通过最小化每个元素，N-1 次操作中的最大数组最小值

给定一个包含N个整数的数组arr[] ，任务是在N-1 次删除操作后找到所有最小元素的最大值。在一个操作中，从数组中删除最小的元素，然后从所有剩余元素中减去它。

例子：

Input: arr[] = {-1, -2, 4, 3, 5}
Output: 4
Explanation: Following are the operations performed:
Operation 1: 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。按照下面的观察，解决问题：

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 and after removing it, the array becomes {a2-a1, a3-a1, a4-a1, a5-a1, …}
Now a2-a1 is the minimum and after removing it, the array becomes {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…
So, res = max(a1, ∑(i=0 to (N-1)) (ai+1 -ai))

因此，最终结果将是两个连续元素的最大差异，如上述证明所示。请按照以下步骤解决问题：

1. 创建一个变量max_value ，用于存储最终答案并使用arr[0]对其进行初始化。
2. 按升序对数组arr[]进行排序。
3. 运行从i=0 到 i<(N-1)的循环，并且在每次迭代中：
   • 在每次迭代中跟踪最小值的最大值（即差arr[i + 1] – arr[i] ）。因此，将max_value和max_value和(arr[i+1] – arr[i])设为最大值。
4. 返回max_value作为最终答案。

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 maxOfAllMins(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);
 
    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()
{
    int arr[] = { -1, -2, 4, 3, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxOfAllMins(arr, N);
    return 0;
}

Java

// Java program for above approach
import java.util.*;
public class GFG {
 
  // Function to find maximum of minimum value
  // of the array in the array
  // in each operation
  static int maxOfAllMins(int[] arr, int n)
  {
 
    // If no operations are done
    int max_value = arr[0];
 
    // Sort the array arr in ascending order
    Arrays.sort(arr);
 
    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[])
  {
    int[] arr = { -1, -2, 4, 3, 5 };
    int N = arr.length;
    System.out.println(maxOfAllMins(arr, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# python3 code for the above approach
 
# Function to find maximum of minimum value
# of the array in the array
# in each operation
def maxOfAllMins(arr, n):
 
    # If no operations are done
    max_value = arr[0]
 
    # Sort the array arr in ascending order
    arr.sort()
 
    for i in range(0, n-1):
        max_value = max(max_value,
                        arr[i + 1] - arr[i])
 
    return max_value
 
# Driver code
if __name__ == "__main__":
 
    arr = [-1, -2, 4, 3, 5]
    N = len(arr)
 
    print(maxOfAllMins(arr, N))
 
# This code is contributed by rakeshsahni

C#

// C# code for the above approach
using System;
class GFG {
 
    // Function to find maximum of minimum value
    // of the array in the array
    // in each operation
    static int maxOfAllMins(int[] arr, int n)
    {
        // If no operations are done
        int max_value = arr[0];
 
        // Sort the array arr in ascending order
        Array.Sort(arr);
 
        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()
    {
        int[] arr = { -1, -2, 4, 3, 5 };
        int N = arr.Length;
        Console.WriteLine(maxOfAllMins(arr, N));
    }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript code for the above approach
 
 
// Function to find maximum of minimum value
// of the array in the array
// in each operation
function maxOfAllMins(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);
 
    for (let i = 0; i < n - 1; i++) {
        max_value = Math.max(max_value,
                        arr[i + 1] - arr[i]);
    }
 
    return max_value;
}
 
// Driver code
 
let arr = [ -1, -2, 4, 3, 5 ];
let N = arr.length
 
document.write(maxOfAllMins(arr, N));
 
// This code is contributed by gfgking.

输出

4

时间复杂度： O(NlogN)
辅助空间： O(1)