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++
#include <bits/stdc++.h>
using namespace std;
int min_after_del(int arr[], int n)
{
int max_value = arr[0];
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;
}
int main()
{
int arr[] = { -1, -2, 4, 3, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << (min_after_del(arr, N));
return 0;
}
|
Java
import java.util.*;
class GFG{
static int min_after_del(int arr[], int n)
{
int max_value = arr[0];
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;
}
public static void main(String[] args)
{
int arr[] = { -1, -2, 4, 3, 5 };
int N = arr.length;
System.out.print(min_after_del(arr, N));
}
}
|
Python3
def min_after_del (arr, n):
max_value = arr[0];
arr.sort();
for i in range(n - 1):
max_value = max(max_value, arr[i + 1] - arr[i]);
return max_value;
arr = [-1, -2, 4, 3, 5];
N = len(arr)
print(min_after_del(arr, N));
|
C#
using System;
public class GFG{
static int min_after_del(int[] arr, int n)
{
int max_value = arr[0];
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;
}
static public void Main (){
int[] arr = { -1, -2, 4, 3, 5 };
int N = arr.Length;
Console.Write(min_after_del(arr, N));
}
}
|
Javascript
<script>
const min_after_del = (arr, n) => {
let max_value = arr[0];
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;
}
let arr = [-1, -2, 4, 3, 5];
let N = arr.length;
document.write(min_after_del(arr, N));
</script>
|
Time Complexity : O(N log(N))
Space Complexity : O(1)
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