Largest Array minimum in N-1 operations by reducing each element by minimum

Given an array arr[] containing N integers, the task is to find the maximum of all minimum elements after N-1 delete operations. In one operation, remove the smallest element from the array and subtract it from all remaining elements. 

Examples:

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

Naive Approach: Remove the minimum element from the array and do subtraction from the remaining elements and keep tracking the maximum of a minimum of the array in each operation while the size of the array is not equal to 0.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using the Greedy Approach. This can be derived mathematically since the minimum element needs to be removed each time and so it is independent of the order of elements in the array. So the array needs to be sorted. Follow the observation below, to solve the problem:

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))

Therefore, the end result is going to be the maximum difference of two consecutive elements, as seen from the above proof. Follow the steps below to solve the problem:

1. Create a variable max_value, to store the final answer and initialise it with arr[0].
2. Sort the array arr[] in ascending order.
3. Run a loop from i=0 to i<(N-1), and in each iteration:
   • Keep track of the maximum of minimum value (i.e the difference arr[i + 1] – arr[i]) on each iteration. So, make max_value, maximum out of the max_value and (arr[i+1] – arr[i]).
4. Return max_value as the final answer.

Below is the implementation of the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
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

<script>
// 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.
</script>

Output

4

Time Complexity: O(NlogN)
Auxiliary Space: O(1)