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 
and min(a) means smallest number in array .
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
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-minint max_min(int a[], int n){ sort(a, a + n); return min(a[n - 2] - a[0], a[n - 1] - a[1]);}// Driver codeint 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 approachimport 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 approachusing 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-minfunction 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)
Auxiliary Space: O(1)
Efficient Approach:
An efficient approach is to do following.
- Find first minimum and second minimum
- Find first maximum and second maximum
- Return the minimum of following two differences.
- First maximum and second minimum
- Second maximum and first minimum
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// function to calculate max-minint 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 codeint 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 approachclass 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 approachusing 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-minfunction 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)
Auxiliary Space: O(1)
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