Find the smallest and second smallest elements in an array

Write an efficient C++/C program to find the 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

Naive approach:

A Simple Solution is to sort the array in increasing order. The first two elements in the sorted array would be the two smallest elements. In this approach, if the smallest element is present more than one time then we will have to use a loop for printing the unique smallest and second smallest elements. 

Below is the implementation of the above approach:

C++

//C++ simple approach to print smallest
//and second smallest element.
#include<bits/stdc++.h>
using namespace std;
int main() {
int arr[]={111, 13, 25, 9, 34, 1};
int n=sizeof(arr)/sizeof(arr[0]);
//sorting the array using
//in-built sort function
sort(arr,arr+n);
//printing the desired element
cout<<"smallest element is "<<arr[0]<<endl;
cout<<"second smallest element is "<<arr[1];
return 0;
}

//this code is contributed by Machhaliya Muhammad

Java

/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

class GFG {
//Java simple approach to print smallest
//and second smallest element.

// Driver Code
public static void main(String args[])
{
    int arr[]={111, 13, 25, 9, 34, 1};
    int n=arr.length;

    // sorting the array using
    // in-built sort function
    Arrays.sort(arr);

    // printing the desired element
    System.out.println("smallest element is "+arr[0]);
    System.out.println("second smallest element is "+arr[1]);
}
}

// This code is contributed by shinjanpatra

Python3

# Python3 simple approach to print smallest
# and second smallest element.

# driver code

arr = [111, 13, 25, 9, 34, 1]
n = len(arr)
# sorting the array using
# in-built sort function
arr.sort()

# printing the desired element
print("smallest element is "+str(arr[0]))
print("second smallest element is "+str(arr[1]))

# This code is contributed by shinjanpatra

C#

// C# simple approach to print smallest
// and second smallest element.
using System;

public class GFG
{
  // Driver Code
  static public void Main()
  {
    int[] arr = {111, 13, 25, 9, 34, 1};
    int n = arr.Length;

    // sorting the array using
    // in-built sort function
    Array.Sort(arr);

    // printing the desired element
    Console.WriteLine("smallest element is "+arr[0]);
    Console.WriteLine("second smallest element is "+arr[1]);
  }
}

// This code is contributed by kothavvsaakash

Javascript

<script>

// JavaScript simple approach to print smallest
// and second smallest element.

// driver code

let arr = [111, 13, 25, 9, 34, 1];
let n = arr.length;
// sorting the array using
// in-built sort function
arr.sort();

// printing the desired element
document.write("smallest element is "+arr[0],"</br>");
document.write("second smallest element is "+arr[1],"</br>");

// This code is contributed by shinjanpatra

</script>

Output

smallest element is 1
second smallest element is 9

Time complexity: O(N*logN)
Auxiliary space: O(1)

An Efficient Approach: A Better Solution is to scan the array twice. In the first traversal find the minimum element. Let this element be x. In the second traversal, find the smallest element greater than x.

Using this method, we can overcome the problem of Method 1 which occurs when the smallest element is present in an array more than one time.
The above solution requires two traversals of the input array. 

C++

// C++ program to find smallest and
// second smallest element in array
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int arr[] = {12, 13, 1, 10, 34, 1};
    int n = sizeof(arr) / sizeof(arr[0]);
    int smallest = INT_MAX;
    // traversing the array to find
    // smallest element.
    for (int i = 0; i < n; i++)
    {
        if (arr[i] < smallest)
        {
            smallest = arr[i];
        }
    }
    cout << "smallest element is: " << smallest << endl;

    int second_smallest = INT_MAX;

    // traversing the array to find second smallest element
    for (int i = 0; i < n; i++)
    {
        if (arr[i] < second_smallest && arr[i] > smallest)
        {
            second_smallest = arr[i];
        }
    }
    cout << "second smallest element is: " << second_smallest << endl;
    return 0;
}

// This code is contributed by Machhaliya Muhamma

Java

// Java program to find smallest and
// second smallest element in array
import java.io.*;
class GFG {
    public static void main(String args[])
    {
        int arr[] = { 12, 13, 1, 10, 34, 1 };
        int n = arr.length;
        int smallest = Integer.MAX_VALUE;
        // traversing the array to find
        // smallest element.
        for (int i = 0; i < n; i++) {
            if (arr[i] < smallest) {
                smallest = arr[i];
            }
        }
        System.out.println("smallest element is: "
                           + smallest);

        int second_smallest = Integer.MAX_VALUE;

        // traversing the array to find second smallest
        // element
        for (int i = 0; i < n; i++) {
            if (arr[i] < second_smallest
                && arr[i] > smallest) {
                second_smallest = arr[i];
            }
        }
        System.out.println("second smallest element is: "
                           + second_smallest);
    }
}

// This code is contributed by Lovely Jain

Python

# python program to find smallest and second smallest element in array

# import the module
import sys

arr = [12, 13, 1, 10, 34, 1]
n = len(arr)
smallest = sys.maxint

# traversing the array to find smallest element.
for i in range(n):
    if(arr[i] < smallest):
        smallest = arr[i]

print('smallest element is: ' + str(smallest))
second_smallest = sys.maxint

# traversing the array to find second smallest element
for i in range(n):
    if(arr[i] < second_smallest and arr[i] > smallest):
        second_smallest = arr[i]

print('second smallest element is: ' + str(second_smallest))

# This code is contributed by lokeshmvs21.

C#

// C# program to find smallest and
// second smallest element in array

using System;

public class GFG
{
  static public void Main ()
  {
    int[] arr = { 12, 13, 1, 10, 34, 1 };
    int n = arr.Length;
    int smallest = Int32.MaxValue;
    // traversing the array to find
    // smallest element.
    for (int i = 0; i < n; i++)
    {
      if (arr[i] < smallest)
      {
        smallest = arr[i];
      }
    }
    Console.WriteLine("smallest element is: " + smallest);

    int second_smallest = Int32.MaxValue;

    // traversing the array to find second smallest
    // element
    for (int i = 0; i < n; i++)
    {
      if (arr[i] < second_smallest && arr[i] > smallest)
      {
        second_smallest = arr[i];
      }
    }
    Console.WriteLine("second smallest element is: " + second_smallest);
  }
}

// This code is contributed by kothavvsaakash

Javascript

<script>

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

function solution( arr, arr_size)
{
  let first = Number.MAX_VALUE,
        second = Number.MAX_VALUE;

  /* There should be atleast two elements */
  if (arr_size < 2)
  {
    document.write(" Invalid Input ");
    return;
  }
   /* find the smallest element */
  for (let i = 0; i < arr_size ; i ++)
  {
    if (arr[i] < first){
      first = arr[i];
    }
  }
   /* find the second smallest element */
   for (let i = 0; i < arr_size ; i ++){
    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;
  solution(arr, n);

</script>

Output

smallest element is: 1
second smallest element is: 10

Time complexity: O(N)
Auxiliary space: O(1)

An Efficient Solution can find the minimum two elements in one traversal. Below is the 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

Below is the implementation of the above approach:

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)
Auxiliary Space: O(1)

Approach: A log(N) approach using Priority_queue data structure. You can read about Priority Queue in more detail here.

C++

#include <bits/stdc++.h>
using namespace std;

int main() {
    int arr[] = {2,5,7,3,9,10,11,1};
      int n = sizeof(arr) / sizeof(arr[0]);
      priority_queue<int,vector<int>,greater<int>>pq;//min heap priority_queue
      //adding elements to priority queue
      for(int i=0;i<n;i++){
          pq.push(arr[i]);
    }
      int t = pq.top();//smallest element will be on top of pq as it is min heap
      pq.pop();//removing first element so second will be on top.
      int w = pq.top();

      cout<<"smallest element : "<<t<<endl;
      cout<<"second smallest element : "<<w<<endl;

    return 0;
}
// this code is contributed by lakshyadeeppatel

Java

import java.util.*;

class GFG {
  public static void main(String[] args)
  {
    int arr[] = { 2, 5, 7, 3, 9, 10, 11, 1 };
    int n = arr.length;

    PriorityQueue<Integer> pq
      = new PriorityQueue<Integer>();

    // adding elements to priority queue
    for (int i = 0; i < n; i++) {
      pq.add(arr[i]);
    }

    int t = pq.peek(); // smallest element will be on
    // top of pq
    pq.poll(); // removing first element so second will
    // be on top.
    int w = pq.peek();

    System.out.println("smallest element : " + t);
    System.out.println("second smallest element : "
                       + w);
  }
}

// This code is contributed by divyansh2212

Output

smallest element : 1
second smallest element : 2

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

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