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Find the first, second and third minimum elements in an array

  • Difficulty Level : Basic
  • Last Updated : 27 Aug, 2021
Geek Week

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).
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

 

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



// 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
Python3



# 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#



// 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
// 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
?>
Javascript



<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>
Output:  


First min = 1
Second min = 4
Third min = 9
 
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Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
DevanshuAgarwal
@DevanshuAgarwal
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