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

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

// 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

Python3

# 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#

// 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

// 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);

Output

First 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

// 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

Time Complexity: O(n)
Auxiliary Space: O(1)