通过一一减少元素来最大化Array的最小元素

给定一个包含N个整数的数组arr[] 。在每个操作中，从数组中选择一个最小整数，并在从剩余元素中减去它后从数组中删除。任务是在任意数量的此类操作之后找到数组的最小值的最大值。

例子：

Input: arr[] = {-1, -2, 4, 3, 5}
Output: 4
Explanation: Following are the operations performed in array  
First operation: remove -2 and subtract it from remaining. Now array arr[] becomes {1, 6, 5, 7} minimum element =1, max minimum element = 1.
Operation 2: remove 1 and subtract it from remaining. Now array arr[] becomes {5, 4, 6} minimum element =4, max minimum element = 4.
Operation 3: remove 4 and subtract it from the remaining. Now arr[] becomes {1, 2} minimum element =1 max minimum element = 4 till now.
Operation 4: remove 1 and subtract it from remaining. Now arr[] becomes {1}. minimum element = 1, max minimum element = 4 till now
Therefore, Maximum minimum element is 4.

Input: arr[] = {-3, -1, -6, -7}
Output: 3

朴素方法：从数组中删除最小元素并从剩余元素中进行减法，并在每次操作中跟踪数组的最小值的最大值，而数组的大小不等于0 。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：上述方法可以通过使用贪心方法进行优化。这可以从数学上推导出来，因为每次都需要删除最小元素，因此它与数组中元素的顺序无关。所以需要对数组进行排序。请遵循以下观察：

Since the Minimum element needs to be removed in each operation.
Consider the array after sorting in increasing order is {a1, a2, a3, a4, a5, …}

• Initially a1 is the minimum after removing it the array is {a2-a1, a3-a1, a4-a1, a5-a1, ..}
• Now a2-a1 is the minimum after removing it the array is{a3-a1-(a2-a1), a4-a1-(a2-a1), ..} which is equal to {a3-a2, a4-a2, a5-a2, …}
• Now a3-a2 is the minimum and it continues so…

res = max(a1, ∑(i=0 to N-1) (ai+1 -ai))

从上述证明中可以看出，最终结果将是连续元素的差异。因为每个步骤中的最小值是相邻元素的差异。

请按照以下步骤解决问题：

• 将初始答案初始化为max_value ，其中0次操作为arr[0] 。
• 按升序对数组arr[]进行排序。
• 在[0, N-1]范围内迭代
• 在每次迭代中跟踪最小值的最大值（即差arr[i + 1] – arr[i]） 。
• 返回最大值。

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to find maximum of minimum value
// of the array in the array
// in each operation
int min_after_del(int arr[], int n)
{
    // If no operations are done
    int max_value = arr[0];
 
    // Sort the array arr in ascending order
    sort(arr, arr + n);
 
    // Traverse the array to check
    // the required condition
    for (int i = 0; i < n - 1; i++) {
        max_value = max(max_value,
                        arr[i + 1] - arr[i]);
    }
 
    return max_value;
}
 
// Driver code
int main()
{
 
    // Initializing array of arr
    int arr[] = { -1, -2, 4, 3, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Call the function and
    // print the answer
    cout << (min_after_del(arr, N));
    return 0;
}

Java

// Java code for the above approach
import java.util.*;
class GFG{
 
// Function to find maximum of minimum value
// of the array in the array
// in each operation
static int min_after_del(int arr[], int n)
{
    // If no operations are done
    int max_value = arr[0];
 
    // Sort the array arr in ascending order
    Arrays.sort(arr);
 
    // Traverse the array to check
    // the required condition
    for (int i = 0; i < n - 1; i++) {
        max_value = Math.max(max_value,
                        arr[i + 1] - arr[i]);
    }
 
    return max_value;
}
 
// Driver code
public static void main(String[] args)
{
 
    // Initializing array of arr
    int arr[] = { -1, -2, 4, 3, 5 };
    int N = arr.length;
 
    // Call the function and
    // print the answer
    System.out.print(min_after_del(arr, N));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python code for the above approach
 
# Function to find maximum of minimum value
# of the array in the array
# in each operation
def min_after_del (arr, n):
 
    # If no operations are done
    max_value = arr[0];
 
    # Sort the array arr in ascending order
    arr.sort();
     
    # Traverse the array to check
    # the required condition
    for i in range(n - 1):
        max_value = max(max_value, arr[i + 1] - arr[i]);
 
    return max_value;
 
# Driver code
# Initializing array of arr
arr = [-1, -2, 4, 3, 5];
N = len(arr)
 
# Call the function and
# print the answer
print(min_after_del(arr, N));
 
# This code is contributed by Saurabh Jaiswal

C#

using System;
 
public class GFG{
 
  // Function to find maximum of minimum value
  // of the array in the array
  // in each operation
  static int min_after_del(int[] arr, int n)
  {
    // If no operations are done
    int max_value = arr[0];
 
    // Sort the array arr in ascending order
    Array.Sort(arr);
 
    // Traverse the array to check
    // the required condition
    for (int i = 0; i < n - 1; i++) {
      max_value = Math.Max(max_value,
                           arr[i + 1] - arr[i]);
    }
 
    return max_value;
  }
 
  // Driver code
  static public void Main (){
 
    int[] arr = { -1, -2, 4, 3, 5 };
    int N = arr.Length;
 
    // Call the function and
    // print the answer
    Console.Write(min_after_del(arr, N));
  }
}
 
// This code is contributed by hrithikgarg03188.

Javascript

    // JavaScript code for the above approach
 
    // Function to find maximum of minimum value
    // of the array in the array
    // in each operation
    const min_after_del = (arr, n) => {
     
        // If no operations are done
        let max_value = arr[0];
 
        // Sort the array arr in ascending order
        arr.sort((a, b) => a - b);
         
        // Traverse the array to check
        // the required condition
        for (let i = 0; i < n - 1; i++) {
            max_value = Math.max(max_value,
                arr[i + 1] - arr[i]);
        }
 
        return max_value;
    }
 
    // Driver code
 
 
    // Initializing array of arr
    let arr = [-1, -2, 4, 3, 5];
    let N = arr.length;
 
    // Call the function and
    // print the answer
    document.write(min_after_del(arr, N));
 
// This code is contributed by rakeshsahni
 

输出

4

时间复杂度： O(N log(N))
空间复杂度： O(1)