Find the smallest and second smallest elements in an array

Write an efficient C program to find smallest and second smallest element in an array.

Example: 

Input:  arr[] = {12, 13, 1, 10, 34, 1}
Output: The smallest element is 1 and 
        second Smallest element is 10

A Simple Solution is to sort the array in increasing order. The first two elements in sorted array would be two smallest elements. Time complexity of this solution is O(n Log n).
A Better Solution is to scan the array twice. In first traversal find the minimum element. Let this element be x. In second traversal, find the smallest element greater than x. Time complexity of this solution is O(n).
The above solution requires two traversals of input array. 
An Efficient Solution can find the minimum two elements in one traversal. Below is complete algorithm.
Algorithm:

1) Initialize both first and second smallest as INT_MAX
   first = second = INT_MAX
2) Loop through all the elements.
   a) If the current element is smaller than first, then update first 
       and second. 
   b) Else if the current element is smaller than second then update 
    second

Implementation:

C++

// C++ program to find smallest and
// second smallest elements
#include <bits/stdc++.h>
using namespace std; /* For INT_MAX */

void print2Smallest(int arr[], int arr_size)
{
    int i, first, second;

    /* There should be atleast two elements */
    if (arr_size < 2)
    {
        cout<<" Invalid Input ";
        return;
    }

    first = second = INT_MAX;
    for (i = 0; i < arr_size ; i ++)
    {
        /* If current element is smaller than first
        then update both first and second */
        if (arr[i] < first)
        {
            second = first;
            first = arr[i];
        }

        /* If arr[i] is in between first and second
        then update second */
        else if (arr[i] < second && arr[i] != first)
            second = arr[i];
    }
    if (second == INT_MAX)
        cout << "There is no second smallest element\n";
    else
        cout << "The smallest element is " << first << " and second "
            "Smallest element is " << second << endl;
}

/* Driver code */
int main()
{
    int arr[] = {12, 13, 1, 10, 34, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    print2Smallest(arr, n);
    return 0;
}

// This is code is contributed by rathbhupendra

C

// C program to find smallest and second smallest elements
#include <stdio.h>
#include <limits.h> /* For INT_MAX */

void print2Smallest(int arr[], int arr_size)
{
    int i, first, second;

    /* There should be atleast two elements */
    if (arr_size < 2)
    {
        printf(" Invalid Input ");
        return;
    }

    first = second = INT_MAX;
    for (i = 0; i < arr_size ; i ++)
    {
        /* If current element is smaller than first
           then update both first and second */
        if (arr[i] < first)
        {
            second = first;
            first = arr[i];
        }

        /* If arr[i] is in between first and second
           then update second  */
        else if (arr[i] < second && arr[i] != first)
            second = arr[i];
    }
    if (second == INT_MAX)
        printf("There is no second smallest element\n");
    else
        printf("The smallest element is %d and second "
               "Smallest element is %d\n", first, second);
}

/* Driver program to test above function */
int main()
{
    int arr[] = {12, 13, 1, 10, 34, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    print2Smallest(arr, n);
    return 0;
}

Java

// Java program to find smallest and second smallest elements
import java.io.*;

class SecondSmallest
{
    /* Function to print first smallest and second smallest
      elements */
    static void print2Smallest(int arr[])
    {
        int first, second, arr_size = arr.length;

        /* There should be atleast two elements */
        if (arr_size < 2)
        {
            System.out.println(" Invalid Input ");
            return;
        }

        first = second = Integer.MAX_VALUE;
        for (int i = 0; i < arr_size ; i ++)
        {
            /* If current element is smaller than first
              then update both first and second */
            if (arr[i] < first)
            {
                second = first;
                first = arr[i];
            }

            /* If arr[i] is in between first and second
               then update second  */
            else if (arr[i] < second && arr[i] != first)
                second = arr[i];
        }
        if (second == Integer.MAX_VALUE)
            System.out.println("There is no second" +
                               "smallest element");
        else
            System.out.println("The smallest element is " +
                               first + " and second Smallest" +
                               " element is " + second);
    }

    /* Driver program to test above functions */
    public static void main (String[] args)
    {
        int arr[] = {12, 13, 1, 10, 34, 1};
        print2Smallest(arr);
    }
}
/*This code is contributed by Devesh Agrawal*/

Python3

# Python program to find smallest and second smallest elements
import math

def print2Smallest(arr):

    # There should be atleast two elements
    arr_size = len(arr)
    if arr_size < 2:
        print ("Invalid Input")
        return

    first = second = math.inf
    for i in range(0, arr_size):

        # If current element is smaller than first then
        # update both first and second
        if arr[i] < first:
            second = first
            first = arr[i]

        # If arr[i] is in between first and second then
        # update second
        elif (arr[i] < second and arr[i] != first):
            second = arr[i];

    if (second == math.inf):
        print ("No second smallest element")
    else:
        print ('The smallest element is',first,'and', \
              ' second smallest element is',second)

# Driver function to test above function
arr = [12, 13, 1, 10, 34, 1]
print2Smallest(arr)

# This code is contributed by Devesh Agrawal

C#

// C# program to find smallest
// and second smallest elements
using System;

class GFG
{

    /* Function to print first smallest
     and second smallest elements */
    static void print2Smallest(int []arr)
    {
        int first, second, arr_size = arr.Length;

        /* There should be atleast two elements */
        if (arr_size < 2)
        {
            Console.Write(" Invalid Input ");
            return;
        }

        first = second = int.MaxValue;

        for (int i = 0; i < arr_size ; i ++)
        {
            /* If current element is smaller than first
            then update both first and second */
            if (arr[i] < first)
            {
                second = first;
                first = arr[i];
            }

            /* If arr[i] is in between first and second
            then update second */
            else if (arr[i] < second && arr[i] != first)
                second = arr[i];
        }
        if (second == int.MaxValue)
            Console.Write("There is no second" +
                            "smallest element");
        else
            Console.Write("The smallest element is " +
                            first + " and second Smallest" +
                            " element is " + second);
    }

    /* Driver program to test above functions */
    public static void Main()
    {
        int []arr = {12, 13, 1, 10, 34, 1};
        print2Smallest(arr);
    }
}

// This code is contributed by Sam007

PHP

<?php
// PHP  program to find smallest and
// second smallest elements

function print2Smallest($arr, $arr_size)
{
    $INT_MAX = 2147483647;

    /* There should be atleast
       two elements */
    if ($arr_size < 2)
    {
        echo(" Invalid Input ");
        return;
    }

    $first = $second = $INT_MAX;
    for ($i = 0; $i < $arr_size ; $i ++)
    {

        /* If current element is
           smaller than first then
           update both first and
           second */
        if ($arr[$i] < $first)
        {
            $second = $first;
            $first = $arr[$i];
        }

        /* If arr[i] is in between
           first and second then
           update second */
        else if ($arr[$i] < $second &&
                 $arr[$i] != $first)
            $second = $arr[$i];
    }
    if ($second == $INT_MAX)
        echo("There is no second smallest element\n");
    else
        echo "The smallest element is ",$first
             ," and second Smallest element is "
                                     , $second;
}

// Driver Code
$arr = array(12, 13, 1, 10, 34, 1);
$n = count($arr);
print2Smallest($arr, $n)

// This code is contributed by Smitha
?>

Javascript

<script>

// Javascript program to find smallest and
// second smallest elements

function print2Smallest( arr, arr_size)
{
    let i, first, second;

    /* There should be atleast two elements */
    if (arr_size < 2)
    {
        document.write(" Invalid Input ");
        return;
    }

    first=Number.MAX_VALUE ;
    second=Number.MAX_VALUE ;
    for (i = 0; i < arr_size ; i ++)
    {
        /* If current element is smaller than first
        then update both first and second */
        if (arr[i] < first)
        {
            second = first;
            first = arr[i];
        }

        /* If arr[i] is in between first and second
        then update second */
        else if (arr[i] < second && arr[i] != first)
            second = arr[i];
    }
    if (second == Number.MAX_VALUE )
        document.write("There is no second smallest element\n");
    else
        document.write("The smallest element is " + first + " and second "+
            "Smallest element is " + second +'\n');
}


    // Driver program

    let arr = [ 12, 13, 1, 10, 34, 1 ];
    let n = arr.length;
    print2Smallest(arr, n);

</script>

Output :

The smallest element is 1 and second Smallest element is 10

The same approach can be used to find the largest and second largest elements in an array.
Time Complexity: O(n)

Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.