Given an array arr[] of N elements and an integer K, the task is to generate an B[] with the following rules:
- Copy elements arr[1…N], N times to array B[].
- Copy elements arr[1…N/2], 2*N times to array B[].
- Copy elements arr[1…N/4], 3*N times to array B[].
- 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:
- 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.
- Take cumulative sum of count array.
- Maintain all elements of arr[] with their count in Array B[] along with their counts and sort them based on element value.
- Traverse through vector and see which element has Kth position in B[] as per their individual counts.
- If K is out of bounds of B[] then return -1.
Below is the implementation of the above approach:
C++
// 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 codeint main(){ int arr[] = { 2, 4, 5, 1 }; int N = sizeof(arr) / sizeof(arr[0]); int K = 13; cout << solve(arr, N, K); return 0;} |
Java
// 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 contributed by// sanjeev2552 |
Python3
# 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 |
C#
// C# implementation of the approachusing 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 |
Javascript
<script>// JavaScript implementation of the approach// Function to return the Kth element in B[]function solve(arr, N, K) { // Initialize the count Array let count_Arr = new Array(N + 1).fill(0); let factor = 1; let size = N; // Reduce N repeatedly to half its value while (size) { let start = 1; let 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 = Math.floor(size / 2); } for (let i = 2; i <= N; i++) count_Arr[i] += count_Arr[i - 1]; // Store each element of Array[] with their count let element = []; for (let i = 0; i < N; i++) { element.push([arr[i], count_Arr[i + 1]]); } // Sort the elements wrt value element.sort((a, b) => a - b); let start = 1; for (let i = 0; i < N; i++) { let end = start + element[i][1] - 1; // If Kth element is in range of element[i] // return element[i] if (K >= start && K <= end) { return element[i][0]; } start += element[i][1]; } // If K is out of bound return -1;}// Driver codelet arr = [2, 4, 5, 1];let N = arr.length;let K = 13;document.write(solve(arr, N, K));// This code is contributed by gfgking</script> |
Output:
2
Time Complexity: O(N * log N)
Auxiliary Space: O(N)
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