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_MAXfirst=second= INT_MAX 2) Loop through all the elements. a) If the current element is smaller thanfirst, then updatefirstandsecond. b) Else if the current element is smaller thansecondthen updatesecond

**Implementation:**

## C/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*/

## Python

# Python program to find smallest and second smallest elements import sys def print2Smallest(arr): # There should be atleast two elements arr_size = len(arr) if arr_size < 2: print "Invalid Input" return first = second = sys.maxint 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 == sys.maxint): 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

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)

**Related Article:**

Minimum and Second minimum elements using minimum comparisons

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