The smallest and second smallest elements

Software Development
#1

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++ 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;

}

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)