# QuickSelect (A Simple Iterative Implementation)

Quickselect is a selection algorithm to find the k-th smallest element in an unordered list. It is related to the quick sort sorting algorithm.

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The QuickSelect algorithm is based QuickSort. The difference is, instead of recurring for both sides (after finding pivot), it recurs only for the part that contains the k-th smallest element. The logic is simple, if index of partitioned element is more than k, then we recur for left part. If index is same as k, we have found the k-th smallest element and we return. If index is less than k, then we recur for right part. This reduces the expected complexity from O(n log n) to O(n), with a worst case of O(n^2).

```function quickSelect(list, left, right, k)

if left = right
return list[left]

Select a pivotIndex between left and right

pivotIndex := partition(list, left, right,
pivotIndex)
if k = pivotIndex
return list[k]
else if k < pivotIndex
right := pivotIndex - 1
else
left := pivotIndex + 1
```

We have discussed a recursive implementation of QuickSelect. In this post, we are going to discuss simple iterative implementation.

## C++

 `// CPP program for iterative implementation of QuickSelect ` `#include ` `using` `namespace` `std; ` ` `  `// Standard Lomuto partition function ` `int` `partition(``int` `arr[], ``int` `low, ``int` `high) ` `{ ` `    ``int` `pivot = arr[high]; ` `    ``int` `i = (low - 1); ` `    ``for` `(``int` `j = low; j <= high - 1; j++) { ` `        ``if` `(arr[j] <= pivot) { ` `            ``i++; ` `            ``swap(arr[i], arr[j]); ` `        ``} ` `    ``} ` `    ``swap(arr[i + 1], arr[high]); ` `    ``return` `(i + 1); ` `} ` ` `  `// Implementation of QuickSelect ` `int` `kthSmallest(``int` `a[], ``int` `left, ``int` `right, ``int` `k) ` `{ ` ` `  `    ``while` `(left <= right) { ` ` `  `        ``// Partition a[left..right] around a pivot ` `        ``// and find the position of the pivot ` `        ``int` `pivotIndex = partition(a, left, right); ` ` `  `        ``// If pivot itself is the k-th smallest element ` `        ``if` `(pivotIndex == k - 1) ` `            ``return` `a[pivotIndex]; ` ` `  `        ``// If there are more than k-1 elements on ` `        ``// left of pivot, then k-th smallest must be ` `        ``// on left side. ` `        ``else` `if` `(pivotIndex > k - 1) ` `            ``right = pivotIndex - 1; ` ` `  `        ``// Else k-th smallest is on right side. ` `        ``else` `            ``left = pivotIndex + 1; ` `    ``} ` `    ``return` `-1; ` `} ` ` `  `// Driver program to test above methods ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 4, 5, 8, 11, 6, 26 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `k = 5; ` `    ``cout << ``"K-th smallest element is "` `         ``<< kthSmallest(arr, 0, n - 1, k); ` `    ``return` `0; ` `} `

## Java

 `// Java program for iterative implementation ` `// of QuickSelect ` `class` `GFG  ` `{ ` `     `  `    ``// Standard Lomuto partition function  ` `    ``static` `int` `partition(``int` `arr[],  ` `                         ``int` `low, ``int` `high)  ` `    ``{  ` `        ``int` `temp; ` `        ``int` `pivot = arr[high];  ` `        ``int` `i = (low - ``1``);  ` `        ``for` `(``int` `j = low; j <= high - ``1``; j++)  ` `        ``{  ` `            ``if` `(arr[j] <= pivot)  ` `            ``{  ` `                ``i++;  ` `                ``temp = arr[i]; ` `                ``arr[i] = arr[j]; ` `                ``arr[j] = temp; ` `            ``}  ` `        ``}  ` `         `  `            ``temp = arr[i + ``1``]; ` `            ``arr[i + ``1``] = arr[high]; ` `            ``arr[high] = temp; ` `             `  `        ``return` `(i + ``1``);  ` `    ``}  ` `     `  `    ``// Implementation of QuickSelect  ` `    ``static` `int` `kthSmallest(``int` `a[], ``int` `left, ` `                           ``int` `right, ``int` `k)  ` `    ``{  ` `        ``while` `(left <= right)  ` `        ``{  ` `     `  `            ``// Partition a[left..right] around a pivot  ` `            ``// and find the position of the pivot  ` `            ``int` `pivotIndex = partition(a, left, right);  ` `     `  `            ``// If pivot itself is the k-th smallest element  ` `            ``if` `(pivotIndex == k - ``1``)  ` `                ``return` `a[pivotIndex];  ` `     `  `            ``// If there are more than k-1 elements on  ` `            ``// left of pivot, then k-th smallest must be  ` `            ``// on left side.  ` `            ``else` `if` `(pivotIndex > k - ``1``)  ` `                ``right = pivotIndex - ``1``;  ` `     `  `            ``// Else k-th smallest is on right side.  ` `            ``else` `                ``left = pivotIndex + ``1``;  ` `        ``}  ` `        ``return` `-``1``;  ` `    ``}  ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{  ` `        ``int` `arr[] = { ``10``, ``4``, ``5``, ``8``, ``11``, ``6``, ``26` `};  ` `        ``int` `n = arr.length;  ` `        ``int` `k = ``5``;  ` `        ``System.out.println(``"K-th smallest element is "` `+  ` `                         ``kthSmallest(arr, ``0``, n - ``1``, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 program for iterative implementation ` `# of QuickSelect  ` ` `  `# Standard Lomuto partition function  ` `def` `partition(arr, low, high) :  ` ` `  `    ``pivot ``=` `arr[high]  ` `    ``i ``=` `(low ``-` `1``)  ` `    ``for` `j ``in` `range``(low, high) :  ` `        ``if` `arr[j] <``=` `pivot :  ` `            ``i ``+``=` `1` `            ``arr[i], arr[j] ``=` `arr[j], arr[i]  ` `         `  `    ``arr[i ``+` `1``], arr[high] ``=` `arr[high], arr[i ``+` `1``]  ` `    ``return` `(i ``+` `1``)  ` ` `  `# Implementation of QuickSelect  ` `def` `kthSmallest(a, left, right, k) :  ` ` `  `    ``while` `left <``=` `right :  ` ` `  `        ``# Partition a[left..right] around a pivot  ` `        ``# and find the position of the pivot  ` `        ``pivotIndex ``=` `partition(a, left, right)  ` ` `  `        ``# If pivot itself is the k-th smallest element  ` `        ``if` `pivotIndex ``=``=` `k ``-` `1` `:  ` `            ``return` `a[pivotIndex]  ` ` `  `        ``# If there are more than k-1 elements on  ` `        ``# left of pivot, then k-th smallest must be  ` `        ``# on left side.  ` `        ``elif` `pivotIndex > k ``-` `1` `: ` `            ``right ``=` `pivotIndex ``-` `1` ` `  `        ``# Else k-th smallest is on right side.  ` `        ``else` `: ` `            ``left ``=` `pivotIndex ``+` `1` `     `  `    ``return` `-``1` ` `  `# Driver Code ` `arr ``=` `[ ``10``, ``4``, ``5``, ``8``, ``11``, ``6``, ``26` `]  ` `n ``=` `len``(arr)  ` `k ``=` `5` `print``(``"K-th smallest element is"``,  ` `       ``kthSmallest(arr, ``0``, n ``-` `1``, k))  ` ` `  `# This code is contributed by ` `# divyamohan123 `

## C#

 `// C# program for iterative implementation ` `// of QuickSelect ` `using` `System; ` `                     `  `class` `GFG  ` `{ ` `     `  `    ``// Standard Lomuto partition function  ` `    ``static` `int` `partition(``int` `[]arr,  ` `                         ``int` `low, ``int` `high)  ` `    ``{  ` `        ``int` `temp; ` `        ``int` `pivot = arr[high];  ` `        ``int` `i = (low - 1);  ` `        ``for` `(``int` `j = low; j <= high - 1; j++)  ` `        ``{  ` `            ``if` `(arr[j] <= pivot)  ` `            ``{  ` `                ``i++;  ` `                ``temp = arr[i]; ` `                ``arr[i] = arr[j]; ` `                ``arr[j] = temp; ` `            ``}  ` `        ``}  ` `         `  `        ``temp = arr[i + 1]; ` `        ``arr[i + 1] = arr[high]; ` `        ``arr[high] = temp; ` `             `  `        ``return` `(i + 1);  ` `    ``}  ` `     `  `    ``// Implementation of QuickSelect  ` `    ``static` `int` `kthSmallest(``int` `[]a, ``int` `left, ` `                           ``int` `right, ``int` `k)  ` `    ``{  ` `        ``while` `(left <= right)  ` `        ``{  ` `     `  `            ``// Partition a[left..right] around a pivot  ` `            ``// and find the position of the pivot  ` `            ``int` `pivotIndex = partition(a, left, right);  ` `     `  `            ``// If pivot itself is the k-th smallest element  ` `            ``if` `(pivotIndex == k - 1)  ` `                ``return` `a[pivotIndex];  ` `     `  `            ``// If there are more than k-1 elements on  ` `            ``// left of pivot, then k-th smallest must be  ` `            ``// on left side.  ` `            ``else` `if` `(pivotIndex > k - 1)  ` `                ``right = pivotIndex - 1;  ` `     `  `            ``// Else k-th smallest is on right side.  ` `            ``else` `                ``left = pivotIndex + 1;  ` `        ``}  ` `        ``return` `-1;  ` `    ``}  ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main (String[] args)  ` `    ``{  ` `        ``int` `[]arr = { 10, 4, 5, 8, 11, 6, 26 };  ` `        ``int` `n = arr.Length;  ` `        ``int` `k = 5;  ` `        ``Console.WriteLine(``"K-th smallest element is "` `+  ` `                        ``kthSmallest(arr, 0, n - 1, k));  ` `    ``}  ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```K-th smallest element is 10
```

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