# 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[0]);     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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