already removed
    // from the subarray
    if (j == 0)
    {
        return dp[i, j] = X;
    }
 
    // Remove current element from the subarray
    int Y = mxSubSum(i - 1, arr, j - 1, dp);
 
    return dp[i, j] = Math.Max(X, Y);
}
 
// Utility function to find the maximum subarray
// sum by removing at most K array elements
static int MaximumSubarraySum(int n, int []arr, int k)
{
     
    // Stores overlapping subproblems
    // of the recurrence relation
    int [,]dp = new int[M, M];
 
    // Initialize dp[,] to -1
    for(int i = 0; i < M; i++)
        for(int j = 0; j < M; j++)
            dp[i, j] = -1;
 
    mxSubSum(n - 1, arr, k, dp);
 
    // Stores maximum subarray sum by
    // removing at most K elements
    int res = 0;
 
    // Calculate maximum element
    // in dp[,]
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j <= k; j++)
        {
             
            // Update res
            res = Math.Max(res, dp[i, j]);
        }
    }
 
    Array.Sort(arr);
     
    // If all array elements are negative
     
    if (arr[n - 1] < 0)
    {
         
        // Update res
        res = arr[n - 1];
    }
    return res;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { -2, 1, 3, -2, 4, -7, 20 };
    int K = 1;
    int N = arr.Length;
     
    Console.WriteLine(MaximumSubarraySum(N, arr, K));
}
}
 
// This code is contributed by AnkThon

Javascript

 
// Javascript program to implement
// the above approach
var M = 100;
 
// Function to find the maximum subarray
// sum greater than or equal to 0 by
// removing K array elements
function mxSubSum(i, arr, j, dp)
{
     
    // Base case
    if (i == 0)
    {
        dp[i][j] = Math.max(0, arr[i]);
        return dp[i][j];
    }
 
    // If overlapping subproblems
    // already occurred
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }
 
    // Include current element in the subarray
    var X = Math.max(
        0, arr[i] + mxSubSum(i - 1, arr, j, dp));
 
    // If K elements already removed
    // from the subarray
    if (j == 0)
    {
        dp[i][j] = X;
        return dp[i][j]
    }
 
    // Remove current element from the subarray
    var Y = mxSubSum(i - 1, arr, j - 1, dp);
 
    dp[i][j] = Math.max(X, Y);
    return dp[i][j]
}
 
// Utility function to find the maximum subarray
// sum by removing at most K array elements
function MaximumSubarraySum(n, arr, k)
{
 
    // Stores overlapping subproblems
    // of the recurrence relation
    var dp = Array.from(Array(M), () => Array(M).fill(-1));
 
    mxSubSum(n - 1, arr, k, dp);
 
    // Stores maximum subarray sum by
    // removing at most K elements
    var res = 0;
 
    // Calculate maximum element
    // in dp[][]
    for(var i = 0; i < n; i++)
    {
        for(var j = 0; j <= k; j++)
        {
             
            // Update res
            res = Math.max(res, dp[i][j]);
        }
    }
 
    // If all array elements are negative
    if (arr.reduce((a, b) => Math.max(a, b)) < 0)
    {
         
        // Update res
        res = arr.reduce((a, b) => Math.max(a, b));
    }
    return res;
}
 
// Driver Code
var arr = [ -2, 1, 3, -2, 4, -7, 20 ];
var K = 1;
var N = arr.length;
 
document.write( MaximumSubarraySum(N, arr, K));
 
// This code is contributed by rrrtnx
 

输出：

26

时间复杂度： O(N * K)
辅助空间： O(N * K)

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