给定一个由 N 个正整数组成的数组。任务是从该数组中准确删除一个元素以最小化max(a) – min(a)并打印可能的最小值(max(a) – min(a)) 。
注意: max(a) 表示数组中最大的数 min(a) 表示数组中最小的数 .
数组中至少有 2 个元素。
例子:
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
简单的方法:从这里可以看出,我们总是要删除数组的最小值或最大值。我们首先对数组进行排序。排序后,如果我们删除最小元素,差异将是 a[n-1] – a[1]。如果我们删除最大元素,差异将是 a[n-2] – a[0]。我们返回这两个差异中的最小值。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include
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
Javascript
C++
// C++ implementation of the above approach
#include
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= 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= 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= 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
= $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
输出:
2
时间复杂度: O(n log n)
有效的方法:
一种有效的方法是执行以下操作。
1) 找到第一个最小值和第二个最小值
2)找到第一个最大值和第二个最大值
3) 返回以下两个差值中的最小值。
…..a) 第一个最大值和第二个最小值
…..b) 第二个最大值和第一个最小值
C++
// C++ implementation of the above approach
#include
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= 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= 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
蟒蛇3
# 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= 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
= $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
输出:
2
时间复杂度: O(n)
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