# K’th Smallest/Largest Element in Unsorted Array | Set 2 (Expected Linear Time)

We recommend reading the following post as a prerequisite of this post.

Given an array and a number k where k is smaller than the size of the array, we need to find the k’th smallest element in the given array. It is given that all array elements are distinct.

Examples:

```Input: arr[] = {7, 10, 4, 3, 20, 15}
k = 3
Output: 7

Input: arr[] = {7, 10, 4, 3, 20, 15}
k = 4
Output: 10```

We have discussed three different solutions here.

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

In this post method 4 is discussed which is mainly an extension of method 3 (QuickSelect) discussed in the previous post. The idea is to randomly pick a pivot element. To implement randomized partition, we use a random function, rand() to generate index between l and r, swap the element at randomly generated index with the last element, and finally call the standard partition process which uses last element as pivot.

Following is an implementation of the above Randomized QuickSelect.

## C/C++

 `// C++ implementation of randomized quickSelect ` `#include ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `int` `randomPartition(``int` `arr[], ``int` `l, ``int` `r); ` ` `  `// This function returns k'th smallest element in arr[l..r] using ` `// QuickSort based method. ASSUMPTION: ELEMENTS IN ARR[] ARE DISTINCT ` `int` `kthSmallest(``int` `arr[], ``int` `l, ``int` `r, ``int` `k) ` `{ ` `    ``// If k is smaller than number of elements in array ` `    ``if` `(k > 0 && k <= r - l + 1) ` `    ``{ ` `        ``// Partition the array around a random element and ` `        ``// get position of pivot element in sorted array ` `        ``int` `pos = randomPartition(arr, l, r); ` ` `  `        ``// If position is same as k ` `        ``if` `(pos-l == k-1) ` `            ``return` `arr[pos]; ` `        ``if` `(pos-l > k-1)  ``// If position is more, recur for left subarray ` `            ``return` `kthSmallest(arr, l, pos-1, k); ` ` `  `        ``// Else recur for right subarray ` `        ``return` `kthSmallest(arr, pos+1, r, k-pos+l-1); ` `    ``} ` ` `  `    ``// If k is more than the number of elements in the array ` `    ``return` `INT_MAX; ` `} ` ` `  `void` `swap(``int` `*a, ``int` `*b) ` `{ ` `    ``int` `temp = *a; ` `    ``*a = *b; ` `    ``*b = temp; ` `} ` ` `  `// Standard partition process of QuickSort().  It considers the last ` `// element as pivot and moves all smaller element to left of it and ` `// greater elements to right. This function is used by randomPartition() ` `int` `partition(``int` `arr[], ``int` `l, ``int` `r) ` `{ ` `    ``int` `x = arr[r], i = l; ` `    ``for` `(``int` `j = l; j <= r - 1; j++) ` `    ``{ ` `        ``if` `(arr[j] <= x) ` `        ``{ ` `            ``swap(&arr[i], &arr[j]); ` `            ``i++; ` `        ``} ` `    ``} ` `    ``swap(&arr[i], &arr[r]); ` `    ``return` `i; ` `} ` ` `  `// Picks a random pivot element between l and r and partitions ` `// arr[l..r] around the randomly picked element using partition() ` `int` `randomPartition(``int` `arr[], ``int` `l, ``int` `r) ` `{ ` `    ``int` `n = r-l+1; ` `    ``int` `pivot = ``rand``() % n; ` `    ``swap(&arr[l + pivot], &arr[r]); ` `    ``return` `partition(arr, l, r); ` `} ` ` `  `// Driver program to test above methods ` `int` `main() ` `{ ` `    ``int` `arr[] = {12, 3, 5, 7, 4, 19, 26}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr), k = 3; ` `    ``cout << ``"K'th smallest element is "` `<< kthSmallest(arr, 0, n-1, k); ` `    ``return` `0; ` `}`

## Java

 `// Java program to find k'th smallest element in expected ` `// linear time ` `class` `KthSmallst ` `{ ` `    ``// This function returns k'th smallest element in arr[l..r] ` `    ``// using QuickSort based method.  ASSUMPTION: ALL ELEMENTS ` `    ``// IN ARR[] ARE DISTINCT ` `    ``int` `kthSmallest(``int` `arr[], ``int` `l, ``int` `r, ``int` `k) ` `    ``{ ` `        ``// If k is smaller than number of elements in array ` `        ``if` `(k > ``0` `&& k <= r - l + ``1``) ` `        ``{ ` `            ``// Partition the array around a random element and ` `            ``// get position of pivot element in sorted array ` `            ``int` `pos = randomPartition(arr, l, r); ` ` `  `            ``// If position is same as k ` `            ``if` `(pos-l == k-``1``) ` `                ``return` `arr[pos]; ` ` `  `            ``// If position is more, recur for left subarray ` `            ``if` `(pos-l > k-``1``) ` `                ``return` `kthSmallest(arr, l, pos-``1``, k); ` ` `  `            ``// Else recur for right subarray ` `            ``return` `kthSmallest(arr, pos+``1``, r, k-pos+l-``1``); ` `        ``} ` ` `  `        ``// If k is more than number of elements in array ` `        ``return` `Integer.MAX_VALUE; ` `    ``} ` ` `  `    ``// Utility method to swap arr[i] and arr[j] ` `    ``void` `swap(``int` `arr[], ``int` `i, ``int` `j) ` `    ``{ ` `        ``int` `temp = arr[i]; ` `        ``arr[i] = arr[j]; ` `        ``arr[j] = temp; ` `    ``} ` ` `  `    ``// Standard partition process of QuickSort().  It considers ` `    ``// the last element as pivot and moves all smaller element  ` `    ``// to left of it and greater elements to right. This function ` `    ``// is used by randomPartition() ` `    ``int` `partition(``int` `arr[], ``int` `l, ``int` `r) ` `    ``{ ` `        ``int` `x = arr[r], i = l; ` `        ``for` `(``int` `j = l; j <= r - ``1``; j++) ` `        ``{ ` `            ``if` `(arr[j] <= x) ` `            ``{ ` `                ``swap(arr, i, j); ` `                ``i++; ` `            ``} ` `        ``} ` `        ``swap(arr, i, r); ` `        ``return` `i; ` `    ``} ` ` `  `    ``// Picks a random pivot element between l and r and  ` `    ``// partitions arr[l..r] arount the randomly picked  ` `    ``// element using partition() ` `    ``int` `randomPartition(``int` `arr[], ``int` `l, ``int` `r) ` `    ``{ ` `        ``int` `n = r-l+``1``; ` `        ``int` `pivot = (``int``)(Math.random()) * (n-``1``); ` `        ``swap(arr, l + pivot, r); ` `        ``return` `partition(arr, l, r); ` `    ``} ` ` `  `    ``// Driver method to test above ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``KthSmallst ob = ``new` `KthSmallst(); ` `        ``int` `arr[] = {``12``, ``3``, ``5``, ``7``, ``4``, ``19``, ``26``}; ` `        ``int` `n = arr.length,k = ``3``; ` `        ``System.out.println(``"K'th smallest element is "``+ ` `                           ``ob.kthSmallest(arr, ``0``, n-``1``, k)); ` `    ``} ` `} ` `/*This code is contributed by Rajat Mishra*/`

## Python3

 `# Python3 implementation of randomized  ` `# quickSelect  ` `import` `random ` ` `  `# This function returns k'th smallest  ` `# element in arr[l..r] using QuickSort ` `# based method. ASSUMPTION: ELEMENTS ` `# IN ARR[] ARE DISTINCT  ` `def` `kthSmallest(arr, l, r, k): ` `     `  `    ``# If k is smaller than number of ` `    ``# elements in array  ` `    ``if` `(k > ``0` `and` `k <``=` `r ``-` `l ``+` `1``): ` `         `  `        ``# Partition the array around a random  ` `        ``# element and get position of pivot  ` `        ``# element in sorted array  ` `        ``pos ``=` `randomPartition(arr, l, r)  ` ` `  `        ``# If position is same as k  ` `        ``if` `(pos ``-` `l ``=``=` `k ``-` `1``):  ` `            ``return` `arr[pos]  ` `        ``if` `(pos ``-` `l > k ``-` `1``): ``# If position is more,  ` `                            ``# recur for left subarray  ` `            ``return` `kthSmallest(arr, l, pos ``-` `1``, k)  ` ` `  `        ``# Else recur for right subarray  ` `        ``return` `kthSmallest(arr, pos ``+` `1``, r,  ` `                           ``k ``-` `pos ``+` `l ``-` `1``) ` ` `  `    ``# If k is more than the number of  ` `    ``# elements in the array  ` `    ``return` `999999999999` ` `  `def` `swap(arr, a, b): ` `    ``temp ``=` `arr[a] ` `    ``arr[a] ``=` `arr[b] ` `    ``arr[b] ``=` `temp ` ` `  `# Standard partition process of QuickSort().  ` `# It considers the last element as pivot and ` `# moves all smaller element to left of it and  ` `# greater elements to right. This function ` `# is used by randomPartition()  ` `def` `partition(arr, l, r): ` `    ``x ``=` `arr[r] ` `    ``i ``=` `l ` `    ``for` `j ``in` `range``(l, r): ` `        ``if` `(arr[j] <``=` `x): ` `            ``swap(arr, i, j)  ` `            ``i ``+``=` `1` `    ``swap(arr, i, r)  ` `    ``return` `i ` ` `  `# Picks a random pivot element between l and r  ` `# and partitions arr[l..r] around the randomly ` `# picked element using partition()  ` `def` `randomPartition(arr, l, r): ` `    ``n ``=` `r ``-` `l ``+` `1` `    ``pivot ``=` `int``(random.random() ``%` `n)  ` `    ``swap(arr, l ``+` `pivot, r)  ` `    ``return` `partition(arr, l, r) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``arr ``=` `[``12``, ``3``, ``5``, ``7``, ``4``, ``19``, ``26``]  ` `    ``n ``=` `len``(arr) ` `    ``k ``=` `3` `    ``print``(``"K'th smallest element is"``,  ` `           ``kthSmallest(arr, ``0``, n ``-` `1``, k)) ` ` `  `# This code is contributed by PranchalK `

## C#

 `// C# program to find k'th smallest  ` `// element in expected linear time  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `// This function returns k'th smallest  ` `// element in arr[l..r] using QuickSort  ` `// based method. ASSUMPTION: ALL ELEMENTS  ` `// IN ARR[] ARE DISTINCT  ` `int` `kthSmallest(``int` `[]arr, ``int` `l, ``int` `r, ``int` `k)  ` `{  ` `    ``// If k is smaller than number  ` `    ``// of elements in array  ` `    ``if` `(k > 0 && k <= r - l + 1)  ` `    ``{  ` `        ``// Partition the array around a  ` `        ``// random element and get position  ` `        ``// of pivot element in sorted array  ` `        ``int` `pos = randomPartition(arr, l, r);  ` ` `  `        ``// If position is same as k  ` `        ``if` `(pos-l == k - 1)  ` `            ``return` `arr[pos];  ` ` `  `        ``// If position is more, recur  ` `        ``// for left subarray  ` `        ``if` `(pos - l > k - 1)  ` `            ``return` `kthSmallest(arr, l, pos - 1, k);  ` ` `  `        ``// Else recur for right subarray  ` `        ``return` `kthSmallest(arr, pos + 1, r,  ` `                           ``k - pos + l - 1);  ` `    ``}  ` ` `  `    ``// If k is more than number of  ` `    ``// elements in array  ` `    ``return` `int``.MaxValue;  ` `}  ` ` `  `// Utility method to swap arr[i] and arr[j]  ` `void` `swap(``int` `[]arr, ``int` `i, ``int` `j)  ` `{  ` `    ``int` `temp = arr[i];  ` `    ``arr[i] = arr[j];  ` `    ``arr[j] = temp;  ` `}  ` ` `  `// Standard partition process of QuickSort().  ` `// It considers the last element as pivot and  ` `// moves all smaller element to left of it  ` `// and greater elements to right. This function  ` `// is used by randomPartition()  ` `int` `partition(``int` `[]arr, ``int` `l, ``int` `r)  ` `{  ` `    ``int` `x = arr[r], i = l;  ` `    ``for` `(``int` `j = l; j <= r - 1; j++)  ` `    ``{  ` `        ``if` `(arr[j] <= x)  ` `        ``{  ` `            ``swap(arr, i, j);  ` `            ``i++;  ` `        ``}  ` `    ``}  ` `    ``swap(arr, i, r);  ` `    ``return` `i;  ` `}  ` ` `  `// Picks a random pivot element between  ` `// l and r and partitions arr[l..r]  ` `// around the randomly picked element ` `// using partition()  ` `int` `randomPartition(``int` `[]arr, ``int` `l, ``int` `r)  ` `{  ` `    ``int` `n = r - l + 1;  ` `    ``Random rnd = ``new` `Random(); ` `    ``int` `rand = rnd.Next(0, 1); ` `    ``int` `pivot = (``int``)(rand * (n - 1));  ` `    ``swap(arr, l + pivot, r);  ` `    ``return` `partition(arr, l, r);  ` `}  ` ` `  `// Driver Code ` `public` `static` `void` `Main()  ` `{  ` `    ``GFG ob = ``new` `GFG();  ` `    ``int` `[]arr = {12, 3, 5, 7, 4, 19, 26};  ` `    ``int` `n = arr.Length,k = 3;  ` `    ``Console.Write(``"K'th smallest element is "``+  ` `            ``ob.kthSmallest(arr, 0, n - 1, k));  ` `}  ` `}  ` ` `  `// This code is contributed by 29AjayKumar `

## PHP

 ` 0 && ``\$k` `<= ``\$r` `- ``\$l` `+ 1)  ` `    ``{  ` `        ``// Partition the array around a random element and  ` `        ``// get position of pivot element in sorted array  ` `        ``\$pos` `= randomPartition(``\$arr``, ``\$l``, ``\$r``);  ` ` `  `        ``// If position is same as k  ` `        ``if` `(``\$pos``-``\$l` `== ``\$k``-1)  ` `            ``return` `\$arr``[``\$pos``];  ` `             `  `        ``// If position is more, recur for left subarray  ` `        ``if` `(``\$pos``-``\$l` `> ``\$k``-1) ` `            ``return` `kthSmallest(``\$arr``, ``\$l``, ``\$pos``-1, ``\$k``);  ` ` `  `        ``// Else recur for right subarray  ` `        ``return` `kthSmallest(``\$arr``, ``\$pos``+1, ``\$r``, ` `                            ``\$k``-``\$pos``+``\$l``-1);  ` `    ``}  ` ` `  `    ``// If k is more than the number of elements in the array  ` `    ``return` `PHP_INT_MAX;  ` `}  ` ` `  `function` `swap(``\$a``, ``\$b``)  ` `{  ` `    ``\$temp` `= ``\$a``;  ` `    ``\$a` `= ``\$b``;  ` `    ``\$b` `= ``\$temp``;  ` `}  ` ` `  `// Standard partition process of QuickSort().  ` `// It considers the last element as pivot ` `// and moves all smaller element to left   ` `// of it and greater elements to right. ` `// This function is used by randomPartition()  ` `function` `partition(``\$arr``, ``\$l``, ``\$r``)  ` `{  ` `    ``\$x` `= ``\$arr``[``\$r``]; ` `    ``\$i` `= ``\$l``;  ` `    ``for` `(``\$j` `= ``\$l``; ``\$j` `<= ``\$r` `- 1; ``\$j``++)  ` `    ``{  ` `        ``if` `(``\$arr``[``\$j``] <= ``\$x``)  ` `        ``{  ` `            ``list(``\$arr``[``\$i``], ``\$arr``[``\$j``])=``array``(``\$arr``[``\$j``],``\$arr``[``\$i``]); ` `            ``//swap(&arr[i], &arr[j]);  ` `            ``\$i``++;  ` `        ``}  ` `    ``}  ` `    ``list(``\$arr``[``\$i``], ``\$arr``[``\$r``])=``array``(``\$arr``[``\$r``],``\$arr``[``\$i``]); ` `    ``//swap(&arr[i], &arr[r]);  ` `    ``return` `\$i``;  ` `}  ` ` `  `// Picks a random pivot element between ` `//  l and r and partitions arr[l..r] around  ` `// the randomly picked element using partition()  ` `function` `randomPartition(``\$arr``, ``\$l``, ``\$r``)  ` `{  ` `    ``\$n` `= ``\$r``-``\$l``+1;  ` `    ``\$pivot` `= rand() % ``\$n``;  ` `     `  `    ``list(``\$arr``[``\$l` `+ ``\$pivot``], ``\$arr``[``\$r``]) =  ` `            ``array``(``\$arr``[``\$r``],``\$arr``[``\$l` `+ ``\$pivot``] ); ` `     `  `    ``//swap(&arr[l + pivot], &arr[r]);  ` `    ``return` `partition(``\$arr``, ``\$l``, ``\$r``);  ` `}  ` ` `  `// Driver program to test the above methods  ` `    ``\$arr` `= ``array``(12, 3, 5, 7, 4, 19, 260);  ` `    ``\$n` `= sizeof(``\$arr``)/sizeof(``\$arr``); ` `    ``\$k` `= 3;  ` `    ``echo` `"K'th smallest element is "` `, ` `            ``kthSmallest(``\$arr``, 0, ``\$n``-1, ``\$k``);  ` `     `  ` `  `// This code is contributed by ajit. ` `?> `

Output:

`K'th smallest element is 5 `

Time Complexity:
The worst case time complexity of the above solution is still O(n2). In worst case, the randomized function may always pick a corner element. The expected time complexity of above randomized QuickSelect is O(n), see CLRS book or MIT video lecture for proof. The assumption in the analysis is, random number generator is equally likely to generate any number in the input range.