Program to find the smallest element among three elements
Last Updated :
23 Mar, 2023
Given three numbers. The task is to find the smallest among the given three numbers.
Examples:
Input: first = 15, second = 16, third = 10
Output: 10
Input: first = 5, second = 3, third = 6
Output: 3
Approach:
- Check if the first element is smaller than or equal to second and third. If yes then print it.
- Else if check if the second element is smaller than or equal to first and third. If yes then print it.
- Else third is the smallest element and print it.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a = 5, b = 7, c = 10;
if (a <= b && a <= c)
cout << a << " is the smallest" ;
else if (b <= a && b <= c)
cout << b << " is the smallest" ;
else
cout << c << " is the smallest" ;
return 0;
}
|
C
#include <stdio.h>
int main()
{
int a = 5, b = 7, c = 10;
if (a <= b && a <= c)
printf ( "%d is the smallest" ,a);
else if (b <= a && b <= c)
printf ( "%d is the smallest" ,b);
else
printf ( "%d is the smallest" ,c);
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void main (String[] args) {
int a = 5 , b = 7 , c = 10 ;
if (a <= b && a <= c)
System.out.println( a + " is the smallest" );
else if (b <= a && b <= c)
System.out.println( b + " is the smallest" );
else
System.out.println( c + " is the smallest" );
}
}
|
Python3
a, b, c = 5 , 7 , 10
if (a < = b and a < = c):
print (a, "is the smallest" )
elif (b < = a and b < = c):
print (b, "is the smallest" )
else :
print (c, "is the smallest" )
|
C#
using System;
class GFG
{
static public void Main ()
{
int a = 5, b = 7, c = 10;
if (a <= b && a <= c)
Console.WriteLine( a + " is the smallest" );
else if (b <= a && b <= c)
Console.WriteLine( b + " is the smallest" );
else
Console.WriteLine( c + " is the smallest" );
}
}
|
PHP
<?php
$a = 5; $b = 7; $c = 10;
if ( $a <= $b && $a <= $c )
echo $a . " is the smallest" ;
else if ( $b <= $a && $b <= $c )
echo $b . " is the smallest" ;
else
echo $c . " is the smallest" ;
|
Javascript
<script>
let a = 5, b = 7, c = 10;
if (a <= b && a <= c)
document.write( a + " is the smallest" );
else if (b <= a && b <= c)
document.write( b + " is the smallest" );
else
document.write( c + " is the smallest" );
</script>
|
Time complexity : O(1)
Auxiliary Space: O(1)
Approach:
1. We define three variables a, b, and c with values 5, 10, and 3 respectively.
2. We use the min() function to find the smallest value among a, b, and c.
3. We assign the smallest value to a new variable min_value.
4. We use f-string to print the value of min_value.
C++
#include<bits/stdc++.h>
using namespace std;
int main()
{
int a=5, b=10, c=3;
int min_value = min({a,b,c});
cout<< "The smallest value is " <<min_value;
}
|
Java
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
int a = 5 , b = 10 , c = 3 ;
int min_value = Arrays.stream( new int [] {a, b, c}).min().getAsInt();
System.out.println( "The smallest value is " + min_value);
}
}
|
Python3
a = 5
b = 10
c = 3
min_value = min (a, b, c)
print (f "The smallest value is {min_value}" )
|
C#
using System;
class Program {
static void Main( string [] args)
{
int a = 5, b = 10, c = 3;
int min_value = Math.Min(Math.Min(a, b), c);
Console.WriteLine( "The smallest value is "
+ min_value);
}
}
|
Javascript
const a = 5;
const b = 10;
const c = 3;
const min_value = Math.min(a, b, c);
console.log(`The smallest value is ${min_value}`);
|
Output
The smallest value is 3
Time complexity:
The time complexity of this program is O(1) since we are only comparing three elements, and the min() function has a constant time complexity.
Space complexity:
The space complexity of this program is O(1) since we are only using a constant amount of memory to store three variables and one additional variable to store the smallest value.
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