# Find the smallest and second smallest elements in an array

Last Updated : 11 Mar, 2024

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.Â

Below is the implementation of the above approach:

## C++

 `// C++ simple approach to print smallest` `// and second smallest element.` `#include ` `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],``"
"``);` `document.write(``"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`

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.Â

## C++

 `// C++ program to find smallest and` `// second smallest element in array` `#include ` `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:Â

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.Â
• 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 ` `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 /* For INT_MAX */` `#include `   `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

 ``

## 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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