Find Maximum and Minimum of two numbers using Absolute function
Last Updated :
25 Sep, 2022
Given two numbers, the task is to print the maximum and minimum of the given numbers using Absolute function.
Examples:
Input: 99, 18
Output: Maximum = 99
Minimum = 18
Input: -10, 20
Output: Maximum = 20
Minimum = -10
Input: -1, -5
Output: Maximum = -1
Minimum = -5
Approach:
This problem can be solved by applying the concept of Absolute Function and BODMAS rule.
[(x + y + abs(x - y)) / 2]
[(x + y - abs(x - y)) / 2]
Let us consider two number x and y where x = 20, y = 70 respectively.
For Maximum:
[(x + y + abs(x - y)) / 2]
=> [(20 + 70)+ abs(20-70)) / 2]
=> 140 / 2 = 70 [MAXIMUM]
For Minimum:
[(x + y - abs(x - y)) / 2]
=> [(20 + 70) - abs(20-70)) / 2]
=> 40 / 2 = 20 [MINIMUM]
C++
#include <bits/stdc++.h>
using namespace std;
int maximum( int x, int y)
{
return ((x + y + abs (x - y)) / 2);
}
int minimum( int x, int y)
{
return ((x + y - abs (x - y)) / 2);
}
int main()
{
int x = 99, y = 18;
cout << "Maximum: " << maximum(x, y) << endl;
cout << "Minimum: " << minimum(x, y) << endl;
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
int maximum( int x, int y)
{
return ((x + y + abs (x - y)) / 2);
}
int minimum( int x, int y)
{
return ((x + y - abs (x - y)) / 2);
}
void main()
{
int x = 99, y = 18;
printf ( "Maximum: %d\n" , maximum(x, y));
printf ( "Minimum: %d\n" , minimum(x, y));
}
|
Java
class GFG
{
static int maximum( int x, int y)
{
return ((x + y + Math.abs(x - y)) / 2 );
}
static int minimum( int x, int y)
{
return ((x + y - Math.abs(x - y)) / 2 );
}
public static void main (String[] args)
{
int x = 99 , y = 18 ;
System.out.println( "Maximum: " + maximum(x, y));
System.out.println( "Minimum: " + minimum(x, y));
}
}
|
C#
using System;
class GFG
{
static int maximum( int x, int y)
{
return ((x + y + Math.Abs(x - y)) / 2);
}
static int minimum( int x, int y)
{
return ((x + y - Math.Abs(x - y)) / 2);
}
public static void Main()
{
int x = 99, y = 18;
Console.WriteLine( "Maximum: " + maximum(x, y));
Console.WriteLine( "Minimum: " + minimum(x, y));
}
}
|
Python3
def maximum(x, y):
return ((x + y + abs (x - y)) / / 2 )
def minimum(x, y):
return ((x + y - abs (x - y)) / / 2 )
x = 99
y = 18
print ( "Maximum:" , maximum(x, y))
print ( "Minimum:" , minimum(x, y))
|
Javascript
<script>
function maximum(x,y)
{
return ((x + y + Math.abs(x - y)) / 2);
}
function minimum(x,y)
{
return ((x + y - Math.abs(x - y)) / 2);
}
let x = 99, y = 18;
document.write( "Maximum: " + maximum(x, y)+ "<br>" );
document.write( "Minimum: " + minimum(x, y));
</script>
|
Output:
Maximum: 99
Minimum: 18
Time complexity: O(1)
Auxiliary space: O(1)
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