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Find the smallest and second smallest elements in an array

Last Updated : 11 Mar, 2024
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Given an array arr[] of size N, find the smallest and second smallest element in an array.

Examples:

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

Input: arr[] = {111, 13, 25, 9, 34, 1}
Output: 1 9
Explanation: The smallest element is 1 and second smallest element is 9.

Approach 1:

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


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




// 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((a, b) => a - b);
 
// 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


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


Output

smallest element is 1
second smallest element is 9

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

Finding the smallest and second smallest elements by traversing the array twice (Two-pass):

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[] = { 111, 13, 25, 9, 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[] = { 111, 13, 25, 9, 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 = [111, 13, 25, 9, 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 = { 111, 13, 25, 9, 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




// 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 = [111, 13, 25, 9, 34, 1];
  let n = arr.length;
  solution(arr, n);


Output

smallest element is: 1
second smallest element is: 9

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

Finding the smallest and second smallest elements by traversing the array twice (One-pass):

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.

  • If the current element is smaller than first, then update first and second
  • 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[] = { 111, 13, 25, 9, 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 <limits.h> /* For INT_MAX */
#include <stdio.h>
 
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[] = { 111, 13, 25, 9, 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[] = { 111, 13, 25, 9, 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 = { 111, 13, 25, 9, 34, 1 };
        print2Smallest(arr);
    }
}
 
// This code is contributed by Sam007


Javascript




// 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 = [ 111, 13, 25, 9, 34, 1 ];
    let n = arr.length;
    print2Smallest(arr, n);


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( 111, 13, 25, 9, 34, 1 );
$n = count($arr);
print2Smallest($arr, $n)
 
// This code is contributed by Smitha
?>


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 = [111, 13, 25, 9, 34, 1]
print2Smallest(arr)
 
# This code is contributed by Devesh Agrawal


Output

The smallest element is 1 and second Smallest element is 9

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)

Related Article: Minimum and Second minimum elements using minimum comparisons



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