Find the Kth smallest element in the sorted generated array

Given an array arr[] of N elements and an integer K, the task is to generate an B[] with the following rules:

  1. Copy elements arr[1…N], N times to array B[].
  2. Copy elements arr[1…N/2], 2*N times to array B[].
  3. Copy elements arr[1…N/4], 3*N times to array B[].
  4. Similarly, until only no element is left to be copied to array B[].

Finally print the Kth smallest element from the array B[]. If K is out of bounds of B[] then return -1.

Examples:



Input: arr[] = {1, 2, 3}, K = 4
Output: 1
{1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 1, 1, 1, 1, 1} is the required array B[]
{1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3} in the sorted form where
1 is the 4th smallest element.

Input: arr[] = {2, 4, 5, 1}, K = 13
Output: 2

Approach:

  1. Maintain a Count_Array where we must store the count of times every element occurs in array B[]. It can be done for range of elements by adding the count at start index and subtracting the same count at end index + 1 location.
  2. Take cumulative sum of count array.
  3. Maintain all elements of arr[] with their count in Array B[] along with their counts and sort them based on element value.
  4. Traverse through vector and see which element has Kth position in B[] as per their individual counts.
  5. If K is out of bounds of B[] then return -1.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the Kth element in B[]
int solve(int Array[], int N, int K)
{
  
    // Initialize the count Array
    int count_Arr[N + 1] = { 0 };
    int factor = 1;
    int size = N;
  
    // Reduce N repeatedly to half its value
    while (size) {
        int start = 1;
        int end = size;
  
        // Add count to start
        count_Arr[1] += factor * N;
  
        // Subtract same count after end index
        count_Arr[end + 1] -= factor * N;
        factor++;
        size /= 2;
    }
  
    for (int i = 2; i <= N; i++)
        count_Arr[i] += count_Arr[i - 1];
  
    // Store each element of Array[] with their count
    vector<pair<int, int> > element;
    for (int i = 0; i < N; i++) {
        element.push_back({ Array[i], count_Arr[i + 1] });
    }
  
    // Sort the elements wrt value
    sort(element.begin(), element.end());
  
    int start = 1;
    for (int i = 0; i < N; i++) {
        int end = start + element[i].second - 1;
  
        // If Kth element is in range of element[i]
        // return element[i]
        if (K >= start && K <= end) {
            return element[i].first;
        }
  
        start += element[i].second;
    }
  
    // If K is out of bound
    return -1;
}
  
// Driver code
int main()
{
    int arr[] = { 2, 4, 5, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 13;
  
    cout << solve(arr, N, K);
  
    return 0;
}

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Java

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// Java implementation of the approach 
import java.util.Vector;
  
class GFG 
{
  
    // Pair class implementation to use Pair
    static class Pair
    {
        private int first;
        private int second;
  
        Pair(int first, int second) 
        {
            this.first = first;
            this.second = second;
        }
  
        public int getFirst() 
        {
            return first;
        }
  
        public int getSecond()
        {
            return second;
        }
    }
  
    // Function to return the Kth element in B[]
    static int solve(int[] Array, int N, int K)
    {
  
        // Initialize the count Array
        int[] count_Arr = new int[N + 2];
        int factor = 1;
        int size = N;
  
        // Reduce N repeatedly to half its value
        while (size > 0
        {
            int start = 1;
            int end = size;
  
            // Add count to start
            count_Arr[1] += factor * N;
  
            // Subtract same count after end index
            count_Arr[end + 1] -= factor * N;
            factor++;
            size /= 2;
        }
  
        for (int i = 2; i <= N; i++)
            count_Arr[i] += count_Arr[i - 1];
  
        // Store each element of Array[]
        // with their count
        Vector<Pair> element = new Vector<>();
        for (int i = 0; i < N; i++) 
        {
            Pair x = new Pair(Array[i], 
                              count_Arr[i + 1]);
            element.add(x);
        }
  
        int start = 1;
        for (int i = 0; i < N; i++) 
        {
            int end = start + element.elementAt(0).getSecond() - 1;
  
            // If Kth element is in range of element[i]
            // return element[i]
            if (K >= start && K <= end)
                return element.elementAt(i).getFirst();
  
            start += element.elementAt(i).getSecond();
        }
  
        // If K is out of bound
        return -1;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 2, 4, 5, 1 };
        int N = arr.length;
        int K = 13;
        System.out.println(solve(arr, N, K));
    }
}    
  
// This code is contiributed by
// sanjeev2552

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Python3

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# Python3 implementation of the approach 
  
# Function to return the Kth element in B[] 
def solve(Array, N, K) : 
  
    # Initialize the count Array 
    count_Arr = [0]*(N + 2) ; 
    factor = 1
    size = N; 
  
    # Reduce N repeatedly to half its value 
    while (size) :
        start = 1
        end = size; 
  
        # Add count to start 
        count_Arr[1] += factor * N; 
  
        # Subtract same count after end index 
        count_Arr[end + 1] -= factor * N; 
        factor += 1
        size //= 2
  
    for i in range(2, N + 1) : 
        count_Arr[i] += count_Arr[i - 1]; 
  
    # Store each element of Array[] with their count 
    element = [];
      
    for i in range(N) :
        element.append(( Array[i], count_Arr[i + 1] )); 
  
    # Sort the elements wrt value 
    element.sort(); 
  
    start = 1
    for i in range(N) :
        end = start + element[i][1] - 1
  
        # If Kth element is in range of element[i] 
        # return element[i] 
        if (K >= start and K <= end) :
            return element[i][0]; 
  
        start += element[i][1]; 
  
    # If K is out of bound 
    return -1
  
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 2, 4, 5, 1 ]; 
    N = len(arr); 
    K = 13
  
    print(solve(arr, N, K)); 
  
    # This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach
using System;
using System.Collections.Generic;
      
class GFG 
{
  
    // Pair class implementation to use Pair
    public class Pair
    {
        public int first;
        public int second;
  
        public Pair(int first, int second) 
        {
            this.first = first;
            this.second = second;
        }
  
        public int getFirst() 
        {
            return first;
        }
  
        public int getSecond()
        {
            return second;
        }
    }
  
    // Function to return the Kth element in B[]
    static int solve(int[] Array, int N, int K)
    {
  
        // Initialize the count Array
        int[] count_Arr = new int[N + 2];
        int factor = 1;
        int size = N;
  
        // Reduce N repeatedly to half its value
        while (size > 0) 
        {
            int end = size;
  
            // Add count to start
            count_Arr[1] += factor * N;
  
            // Subtract same count after end index
            count_Arr[end + 1] -= factor * N;
            factor++;
            size /= 2;
        }
  
        for (int i = 2; i <= N; i++)
            count_Arr[i] += count_Arr[i - 1];
  
        // Store each element of Array[]
        // with their count
        List<Pair> element = new List<Pair>();
        for (int i = 0; i < N; i++) 
        {
            Pair x = new Pair(Array[i], 
                              count_Arr[i + 1]);
            element.Add(x);
        }
  
        int start = 1;
        for (int i = 0; i < N; i++) 
        {
            int end = start + element[0].getSecond() - 1;
  
            // If Kth element is in range of element[i]
            // return element[i]
            if (K >= start && K <= end)
                return element[i].getFirst();
  
            start += element[i].getSecond();
        }
  
        // If K is out of bound
        return -1;
    }
  
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 2, 4, 5, 1 };
        int N = arr.Length;
        int K = 13;
        Console.WriteLine(solve(arr, N, K));
    }
}
  
// This code is contributed by Rajput-Ji

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Output:

2


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