Find Kth smallest number among multiples of all elements of array

Last Updated : 18 Apr, 2024

Consider an array arr[] of size N consisting of prime numbers and an integer K. The task is to find the Kth smallest number among the multiples of all elements of the array.

Constraints:

1 <= N <= 10
1 < arr[i] < 30
1<=K<=1e12

Examples:

Input: N = 3, arr[] = {3, 5, 7}, K=5
Output: 9
Explanation:

arr[0] has multiples 3, 6, 9, 12, 15… so on.
arr[1] has multiples 5, 10, 15, 20, 25… so on.
arr[2] has multiples 7, 14, 21, 28, 35… so on.
If we arrange all the multiples in ascending order, then we’ll get: 3, 5, 6, 7, 9, 10, 12… so on. Among all these multiples, the 5th smallest multiple is 9.

Input: N = 2, arr[] = {7, 11}, K = 3
Output: 14
Explanation:

arr[0] has mutiples 7, 14, 21, 28… so on.
arr[1] has multiples 11, 22, 33, 44… so on.
If we arrange all the multiples in ascending order, then we’ll get 7, 11, 14, 21, 22, 28… so on. Among all these multiples, the 3rd smallest multiple is 14.

Naive Approach:

Initially we will be having a count of numbers (cnt) initialized to zero, we will be start iterating from 2 and for each integer we will check if it is divisible by any of the numbers in the array, if it is then the count(cnt) will be incremented by 1. When the value of count is equal to K then we will break the loop and returning the answer.

C++

#include <bits/stdc++.h>
using namespace std;

long long int solve(vector<int>& arr, long long int& K)
{
    // initialise the count (cnt) = 0
    long long int cnt = 0;
    long long int i = 2;
    while (cnt < K) {
        for (int j = 0; j < arr.size(); j++) {
            if (i % arr[j] == 0) {
                cnt++;
                if (cnt == K) {
                    return i;
                    break;
                }
            }
        }
        i++;
    }
    return 0;
}

int main()
{
    long long int K = 5;
    vector<int> arr = { 3, 5, 7 };
    cout << solve(arr, K) << endl;
    return 0;
}

Java

import java.util.ArrayList;

public class Main {

    static long solve(ArrayList<Integer> arr, long K)
    {
        // initialize the count (cnt) = 0
        long cnt = 0;
        long i = 2;

        while (cnt < K) {
            for (int j = 0; j < arr.size(); j++) {
                if (i % arr.get(j) == 0) {
                    cnt++;
                    if (cnt == K) {
                        return i;
                    }
                }
            }
            i++;
        }
        return 0;
    }

    public static void main(String[] args)
    {
        long K = 5;
        ArrayList<Integer> arr = new ArrayList<>();
        arr.add(3);
        arr.add(5);
        arr.add(7);

        System.out.println(solve(arr, K));
    }
}

// This code is contributed by shivamgupta0987654321

Python3

# Python Implementation


def solve(arr, K):
    # initialise the count (cnt) = 0
    cnt = 0
    i = 2
    while cnt < K:
        for j in range(len(arr)):
            if i % arr[j] == 0:
                cnt += 1
                if cnt == K:
                    return i
                    break
        i += 1
    return 0


if __name__ == "__main__":
    K = 5
    arr = [3, 5, 7]
    print(solve(arr, K))


# This code is contributed by Tapesh(tapeshdu420)

using System;
using System.Collections.Generic;

public class GFG {
    static long Solve(List<int> arr, long K)
    {
        // initialize the count (cnt) = 0
        long cnt = 0;
        long i = 2;

        while (cnt < K) {
            foreach(int num in arr)
            {
                if (i % num == 0) {
                    cnt++;
                    if (cnt == K) {
                        return i;
                    }
                }
            }
            i++;
        }
        return 0;
    }

    public static void Main(string[] args)
    {
        long K = 5;
        List<int> arr = new List<int>();
        arr.Add(3);
        arr.Add(5);
        arr.Add(7);

        Console.WriteLine(Solve(arr, K));
    }
}

JavaScript

// Javascript program for the above approach

// Function to find the Kth number that is divisible by any number in the given array
function solve(arr, K) {
    let cnt = 0; // Initialize count of divisible numbers found
    let i = 2; // Start checking from number 2 (since 1 is divisible by all numbers)
    while (cnt < K) { // Continue until K divisible numbers are found
        for (let j = 0; j < arr.length; j++) { // Iterate over each number in the array
            if (i % arr[j] === 0) { // Check if the current number is divisible by any number in the array
                cnt++; // Increment count if divisible
                if (cnt === K) { // If K divisible numbers are found, return the current number
                    return i;
                }
            }
        }
        i++; // Move to the next number to check
    }
    return 0; // If K divisible numbers are not found, return 0
}

const K = 5; // Define the value of K
const arr = [3, 5, 7]; // Define the array of numbers to check for divisibility
console.log(solve(arr, K)); // Call the solve function with the given parameters and print the result

// This code is contributed by Susobhan Akhuli

Output

Time Complexity: O(N * K)
Auxiliary Space: O(1).