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Maximize the minimum element of Array by reducing elements one by one

  • Last Updated : 31 Jan, 2022

Given an array arr[] containing N integers. In each operation, a minimum integer is chosen from the array and deleted from the array after subtracting it from the remaining elements. The task is to find the maximum of minimum values of the array after any number of such operations. 

Examples:

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

 

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:

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))

The end result is going to be a difference of consecutive elements as seen from the above proof. Because the minimum in each step is a difference of the adjacent elements.

Follow the steps below to solve the problem:

  • Initialize the initial answer as max_value with 0 operations as arr[0].
  • Sort the array arr[] in ascending order.
  • Iterate in the range of [0, N-1]
  • Keep track of the maximum of minimum value (i.e the difference arr[i + 1] – arr[i]) on each iteration.
  • Return max_value.

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 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




<script>
    // 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
 
</script>

 
 

Output
4

 

Time Complexity : O(N log(N))
Space Complexity : O(1)

 


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