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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Find More Info here to that Topic: geeksforgeeks.org/find-the-kth-smallest-element-in-the-sorted-generated-array/ […]
… [Trackback]
[…] Find More on on that Topic: geeksforgeeks.org/find-the-kth-smallest-element-in-the-sorted-generated-array/ […]
… [Trackback]
[…] There you can find 96292 additional Info to that Topic: geeksforgeeks.org/find-the-kth-smallest-element-in-the-sorted-generated-array/ […]
… [Trackback]
[…] Read More on on that Topic: geeksforgeeks.org/find-the-kth-smallest-element-in-the-sorted-generated-array/ […]
… [Trackback]
[…] Read More on on that Topic: geeksforgeeks.org/find-the-kth-smallest-element-in-the-sorted-generated-array/ […]