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Remove exactly one element from the array such that max – min is minimum

  • Difficulty Level : Basic
  • Last Updated : 28 May, 2021

Given an array consisting of N positive integer numbers. The task is to remove exactly one element from this array to minimize max(a) – min(a) and print the minimum possible (max(a) – min(a)).
Note: max(a) means largest number in array a   and min(a) means smallest number in array a
There are at least 2 elements in the array.
Examples: 
 

Input: arr[] = {1, 3, 3, 7}
Output: 2
Remove 7, then max(a) will be 3 and min(a) will be 1.
So our answer will be 3-1 = 2.

Input: arr[] = {1, 1000}
Output: 0
Remove either 1 or 1000, then our answer will 1-1 =0 or
1000-1000=0

 

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Simple Approach: Here it can be seen that we always have to remove either minimum or maximum of the array. We first sort the array. After sorting, if we remove minimum element, the difference would be a[n-1] – a[1]. And if we remove the maximum element, difference would be a[n-2] – a[0]. We return minimum of these two differences.
Below is the implementation of the above approach: 
 



C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate max-min
int max_min(int a[], int n)
{
    sort(a, a + n);
 
    return min(a[n - 2] - a[0], a[n - 1] - a[1]);
}
 
// Driver code
int main()
{
    int a[] = { 1, 3, 3, 7 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << max_min(a, n);
    return 0;
}

Java




// Java implementation of the above approach
 
import java.util.*;
class GFG
{
    // function to calculate max-min
    static int max_min(int a[], int n)
    {
        Arrays.sort(a);
     
        return Math.min(a[n - 2] - a[0], a[n - 1] - a[1]);
    }
     
    // Driver code
    public static void main(String []args)
    {
        int a[] = { 1, 3, 3, 7 };
        int n = a.length;
     
        System.out.println(max_min(a, n));
     
    }
}
 
// This code is contributed
// by ihritik

Python3




# Python3 implementation of the
# above approach
 
# function to calculate max-min
def max_min(a, n):
    a.sort()
    return min(a[n - 2] - a[0],
               a[n - 1] - a[1])
 
# Driver code
a = [1, 3, 3, 7]
n = len(a)
print(max_min(a, n))
 
# This code is contributed
# by sahishelangia

C#




// C# implementation of the above approach
 
using System;
class GFG
{
    // function to calculate max-min
    static int max_min(int []a, int n)
    {
        Array.Sort(a);
     
        return Math.Min(a[n - 2] - a[0], a[n - 1] - a[1]);
    }
     
    // Driver code
    public static void Main()
    {
        int []a = { 1, 3, 3, 7 };
        int n = a.Length;
     
        Console.WriteLine(max_min(a, n));
     
    }
}
 
// This code is contributed
// by ihritik

PHP




<?php
// PHP implementation of the above approach
 
// function to calculate max-min
function max_min(&$a, $n)
{
    sort($a);
 
    return min($a[$n - 2] - $a[0],
               $a[$n - 1] - $a[1]);
}
 
// Driver code
$a = array(1, 3, 3, 7);
$n = sizeof($a);
 
echo(max_min($a, $n));
 
// This code is contributed by Shivi_Aggarwal
?>

Javascript




<script>
 
// Javascript program of the above approach
 
 // function to calculate max-min
    function max_min(a, n)
    {
        a.sort();
     
        return Math.min(a[n - 2] - a[0], a[n - 1] - a[1]);
    }
 
// Driver code
 
        let a = [ 1, 3, 3, 7 ];
        let n = a.length;
     
        document.write(max_min(a, n));
 
</script>
Output: 
2

 

Time Complexity: O(n log n)
Efficient Approach: 
An efficient approach is to do following. 
1) Find first minimum and second minimum 
2) Find first maximum and second maximum 
3) Return the minimum of following two differences. 
…..a) First maximum and second minimum 
…..b) Second maximum and first minimum
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// function to calculate max-min
int max_min(int a[], int n)
{
    // There should be at-least two elements
    if (n <= 1)
      return INT_MAX;
 
    // To store first and second minimums
    int f_min = a[0], s_min = INT_MAX;
 
    // To store first and second maximums
    int f_max = a[0], s_max = INT_MIN;
 
    for (int i = 1; i<n ;i++)
    {
        if (a[i] <= f_min)
        {
           s_min = f_min;
           f_min = a[i];
        }
        else if (a[i] < s_min)
        {
           s_min = a[i];
        }
 
        if (a[i] >= f_max)
        {
           s_max = f_max;
           f_max = a[i];
        }
        else if (a[i] > s_max)
        {
           s_max = a[i];
        }
    }
 
    return min((f_max - s_min), (s_max - f_min));
}
 
// Driver code
int main()
{
    int a[] = { 1, 3, 3, 7 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << max_min(a, n);
    return 0;
}

Java




// Java implementation of the above approach
 
class GFG
{
    // function to calculate max-min
    static int max_min(int a[], int n)
    {
        // There should be at-least two elements
        if (n <= 1)
        return Integer.MAX_VALUE;
     
        // To store first and second minimums
        int f_min = a[0], s_min = Integer.MAX_VALUE;
     
        // To store first and second maximums
        int f_max = a[0], s_max = Integer.MIN_VALUE;
     
        for (int i = 1; i<n ;i++)
        {
            if (a[i] <= f_min)
            {
            s_min = f_min;
            f_min = a[i];
            }
            else if (a[i] < s_min)
            {
            s_min = a[i];
            }
     
            if (a[i] >= f_max)
            {
            s_max = f_max;
            f_max = a[i];
            }
            else if (a[i] > s_max)
            {
            s_max = a[i];
            }
        }
     
        return Math.min((f_max - s_min), (s_max - f_min));
    }
     
    // Driver code
    public static void main(String []args)
    {
        int a[] = { 1, 3, 3, 7 };
        int n = a.length;
     
        System.out.println(max_min(a, n));
     
    }
 
}
 
// This code is contributed
// by ihritik

Python3




# Python3 implementation of the
# above approach
import sys
 
# function to calculate max-min
def max_min(a, n) :
     
    # There should be at-least two elements
    if (n <= 1) :
        return sys.maxsize
 
    # To store first and second minimums
    f_min = a[0]
    s_min = sys.maxsize
 
    # To store first and second maximums
    f_max = a[0]
    s_max = -(sys.maxsize - 1)
 
    for i in range(n) :
         
        if (a[i] <= f_min) :
            s_min = f_min
            f_min = a[i]
         
        elif (a[i] < s_min) :
            s_min = a[i]
 
        if (a[i] >= f_max) :
            s_max = f_max
            f_max = a[i]
     
        elif (a[i] > s_max) :
            s_max = a[i]
 
    return min((f_max - s_min), (s_max - f_min))
 
# Driver code
if __name__ == "__main__" :
    a = [ 1, 3, 3, 7 ]
    n = len(a)
 
    print(max_min(a, n))
 
# This code is contributed by Ryuga

C#




// C# implementation of the above approach
using System;
class GFG
{
    // function to calculate max-min
    static int max_min(int []a, int n)
    {
        // There should be at-least two elements
        if (n <= 1)
        return Int32.MaxValue;
     
        // To store first and second minimums
        int f_min = a[0], s_min = Int32.MaxValue;
     
        // To store first and second maximums
        int f_max = a[0], s_max = Int32.MinValue;
     
        for (int i = 1; i<n ;i++)
        {
            if (a[i] <= f_min)
            {
            s_min = f_min;
            f_min = a[i];
            }
            else if (a[i] < s_min)
            {
            s_min = a[i];
            }
     
            if (a[i] >= f_max)
            {
            s_max = f_max;
            f_max = a[i];
            }
            else if (a[i] > s_max)
            {
            s_max = a[i];
            }
        }
     
        return Math.Min((f_max - s_min), (s_max - f_min));
    }
     
    // Driver code
    public static void Main()
    {
        int []a = { 1, 3, 3, 7 };
        int n = a.Length;
     
        Console.WriteLine(max_min(a, n));
     
    }
 
}
 
// This code is contributed
// by ihritik

PHP




<?php
// PHP implementation of the above approach
// function to calculate max-min
function max_min($a, $n)
{
    // There should be at-least
    // two elements
    if ($n <= 1)
    return PHP_INT_MAX;
 
    // To store first and second minimums
    $f_min = $a[0];
    $s_min = PHP_INT_MAX;
 
    // To store first and second maximums
    $f_max = $a[0];
    $s_max = ~PHP_INT_MAX;
 
    for ($i = 1; $i < $n ;$i++)
    {
        if ($a[$i] <= $f_min)
        {
            $s_min = $f_min;
            $f_min = $a[$i];
        }
        else if ($a[$i] < $s_min)
        {
            $s_min = $a[$i];
        }
 
        if ($a[$i] >= $f_max)
        {
            $s_max = $f_max;
            $f_max = $a[$i];
        }
        else if ($a[$i] > $s_max)
        {
            $s_max = $a[$i];
        }
    }
 
    return min(($f_max - $s_min),
               ($s_max - $f_min));
}
 
// Driver code
$a = array ( 1, 3, 3, 7 );
$n = sizeof($a);
 
echo(max_min($a, $n));
 
// This code is contributed
// by Mukul Singh
?>

Javascript




<script>
 
    // JavaScript implementation of the above approach
     
    // function to calculate max-min
    function max_min(a, n)
    {
        // There should be at-least two elements
        if (n <= 1)
            return Number.MAX_VALUE;
      
        // To store first and second minimums
        let f_min = a[0], s_min = Number.MAX_VALUE;
      
        // To store first and second maximums
        let f_max = a[0], s_max = Number.MIN_VALUE;
      
        for (let i = 1; i<n ;i++)
        {
            if (a[i] <= f_min)
            {
              s_min = f_min;
              f_min = a[i];
            }
            else if (a[i] < s_min)
            {
                s_min = a[i];
            }
      
            if (a[i] >= f_max)
            {
              s_max = f_max;
              f_max = a[i];
            }
            else if (a[i] > s_max)
            {
                s_max = a[i];
            }
        }
      
        return Math.min((f_max - s_min), (s_max - f_min));
    }
     
    let a = [ 1, 3, 3, 7 ];
    let n = a.length;
 
    document.write(max_min(a, n));
 
</script>
Output: 
2

 

Time Complexity: O(n)
 




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