Count of sequence of length K in range [1, N] where every element is a multiple of its previous one

Given two integers N and K, the task is to find the count of sequences of K elements from the range [1, N] where every element is a multiple of the previous element.

Example:

Input: N = 4, K = 3 
Output: 13
Explanation: The sequences that can be made from the integers 1, 2, 3, 4 having 3 elements are: {1, 1, 1}, {2, 2, 2}, {3, 3, 3}, {4, 4, 4}, {1, 1, 2}, {1, 2, 2}, {1, 2, 4}, {1, 1, 3}, {1, 3, 3}, {1, 1, 4}, {1, 4, 4}, {2, 2, 4}, and {2, 4, 4}.

Input: N = 9, K = 5 
Output: 111

Approach: The given problem can be solved using recursion with memoization. Follow the below steps to solve the problem:

• Create a 2D array dp[][] which stores the memorized states where dp[i][j] represents the count of sequences of length i having j as their first element.
• Create a recursive function countSequenceUtil(), that takes the length of the sequence and the starting element as arguments, sets the next element as a multiple of the current element, and recursively calls for the remaining sequence.
• Store the answer for the calculated states in the dp[][] array and if for some state the value is already calculated, return it.
• Create a function countSequence() which iterates through all the possible starting elements of the sequence and calls the recursive function to calculate the sequences of K elements with that starting element.
• Maintain the sum of the calculated count for each starting element in a variable ans which is the required value.

Below is the implementation of the above approach:

C++

// C++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;

// Initialize the dp matrix
int static dp[1001][1001];

// Function to find the count of sequences of K
// elements with first element as m where every
// element is a multiple of the previous one
int countSequenceUtil(int k, int m, int n)
{
    // Base case
    if (k == 1) {
        return 1;
    }

    // If the value already exists
    // in the DP then return it
    if (dp[k][m] != -1) {
        return dp[k][m];
    }

    // Variable to store the count
    int res = 0;

    for (int i = 1; i <= (n / m); i++) {

        // Recursive Call
        res += countSequenceUtil(k - 1,
                                 m * i, n);
    }

    // Store the calculated
    // answer and return it
    return dp[k][m] = res;
}

// Function to find count of sequences of K
// elements in the range [1, n] where every
// element is a multiple of the previous one
int countSequence(int N, int K)
{
    // Initializing all values
    // of dp with -1
    memset(dp, -1, sizeof(dp));

    // Variable to store
    // the total count
    int ans = 0;

    // Iterate from 1 to N
    for (int i = 1; i <= N; i++) {
        ans += countSequenceUtil(K, i, N);
    }

    // Return ans
    return ans;
}

// Driver Code
int main()
{
    int N = 9;
    int K = 5;

    cout << countSequence(N, K);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;

class GFG {

// Initialize the dp matrix
static int dp[][] = new int[1001][1001];

// Function to find the count of sequences of K
// elements with first element as m where every
// element is a multiple of the previous one
static int countSequenceUtil(int k, int m, int n)
{
    // Base case
    if (k == 1) {
        return 1;
    }

    // If the value already exists
    // in the DP then return it
    if (dp[k][m] != -1) {
        return dp[k][m];
    }

    // Variable to store the count
    int res = 0;

    for (int i = 1; i <= (n / m); i++) {

        // Recursive Call
        res += countSequenceUtil(k - 1,
                                 m * i, n);
    }

    // Store the calculated
    // answer and return it
    return dp[k][m] = res;
}

// Function to find count of sequences of K
// elements in the range [1, n] where every
// element is a multiple of the previous one
static int countSequence(int N, int K)
{

    // Initializing all values
    // of dp with -1
     for(int i=0;i<dp.length;i++)
     {
       for(int j=0;j<dp[i].length;j++)
       {
         dp[i][j]=-1;
       }
     }

    // Variable to store
    // the total count
    int ans = 0;

    // Iterate from 1 to N
    for (int i = 1; i <= N; i++) {
        ans += countSequenceUtil(K, i, N);
    }

    // Return ans
    return ans;
}

// Driver Code
    public static void main (String[] args) {
      int N = 9;
    int K = 5;


        System.out.println(countSequence(N, K));
    }
}

// This code is contributed by Potta Lokesh

Python3

# Python implementation for the above approach

# Initialize the dp matrix
dp = [[-1 for i in range(1001)] for j in range(1001)]

# Function to find the count of sequences of K
# elements with first element as m where every
# element is a multiple of the previous one
def countSequenceUtil(k, m, n):

    # Base case
    if (k == 1):
        return 1

    # If the value already exists
    # in the DP then return it
    if (dp[k][m] != -1):
        return dp[k][m]

    # Variable to store the count
    res = 0

    for i in range(1, (n // m) + 1):

        # Recursive Call
        res += countSequenceUtil(k - 1,
                                 m * i, n)

    # Store the calculated
    # answer and return it
    dp[k][m] = res

    return dp[k][m]

# Function to find count of sequences of K
# elements in the range [1, n] where every
# element is a multiple of the previous one
def countSequence(N, K):

    # Variable to store
    # the total count
    ans = 0

    # Iterate from 1 to N
    for i in range(1, N + 1):
        ans += countSequenceUtil(K, i, N)

    # Return ans
    return ans

# Driver Code
N = 9
K = 5

print(countSequence(N, K))

# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
class GFG {

    // Initialize the dp matrix
    static int[, ] dp = new int[1001, 1001];

    // Function to find the count of sequences of K
    // elements with first element as m where every
    // element is a multiple of the previous one
    static int countSequenceUtil(int k, int m, int n)
    {

        // Base case
        if (k == 1) {
            return 1;
        }

        // If the value already exists
        // in the DP then return it
        if (dp[k, m] != -1) {
            return dp[k, m];
        }

        // Variable to store the count
        int res = 0;

        for (int i = 1; i <= (n / m); i++) {

            // Recursive Call
            res += countSequenceUtil(k - 1, m * i, n);
        }

        // Store the calculated
        // answer and return it
        return dp[k, m] = res;
    }

    // Function to find count of sequences of K
    // elements in the range [1, n] where every
    // element is a multiple of the previous one
    static int countSequence(int N, int K)
    {

        // Initializing all values
        // of dp with -1
        for (int i = 0; i < dp.GetLength(0); i++) {
            for (int j = 0; j < dp.GetLength(1); j++) {
                dp[i, j] = -1;
            }
        }

        // Variable to store
        // the total count
        int ans = 0;

        // Iterate from 1 to N
        for (int i = 1; i <= N; i++) {
            ans += countSequenceUtil(K, i, N);
        }

        // Return ans
        return ans;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int N = 9;
        int K = 5;

        Console.WriteLine(countSequence(N, K));
    }
}

// This code is contributed by ukasp.

Javascript

<script>
    // JavaScript implementation for the above approach

    // Initialize the dp matrix
    let dp = new Array(1001).fill(-1).map(() => new Array(1001).fill(-1));

    // Function to find the count of sequences of K
    // elements with first element as m where every
    // element is a multiple of the previous one
    const countSequenceUtil = (k, m, n) => {

        // Base case
        if (k == 1) {
            return 1;
        }

        // If the value already exists
        // in the DP then return it
        if (dp[k][m] != -1) {
            return dp[k][m];
        }

        // Variable to store the count
        let res = 0;

        for (let i = 1; i <= parseInt(n / m); i++) {

            // Recursive Call
            res += countSequenceUtil(k - 1,
                m * i, n);
        }

        // Store the calculated
        // answer and return it
        return dp[k][m] = res;
    }

    // Function to find count of sequences of K
    // elements in the range [1, n] where every
    // element is a multiple of the previous one
    const countSequence = (N, K) => {

        // Variable to store
        // the total count
        let ans = 0;

        // Iterate from 1 to N
        for (let i = 1; i <= N; i++) {
            ans += countSequenceUtil(K, i, N);
        }

        // Return ans
        return ans;
    }

    // Driver Code
    let N = 9;
    let K = 5;

    document.write(countSequence(N, K));

    // This code is contributed by rakeshsahni

</script>

Output

111

Time Complexity: O(N*K*log N)
Auxiliary Space: O(N*K)

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

• Create a table to store the solution of the subproblems.
• Initialize the table with base cases
• Fill up the table iteratively
• Return the final solution

Implementation :

C++

// C++ program for above approach

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

// Function to find the count of sequences of K
// elements with first element as m where every
// element is a multiple of the previous one
int countSequence(int N, int K)
{
    // Initialize the dp matrix
    int dp[K + 1][N + 1];

    // Base case
    for (int i = 1; i <= N; i++) {
        dp[1][i] = 1;
    }

    // Tabulate the dp matrix
    for (int k = 2; k <= K; k++) {
        for (int m = 1; m <= N; m++) {
            dp[k][m] = 0;
            for (int i = 1; i <= (N / m); i++) {
                dp[k][m] += dp[k - 1][m * i];
            }
        }
    }

    // Compute the final result
    int ans = 0;
    for (int i = 1; i <= N; i++) {
        ans += dp[K][i];
    }

    // Return the answer
    return ans;
}

// Driver Code
int main()
{
    int N = 9;
    int K = 5;

    cout << countSequence(N, K);

    return 0;
}
// this code is contributed by bhardwajji

Java

import java.util.*;

public class Main {
    // Function to find the count of sequences of K
    // elements with first element as m where every
    // element is a multiple of the previous one
    static int countSequence(int N, int K)
    {
        // Initialize the dp matrix
        int[][] dp = new int[K + 1][N + 1];

        // Base case
        for (int i = 1; i <= N; i++) {
            dp[1][i] = 1;
        }

        // Tabulate the dp matrix
        for (int k = 2; k <= K; k++) {
            for (int m = 1; m <= N; m++) {
                dp[k][m] = 0;
                for (int i = 1; i <= (N / m); i++) {
                    dp[k][m] += dp[k - 1][m * i];
                }
            }
        }

        // Compute the final result
        int ans = 0;
        for (int i = 1; i <= N; i++) {
            ans += dp[K][i];
        }

        // Return the answer
        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 9;
        int K = 5;

        System.out.println(countSequence(N, K));
    }
}
// This code is contributed by Gaurav_Arora

Python3

def countSequence(N, K):
    # Initialize the dp matrix
    dp = [[0 for i in range(N + 1)] for j in range(K + 1)]

    # Base case
    for i in range(1, N + 1):
        dp[1][i] = 1

    # Tabulate the dp matrix
    for k in range(2, K + 1):
        for m in range(1, N + 1):
            dp[k][m] = 0
            for i in range(1, (N // m) + 1):
                dp[k][m] += dp[k - 1][m * i]

    # Compute the final result
    ans = 0
    for i in range(1, N + 1):
        ans += dp[K][i]

    # Return the answer
    return ans

# Driver Code
if __name__ == "__main__":
    N = 9
    K = 5
    print(countSequence(N, K))

C#

// C# program for above approach

using System;

public class GFG
{
    // Function to find the count of sequences of K
    // elements with first element as m where every
    // element is a multiple of the previous one
    static int CountSequence(int N, int K)
    {
        // Initialize the dp matrix
        int[,] dp = new int[K + 1, N + 1];

        // Base case
        for (int i = 1; i <= N; i++)
        {
            dp[1, i] = 1;
        }

        // Tabulate the dp matrix
        for (int k = 2; k <= K; k++)
        {
            for (int m = 1; m <= N; m++)
            {
                dp[k, m] = 0;
                for (int i = 1; i <= (N / m); i++)
                {
                    dp[k, m] += dp[k - 1, m * i];
                }
            }
        }

        // Compute the final result
        int ans = 0;
        for (int i = 1; i <= N; i++)
        {
            ans += dp[K, i];
        }

        // Return the answer
        return ans;
    }

    // Driver Code
    public static void Main(string[] args)
    {
        int N = 9;
        int K = 5;

        Console.WriteLine(CountSequence(N, K));
    }
}

Javascript

// Function to find the count of sequences of K
// elements with first element as m where every
// element is a multiple of the previous one
function countSequence(N, K) {
    // Initialize the dp matrix
    let dp = new Array(K + 1);
    for (let i = 0; i <= K; i++) {
        dp[i] = new Array(N + 1);
    }

    // Base case
    for (let i = 1; i <= N; i++) {
        dp[1][i] = 1;
    }

    // Tabulate the dp matrix
    for (let k = 2; k <= K; k++) {
        for (let m = 1; m <= N; m++) {
            dp[k][m] = 0;
            for (let i = 1; i <= (N / m); i++) {
                dp[k][m] += dp[k - 1][m * i];
            }
        }
    }

    // Compute the final result
    let ans = 0;
    for (let i = 1; i <= N; i++) {
        ans += dp[K][i];
    }

    // Return the answer
    return ans;
}

// Driver Code
let N = 9;
let K = 5;

console.log(countSequence(N, K));

Output

111

Time Complexity: O(N*K*log N)
Auxiliary Space: O(N*K)