Ishaan’s position for internship selection

Problem Statement :-

• Ishaan wants to intern at GeeksForGeeks but for that, he has to go through a test. There are n candidates applying for the internship including Ishaan and only one is to be selected. Since he wants to qualify he asks you to help him. The test is as follows.
  The candidates are asked to stand in a line at positions 1 to n and given a number k. Now, every kth candidate remains and the rest are eliminated. This is repeated until the number of candidates is less than k.
  Out of the remaining candidates, the one standing at the smallest position is selected. You need to tell Ishaan at the position he must stand to get selected.

Examples:

Input: n = 30, k = 3
Output: 27
Explanation: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
3 6 9 12 15 18 21 24 27 30
9 18 27
27

Input: n = 18, k = 3
Output: 9
Explanation:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 183 6 9 12 15 189 18 (less than k)9

Approach: To solve the problem follow the below idea:

Initialize i with 1 and iterate it i*k times and when i*k becomes greater than n means you have elements less than k, immediately break the loop and return i.

Below is the implementation of the above approach:

C++14

// C++ code for the above approach:
#include <iostream>
using namespace std;

int getCandidate(int n, int k)
{
    int i = 1;
    for (i = 1; i< n; i = i * k) {
        if (i * k> n) {
            break;
        }
    }
    return i;
}

// Driver code
int main()
{

    int n = 30, k = 3;

    // Function call
    cout<<"Ishan must stand at position number -";
    cout<< getCandidate(n, k)<< endl;
}

C

#include <stdio.h>

int getCandidate(int n, int k) {
    int i;
    for (i = 1; i < n; i = i * k) {
        if (i * k > n) {
            break;
        }
    }
    return i;
}

int main() {
    int n = 30, k = 3;
    printf("Ishan must stand at position number - %d\n", getCandidate(n, k));
    return 0;
}

Java

import java.util.*;

public class Main {
    public static int getCandidate(int n, int k) {
        int i = 1;
        for (i = 1; i < n; i = i * k) {
            if (i * k > n) {
                break;
            }
        }
        return i;
    }

    public static void main(String[] args) {
        int n = 30, k = 3;

        System.out.print("Ishan must stand at position number - ");
        System.out.println(getCandidate(n, k));
    }
}

Python3

def getCandidate(n, k):
    i = 1
    while i < n:
        if i * k > n:
            break
        i = i * k
    return i

n = 30
k = 3
print("Ishan must stand at position number - ")
print(getCandidate(n, k))

C#

// C# Implementation:
using System;

public class GFG {
    // Function to get candidate
    public static int GetCandidate(int n, int k)
    {
        int i = 1;
        for (i = 1; i < n; i = i * k) {
            if (i * k > n) {
                break;
            }
        }
        return i;
    }

    // driver code
    public static void Main(string[] args)
    {
        int n = 30, k = 3;

        Console.Write(
            "Ishan must stand at position number - ");

        // function call
        Console.WriteLine(GetCandidate(n, k));
    }
}
// This code is contributed by Sakshi

Javascript

class Main
{
    static getCandidate(n, k)
    {
        var i = 1;
        for (i = 1; i < this.n; i = i * this.k)
        {
            if (i * this.k > this.n)
            {
                break;
            }
        }
        return i;
    }
    static main(args)
    {
        var n = 30;
        var k = 3;
        console.log("Ishan must stand at position number - ");
        console.log(Main.getCandidate(n, k));
    }
}
Main.main([]);

Output

Ishan must stand at position number -27

Time complexity: O(log n)
Auxiliary Space: O(1)