最小化使所有数组元素大于或等于零所需的成本

给定一个由N 个整数和一个整数X组成的数组arr[] ，任务是通过执行以下操作任意次数来找到使所有数组元素大于或等于0所需的最小成本：

• 将任何数组元素增加 1。成本 = 1。
• 将所有数组元素增加 1。成本 = X。

例子：

Input: arr[] = {-1, -3, 3, 4, 5}, X = 2
Output: 4
Explanation:
Increment arr[0] by 1. The array arr[] modifies to {0, -3, 3, 4, 5}. Cost = 1.
Increment arr[1] by 1 thrice. The array arr[] modifies to {0, 0, 3, 4, 5}. Therefore, Cost = 4.
Hence, the total cost required is 4.

Input: arr[] = {-3, -2, -1, -5, 7}, X = 2
Output: 8   

方法：想法是使用贪心方法来解决问题。请按照以下步骤解决问题：

• 按升序对数组arr[]进行排序。
• 初始化一个辅助向量，比如列表，来存储负数组元素。
• 初始化一个变量cost = 0，以存储使当前数组元素为0所需的成本，另一个变量min_cost = INT_MAX存储使所有数组元素>= 0的最终最小成本。
• 遍历数组arr[]并尝试通过应用合适的操作来转换列表中的所有数组元素>= 0并相应地更新min_cost 。
• 打印min_cost的值作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the minimum
// cost to make all array elements
// greater than or equal to 0
void minCost(int arr[], int N, int X)
{
    // Sort the array in
    // ascending order
    sort(arr, arr + N);
 
    int sum = 0;
 
    // Stores the cost to make
    // current array element >= 0
    int cost = 0;
 
    // Stores the cost to make
    // all array elements >= 0
    int min_cost = INT_MAX;
 
    // Traverse the array and insert all the
    // elements which are < 0
    for (int i = 0; i < N; i++) {
 
        // If current array element
        // is negative
        if (arr[i] < 0) {
 
            // Cost to make all array
            // elements >= 0
            cost = abs(arr[i]) * X
                   + (sum - abs(arr[i]) * i);
            sum += abs(arr[i]);
 
            // Update curr if ans is minimum
            min_cost = min(min_cost, cost);
        }
    }
 
    // Print the minimum cost
    cout << min_cost;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { -1, -3, -2, 4, -1 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given value of X
    int X = 2;
 
    // Function call to find minimum
    // cost to make all array elements >= 0
    minCost(arr, N, X);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;
public class GFG
{   
 
  // Function to find the minimum
  // cost to make all array elements
  // greater than or equal to 0
  static void minCost(int arr[], int N, int X)
  {
 
    // Sort the array in
    // ascending order
    Arrays.sort(arr) ;
 
    int sum = 0;
 
    // Stores the cost to make
    // current array element >= 0
    int cost = 0;
 
    int INT_MAX = Integer.MAX_VALUE;
 
    // Stores the cost to make
    // all array elements >= 0
    int min_cost = INT_MAX;
 
    // Traverse the array and insert all the
    // elements which are < 0
    for (int i = 0; i < N; i++) {
 
      // If current array element
      // is negative
      if (arr[i] < 0) {
 
        // Cost to make all array
        // elements >= 0
        cost = Math.abs(arr[i]) * X
          + (sum - Math.abs(arr[i]) * i);
        sum += Math.abs(arr[i]);
 
        // Update curr if ans is minimum
        min_cost = Math.min(min_cost, cost);
      }
    }
 
    // Print the minimum cost
    System.out.print(min_cost);
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    // Given array
    int arr[] = { -1, -3, -2, 4, -1 };
 
    // Size of the array
    int N = arr.length;
 
    // Given value of X
    int X = 2;
 
    // Function call to find minimum
    // cost to make all array elements >= 0
    minCost(arr, N, X);
 
  }
 
}
 
// This code is contributed by AnkThon

Python3

# Python3 program for the above approach
import sys
 
# Function to find the minimum
# cost to make all array of elements
# greater than or equal to 0
def mincost(arr, N, X):
   
    # sort the array in
    # ascending order
    arr.sort()
    sum = 0
     
    # stores the count to make
    # current array element >=0
    cost = 0
     
    # stores the cost to make
    # all array elements >=0
    min_cost = sys.maxsize
     
    # Traverse the array and insert all the
    # elements which are <=0
    for i in range(0, N):
       
        # if current array element
        # is negative
        if (arr[i] < 0):
           
            # cost to make all array
            # elements >=0
            cost = abs(arr[i]) * x + (sum - abs(arr[i]) * i)
            sum += abs(arr[i])
             
            # update curr if ans is minimum
            min_cost = min(min_cost,cost)
     
    # return minimum cost
    return min_cost
 
# Driver code
arr = [-1, -3, -2, 4, -1]
 
# size of the array
N = len(arr)
 
# Given value of x
x = 2
 
# Function call to find minimum
# cost to make all array elements >=0
print(mincost(arr, N, x))
 
# This code is contributed by Virusbuddah

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to find the minimum
// cost to make all array elements
// greater than or equal to 0
static void minCost(int[] arr, int N, int X)
{
 
    // Sort the array in
    // ascending order
    Array.Sort(arr) ;
     
    int sum = 0;
     
    // Stores the cost to make
    // current array element >= 0
    int cost = 0;
     
    //int INT_MAX = Int32.MaxValue;
     
    // Stores the cost to make
    // all array elements >= 0
    int min_cost = Int32.MaxValue;
     
    // Traverse the array and insert all the
    // elements which are < 0
    for(int i = 0; i < N; i++)
    {
     
        // If current array element
        // is negative
        if (arr[i] < 0)
        {
         
            // Cost to make all array
            // elements >= 0
            cost = Math.Abs(arr[i]) * X + 
            (sum - Math.Abs(arr[i]) * i);
            sum += Math.Abs(arr[i]);
             
            // Update curr if ans is minimum
            min_cost = Math.Min(min_cost, cost);
        }
    }
     
    // Print the minimum cost
    Console.Write(min_cost);
}
 
// Driver Code
static public void Main ()
{
     
    // Given array
    int[] arr = { -1, -3, -2, 4, -1 };
 
    // Size of the array
    int N = arr.Length;
 
    // Given value of X
    int X = 2;
 
    // Function call to find minimum
    // cost to make all array elements >= 0
    minCost(arr, N, X);
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript

// javascript program for the above approach
 
  // Function to find the minimum
  // cost to make all array elements
  // greater than or equal to 0
  function minCost(arr , N , X)
  {
 
    // Sort the array in
    // ascending order
    arr.sort() ;
 
    var sum = 0;
 
    // Stores the cost to make
    // current array element >= 0
    var cost = 0;
 
    var INT_MAX = Number.MAX_VALUE;
 
    // Stores the cost to make
    // all array elements >= 0
    var min_cost = INT_MAX;
 
    // Traverse the array and insert all the
    // elements which are < 0
    for (i = 0; i < N; i++) {
 
      // If current array element
      // is negative
      if (arr[i] < 0) {
 
        // Cost to make all array
        // elements >= 0
        cost = Math.abs(arr[i]) * X
          + (sum - Math.abs(arr[i]) * i);
        sum += Math.abs(arr[i]);
 
        // Update curr if ans is minimum
        min_cost = Math.min(min_cost, cost);
      }
    }
 
    // Prvar the minimum cost
    document.write(min_cost);
  }
 
  // Driver Code
//Given array
  var arr = [ -1, -3, -2, 4, -1 ];
 
  // Size of the array
  var N = arr.length;
 
  // Given value of X
  var X = 2;
 
  // Function call to find minimum
  // cost to make all array elements >= 0
  minCost(arr, N, X);
 
// This code is contributed by 29AjayKumar

输出：

5

时间复杂度： O(N * logN)
辅助空间： O(1)

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