Find the first, second and third minimum elements in an array
Last Updated :
09 Mar, 2023
Find the first, second and third minimum elements in an array in O(n).
Examples:
Input : 9 4 12 6
Output : First min = 4
Second min = 6
Third min = 9
Input : 4 9 1 32 12
Output : First min = 1
Second min = 4
Third min = 9
First approach : First we can use normal method that is sort the array and then print first, second and third element of the array. Time complexity of this solution is O(n Log n).
C++
// CPP program to find the first, second
// and third minimum element in an array
#include<bits/stdc++.h>
using namespace std;
int Print3Smallest(int array[], int n)
{
// Sorting the array
sort(array,array+n);
cout << "First min = " << array[0] << "\n";
cout << "Second min = " << array[1] << "\n";
cout << "Third min = " << array[2] << "\n";
}
// Driver code
int main()
{
int array[] = {4, 9, 1, 32, 12};
int n = sizeof(array) / sizeof(array[0]);
Print3Smallest(array, n);
return 0;
}
// Code submitted by Pushpesh Raj
// Java program to find the first, second
// and third minimum element in an array
import java.util.Arrays;
public class Main {
static void Print3Smallest(int array[], int n) {
// Sorting the array
Arrays.sort(array);
System.out.println("First min = " + array[0]);
System.out.println("Second min = " + array[1]);
System.out.println("Third min = " + array[2]);
}
// Driver code
public static void main(String[] args) {
int array[] = {4, 9, 1, 32, 12};
int n = array.length;
Print3Smallest(array, n);
}
}
// this code is contributed by bhardwajji
# Python program to find the first, second
# and third minimum element in an array
def print_3_smallest(array):
# Sorting the array
array.sort()
print("First min =", array[0])
print("Second min =", array[1])
print("Third min =", array[2])
# Driver code
if __name__ == '__main__':
array = [4, 9, 1, 32, 12]
n = len(array)
print_3_smallest(array)
// C# program to find the first, second
// and third minimum element in an array
using System;
using System.Linq;
class Program
{
static void Print3Smallest(int[] array, int n)
{
// Sorting the array
Array.Sort(array);
Console.WriteLine("First min = " + array[0]);
Console.WriteLine("Second min = " + array[1]);
Console.WriteLine("Third min = " + array[2]);
}
// Driver code
static void Main()
{
int[] array = {4, 9, 1, 32, 12};
int n = array.Length;
Print3Smallest(array, n);
}
}
// JavaScript program to find the first,
// second and third minimum element in an
// array.
function Print3Smallest(array,n){
// Sorting the array
array.sort((a, b) => a - b);
console.log('First min = ' + array[0]);
console.log('Second min = ' + array[1]);
console.log('Third min = ' + array[2]);
}
// Driver code
let array = [4, 9, 1, 32, 12];
Print3Smallest(array,array.length);
OutputFirst min = 1
Second min = 4
Third min = 9
Second approach : Time complexity of this solution is O(n).
Algorithm:
- First take an element
- then if array[index] < Firstelement
- Thirdelement = Secondelement
- Secondelement = Firstelement
- Firstelement = array[index]
- else if array[index] < Secondelement
- Thirdelement = Secondelement
- Secondelement = array[index]
- else if array[index] < Thirdelement
- Thirdelement = array[index]
- then print all the element
Implementation:
C++
// CPP program to find the first, second
// and third minimum element in an array
#include<bits/stdc++.h>
#define MAX 100000
using namespace std;
int Print3Smallest(int array[], int n)
{
int firstmin = MAX, secmin = MAX, thirdmin = MAX;
for (int i = 0; i < n; i++)
{
/* Check if current element is less than
firstmin, then update first, second and
third */
if (array[i] < firstmin)
{
thirdmin = secmin;
secmin = firstmin;
firstmin = array[i];
}
/* Check if current element is less than
secmin then update second and third */
else if (array[i] < secmin)
{
thirdmin = secmin;
secmin = array[i];
}
/* Check if current element is less than
then update third */
else if (array[i] < thirdmin)
thirdmin = array[i];
}
cout << "First min = " << firstmin << "\n";
cout << "Second min = " << secmin << "\n";
cout << "Third min = " << thirdmin << "\n";
}
// Driver code
int main()
{
int array[] = {4, 9, 1, 32, 12};
int n = sizeof(array) / sizeof(array[0]);
Print3Smallest(array, n);
return 0;
}
// Java program to find the first, second
// and third minimum element in an array
import java.util.*;
public class GFG
{
static void Print3Smallest(int array[], int n)
{
int firstmin = Integer.MAX_VALUE;
int secmin = Integer.MAX_VALUE;
int thirdmin = Integer.MAX_VALUE;
for (int i = 0; i < n; i++)
{
/* Check if current element is less than
firstmin, then update first, second and
third */
if (array[i] < firstmin)
{
thirdmin = secmin;
secmin = firstmin;
firstmin = array[i];
}
/* Check if current element is less than
secmin then update second and third */
else if (array[i] < secmin)
{
thirdmin = secmin;
secmin = array[i];
}
/* Check if current element is less than
then update third */
else if (array[i] < thirdmin)
thirdmin = array[i];
}
System.out.println("First min = " + firstmin );
System.out.println("Second min = " + secmin );
System.out.println("Third min = " + thirdmin );
}
// Driver code
public static void main(String[] args)
{
int array[] = {4, 9, 1, 32, 12};
int n = array.length;
Print3Smallest(array, n);
}
}
// This code is contributed by
// Sam007
# A Python program to find the first,
# second and third minimum element
# in an array
MAX = 100000
def Print3Smallest(arr, n):
firstmin = MAX
secmin = MAX
thirdmin = MAX
for i in range(0, n):
# Check if current element
# is less than firstmin,
# then update first,second
# and third
if arr[i] < firstmin:
thirdmin = secmin
secmin = firstmin
firstmin = arr[i]
# Check if current element is
# less than secmin then update
# second and third
elif arr[i] < secmin:
thirdmin = secmin
secmin = arr[i]
# Check if current element is
# less than,then update third
elif arr[i] < thirdmin:
thirdmin = arr[i]
print("First min = ", firstmin)
print("Second min = ", secmin)
print("Third min = ", thirdmin)
# driver program
arr = [4, 9, 1, 32, 12]
n = len(arr)
Print3Smallest(arr, n)
# This code is contributed by Shrikant13.
// C# program to find the first, second
// and third minimum element in an array
using System;
class GFG
{
static void Print3Smallest(int []array, int n)
{
int firstmin = int.MaxValue;
int secmin = int.MaxValue;
int thirdmin = int.MaxValue;
for (int i = 0; i < n; i++)
{
/* Check if current element is less than
firstmin, then update first, second and
third */
if (array[i] < firstmin)
{
thirdmin = secmin;
secmin = firstmin;
firstmin = array[i];
}
/* Check if current element is less than
secmin then update second and third */
else if (array[i] < secmin)
{
thirdmin = secmin;
secmin = array[i];
}
/* Check if current element is less than
then update third */
else if (array[i] < thirdmin)
thirdmin = array[i];
}
Console.WriteLine("First min = " + firstmin );
Console.WriteLine("Second min = " + secmin );
Console.WriteLine("Third min = " + thirdmin );
}
// Driver code
static void Main()
{
int []array = new int[]{4, 9, 1, 32, 12};
int n = array.Length;
Print3Smallest(array, n);
}
}
// This code is contributed by Sam007
<?php
// php program to find the first, second
// and third minimum element in an array
function Print3Smallest($array, $n)
{
$MAX = 100000;
$firstmin = $MAX;
$secmin = $MAX;
$thirdmin = $MAX;
for ($i = 0; $i < $n; $i++)
{
/* Check if current element is less than
firstmin, then update first, second and
third */
if ($array[$i] < $firstmin)
{
$thirdmin = $secmin;
$secmin = $firstmin;
$firstmin = $array[$i];
}
/* Check if current element is less than
secmin then update second and third */
else if ($array[$i] < $secmin)
{
$thirdmin = $secmin;
$secmin = $array[$i];
}
/* Check if current element is less than
then update third */
else if ($array[$i] < $thirdmin)
$thirdmin = $array[$i];
}
echo "First min = ".$firstmin."\n";
echo "Second min = ".$secmin."\n";
echo "Third min = ".$thirdmin."\n";
}
// Driver code
$array= array(4, 9, 1, 32, 12);
$n = sizeof($array) / sizeof($array[0]);
Print3Smallest($array, $n);
// This code is contributed by mits
?>
<script>
// Javascript program to find the first, second
// and third minimum element in an array
let MAX = 100000
function Print3Smallest( array, n)
{
let firstmin = MAX, secmin = MAX, thirdmin = MAX;
for (let i = 0; i < n; i++)
{
/* Check if current element is less than
firstmin, then update first, second and
third */
if (array[i] < firstmin)
{
thirdmin = secmin;
secmin = firstmin;
firstmin = array[i];
}
/* Check if current element is less than
secmin then update second and third */
else if (array[i] < secmin)
{
thirdmin = secmin;
secmin = array[i];
}
/* Check if current element is less than
then update third */
else if (array[i] < thirdmin)
thirdmin = array[i];
}
document.write("First min = " + firstmin + "</br>");
document.write("Second min = " + secmin + "</br>");
document.write("Third min = " + thirdmin + "</br>");
}
// Driver program
let array = [4, 9, 1, 32, 12];
let n = array.length;
Print3Smallest(array, n);
</script>
OutputFirst min = 1
Second min = 4
Third min = 9
Time Complexity: O(n)
Auxiliary Space: O(1)
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