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# kth smallest/largest in a small range unsorted array

• Difficulty Level : Easy
• Last Updated : 11 May, 2021

Find kth smallest or largest element in an unsorted array, where k<=size of array. It is given that elements of array are in small range.
Examples:

```Input : arr[] = {3, 2, 9, 5, 7, 11, 13}
k = 5
Output: 9

Input : arr[] = {16, 8, 9, 21, 43}
k = 3
Output: 16

Input : arr[] = {50, 50, 40}
k = 2
Output: 50```

As the given array elements are in small range, we can direct index table to do something similar to counting sort. We store counts of elements, then we traverse the count array and print k-th element.
Following is the implementation of the above algorithm

## C++

 `// C++ program of kth smallest/largest in``// a small range unsorted array``#include ``using` `namespace` `std;``#define maxs 1000001` `int` `kthSmallestLargest(``int``* arr, ``int` `n, ``int` `k)``{``    ``int` `max_val = *max_element(arr, arr+n);``    ``int` `hash[max_val+1] = { 0 };` `    ``// Storing counts of elements``    ``for` `(``int` `i = 0; i < n; i++)``        ``hash[arr[i]]++;   ``    ` `    ``// Traverse hash array build above until``    ``// we reach k-th smallest element.``    ``int` `count = 0;``    ``for` `(``int` `i=0; i <= max_val; i++)``    ``{``        ``while` `(hash[i] > 0)``        ``{``           ``count++;``           ``if` `(count == k)``              ``return` `i;``           ``hash[i]--;``        ``}``    ``}``    ` `    ``return` `-1;``}` `int` `main()``{``    ``int` `arr[] = { 11, 6, 2, 9, 4, 3, 16 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]), k = 3;``    ``cout << ``"kth smallest number is: "``         ``<< kthSmallestLargest(arr, n, k) << endl;``    ``return` `0;``}`

## Java

 `// Java program of kth smallest/largest in``// a small range unsorted array``import` `java.util.Arrays;` `class` `GFG``{` `    ``static` `int` `maxs = ``1000001``;` `    ``static` `int` `kthSmallestLargest(``int``[] arr, ``int` `n, ``int` `k)``    ``{``        ``int` `max_val = Arrays.stream(arr).max().getAsInt();``        ``int` `hash[] = ``new` `int``[max_val + ``1``];` `        ``// Storing counts of elements``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``hash[arr[i]]++;``        ``}` `        ``// Traverse hash array build above until``        ``// we reach k-th smallest element.``        ``int` `count = ``0``;``        ``for` `(``int` `i = ``0``; i <= max_val; i++)``        ``{``            ``while` `(hash[i] > ``0``)``            ``{``                ``count++;``                ``if` `(count == k)``                ``{``                    ``return` `i;``                ``}``                ``hash[i]--;``            ``}``        ``}` `        ``return` `-``1``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = {``11``, ``6``, ``2``, ``9``, ``4``, ``3``, ``16``};``        ``int` `n = arr.length, k = ``3``;``        ``System.out.println(``"kth smallest number is: "``                ``+ kthSmallestLargest(arr, n, k));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python 3 program of kth smallest/largest``# in a small range unsorted array` `def` `kthSmallestLargest(arr, n, k):``    ``max_val ``=` `arr[``0``]``    ``for` `i ``in` `range``(``len``(arr)):``        ``if` `(arr[i] > max_val):``            ``max_val ``=` `arr[i]``    ``hash` `=` `[``0` `for` `i ``in` `range``(max_val ``+` `1``)]` `    ``# Storing counts of elements``    ``for` `i ``in` `range``(n):``        ``hash``[arr[i]] ``+``=` `1``    ` `    ``# Traverse hash array build above until``    ``# we reach k-th smallest element.``    ``count ``=` `0``    ``for` `i ``in` `range``(max_val ``+` `1``):``        ``while` `(``hash``[i] > ``0``):``            ``count ``+``=` `1``            ``if` `(count ``=``=` `k):``                ``return` `i``            ``hash``[i] ``-``=` `1``        ` `    ``return` `-``1` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``11``, ``6``, ``2``, ``9``, ``4``, ``3``, ``16``]``    ``n ``=` `len``(arr)``    ``k ``=` `3``    ``print``(``"kth smallest number is:"``,``           ``kthSmallestLargest(arr, n, k))``    ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program of kth smallest/largest in``// a small range unsorted array``using` `System;``using` `System.Linq;` `class` `GFG``{` `    ``static` `int` `maxs = 1000001;` `    ``static` `int` `kthSmallestLargest(``int``[] arr, ``int` `n, ``int` `k)``    ``{``        ``int` `max_val = arr.Max();``        ``int` `[]hash = ``new` `int``[max_val + 1];` `        ``// Storing counts of elements``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``hash[arr[i]]++;``        ``}` `        ``// Traverse hash array build above until``        ``// we reach k-th smallest element.``        ``int` `count = 0;``        ``for` `(``int` `i = 0; i <= max_val; i++)``        ``{``            ``while` `(hash[i] > 0)``            ``{``                ``count++;``                ``if` `(count == k)``                ``{``                    ``return` `i;``                ``}``                ``hash[i]--;``            ``}``        ``}``        ``return` `-1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = {11, 6, 2, 9, 4, 3, 16};``        ``int` `n = arr.Length, k = 3;``        ``Console.WriteLine(``"kth smallest number is: "``                ``+ kthSmallestLargest(arr, n, k));``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## PHP

 ` 0)``        ``{``           ``\$count``++;``           ``if` `(``\$count` `== ``\$k``)``              ``return` `\$i``;``           ``\$hash``[``\$i``]--;``        ``}``    ``}``     ` `    ``return` `-1;``}`` `  `    ``\$arr` `= ``array` `( 11, 6, 2, 9, 4, 3, 16 );``    ``\$n` `= sizeof(``\$arr``) / sizeof(``\$arr``[0]);``    ``\$k` `= 3;``    ``echo` `"kth smallest number is: "``         ``. kthSmallestLargest(``\$arr``, ``\$n``, ``\$k``).``"\n"``;``    ``return` `0;``?>`

## Javascript

 ``
Output:
`kth smallest number is: 4`

Time Complexity: O(n + max_val)

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