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Count of arrays in which all adjacent elements are such that one of them divide the another

  • Difficulty Level : Expert
  • Last Updated : 22 Jun, 2021
Geek Week

Given two positive integer n and n. The task is to find the number of arrays of size n that can be formed such that : 

  1. Each element is in range [1, m]
  2. All adjacent element are such that one of them divide the another i.e element Ai divides Ai + 1 or Ai + 1 divides Ai + 2.

Examples:  

Input : n = 3, m = 3.
Output : 17
{1,1,1}, {1,1,2}, {1,1,3}, {1,2,1}, 
{1,2,2}, {1,3,1}, {1,3,3}, {2,1,1},
{2,1,2}, {2,1,3}, {2,2,1}, {2,2,2},
{3,1,1}, {3,1,2}, {3,1,3}, {3,3,1}, 
{3,3,3} are possible arrays.

Input : n = 1, m = 10.
Output : 10 

We try to find number of possible values at each index of the array. First, at index 0 all values are possible from 1 to m. Now observe for each index, we can reach either to its multiple or its factor. So, precompute that and store it for all the elements. Hence for each position i, ending with integer x, we can go to next position i + 1, with the array ending in integer with multiple of x or factor of x. Also, multiple of x or factor of x must be less than m. 

So, we define an 2D array dp[i][j], which is number of possible array (divisible adjacent element) of size i with j as its first index element.  

1 <= i <= m, dp[1][m] = 1.
for 1 <= j <= m and 2 <= i <= n
  dp[i][j] = dp[i-1][j] + number of factor 
             of previous element less than m 
             + number of multiples of previous
             element less than m.

Below is the implementation of this approach: 



C++




// C++ program to count number of arrays
// of size n such that every element is
// in range [1, m] and adjacen are
// divisible
#include <bits/stdc++.h>
#define  MAX 1000
using namespace std;
 
int numofArray(int n, int m)
{
    int dp[MAX][MAX];
 
    // For storing factors.
    vector<int> di[MAX];
 
    // For storing multiples.
    vector<int> mu[MAX];
 
    memset(dp, 0, sizeof dp);
    memset(di, 0, sizeof di);
    memset(mu, 0, sizeof mu);
 
    // calculating the factors and multiples
    // of elements [1...m].
    for (int i = 1; i <= m; i++)
    {
        for (int j = 2*i; j <= m; j += i)
        {
            di[j].push_back(i);
            mu[i].push_back(j);
        }
        di[i].push_back(i);
    }
 
    // Initialising for size 1 array for
    // each i <= m.
    for (int i = 1; i <= m; i++)
        dp[1][i] = 1;
 
    // Calculating the number of array possible
    // of size i and starting with j.
    for (int i = 2; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            dp[i][j] = 0;
 
            // For all previous possible values.
            // Adding number of factors.
            for (auto x:di[j])
                dp[i][j] += dp[i-1][x];
 
            // Adding number of multiple.
            for (auto x:mu[j])
                dp[i][j] += dp[i-1][x];
        }
    }
 
    // Calculating the total count of array
    // which start from [1...m].
    int ans = 0;
    for (int i = 1; i <= m; i++)
    {
        ans += dp[n][i];
        di[i].clear();
        mu[i].clear();
    }
 
    return ans;
}
 
// Driven Program
int main()
{
    int n = 3, m = 3;
    cout << numofArray(n, m) << "\n";
    return 0;
}

Java




// Java program to count number of arrays
// of size n such that every element is
// in range [1, m] and adjacen are
// divisible
import java.util.*;
 
class GFG
{
static int MAX = 1000;
 
static int numofArray(int n, int m)
{
    int [][]dp = new int[MAX][MAX];
 
    // For storing factors.
    Vector<Integer> []di = new Vector[MAX];
 
    // For storing multiples.
    Vector<Integer> []mu = new Vector[MAX];
 
    for(int i = 0; i < MAX; i++)
    {
        for(int j = 0; j < MAX; j++)
        {
            dp[i][j] = 0;
        }
    }
    for(int i = 0; i < MAX; i++)
    {
        di[i] = new Vector<>();
        mu[i] = new Vector<>();
    }
     
    // calculating the factors and multiples
    // of elements [1...m].
    for (int i = 1; i <= m; i++)
    {
        for (int j = 2 * i; j <= m; j += i)
        {
            di[j].add(i);
            mu[i].add(j);
        }
        di[i].add(i);
    }
 
    // Initialising for size 1 array for
    // each i <= m.
    for (int i = 1; i <= m; i++)
        dp[1][i] = 1;
 
    // Calculating the number of array possible
    // of size i and starting with j.
    for (int i = 2; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            dp[i][j] = 0;
 
            // For all previous possible values.
            // Adding number of factors.
            for (Integer x:di[j])
                dp[i][j] += dp[i - 1][x];
 
            // Adding number of multiple.
            for (Integer x:mu[j])
                dp[i][j] += dp[i - 1][x];
        }
    }
 
    // Calculating the total count of array
    // which start from [1...m].
    int ans = 0;
    for (int i = 1; i <= m; i++)
    {
        ans += dp[n][i];
        di[i].clear();
        mu[i].clear();
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 3, m = 3;
    System.out.println(numofArray(n, m));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program to count the number of
# arrays of size n such that every element is
# in range [1, m] and adjacent are divisible
 
MAX = 1000
 
def numofArray(n, m):
  
    dp = [[0 for i in range(MAX)] for j in range(MAX)]
 
    # For storing factors.
    di = [[] for i in range(MAX)]
 
    # For storing multiples.
    mu = [[] for i in range(MAX)]
 
    # calculating the factors and multiples
    # of elements [1...m].
    for i in range(1, m+1):
      
        for j in range(2*i, m+1, i):
          
            di[j].append(i)
            mu[i].append(j)
          
        di[i].append(i)
 
    # Initialising for size 1 array for each i <= m.
    for i in range(1, m+1):
        dp[1][i] = 1
 
    # Calculating the number of array possible
    # of size i and starting with j.
    for i in range(2, n+1):
      
        for j in range(1, m+1):
          
            dp[i][j] = 0
 
            # For all previous possible values.
            # Adding number of factors.
            for x in di[j]:
                dp[i][j] += dp[i-1][x]
 
            # Adding number of multiple.
            for x in mu[j]:
                dp[i][j] += dp[i-1][x]
          
    # Calculating the total count of array
    # which start from [1...m].
    ans = 0
    for i in range(1, m+1):
      
        ans += dp[n][i]
        di[i].clear()
        mu[i].clear()
     
    return ans
  
# Driven Program
if __name__ == "__main__":
  
    n = m = 3
    print(numofArray(n, m))
     
# This code is contributed by Rituraj Jain

C#




// C# program to count number of arrays
// of size n such that every element is
// in range [1, m] and adjacen are
// divisible
using System;
using System.Collections.Generic;                
     
class GFG
{
static int MAX = 1000;
 
static int numofArray(int n, int m)
{
    int [,]dp = new int[MAX, MAX];
 
    // For storing factors.
    List<int> []di = new List<int>[MAX];
 
    // For storing multiples.
    List<int> []mu = new List<int>[MAX];
 
    for(int i = 0; i < MAX; i++)
    {
        for(int j = 0; j < MAX; j++)
        {
            dp[i, j] = 0;
        }
    }
    for(int i = 0; i < MAX; i++)
    {
        di[i] = new List<int>();
        mu[i] = new List<int>();
    }
     
    // calculating the factors and multiples
    // of elements [1...m].
    for (int i = 1; i <= m; i++)
    {
        for (int j = 2 * i; j <= m; j += i)
        {
            di[j].Add(i);
            mu[i].Add(j);
        }
        di[i].Add(i);
    }
 
    // Initialising for size 1 array for
    // each i <= m.
    for (int i = 1; i <= m; i++)
        dp[1, i] = 1;
 
    // Calculating the number of array possible
    // of size i and starting with j.
    for (int i = 2; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            dp[i, j] = 0;
 
            // For all previous possible values.
            // Adding number of factors.
            foreach (int x in di[j])
                dp[i, j] += dp[i - 1, x];
 
            // Adding number of multiple.
            foreach (int x in mu[j])
                dp[i, j] += dp[i - 1, x];
        }
    }
 
    // Calculating the total count of array
    // which start from [1...m].
    int ans = 0;
    for (int i = 1; i <= m; i++)
    {
        ans += dp[n, i];
        di[i].Clear();
        mu[i].Clear();
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 3, m = 3;
    Console.WriteLine(numofArray(n, m));
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript program to count number of arrays
// of size n such that every element is
// in range [1, m] and adjacen are
// divisible
let MAX = 1000;
 
function numofArray(n, m)
{
    let dp = new Array(MAX);
  
    // For storing factors.
    let di = new Array(MAX);
  
    // For storing multiples.
    let mu = new Array(MAX);
  
    for(let i = 0; i < MAX; i++)
    {
        dp[i] = new Array(MAX);
        for(let j = 0; j < MAX; j++)
        {
            dp[i][j] = 0;
        }
    }
    for(let i = 0; i < MAX; i++)
    {
        di[i] = [];
        mu[i] = [];
    }
      
    // Calculating the factors and multiples
    // of elements [1...m].
    for(let i = 1; i <= m; i++)
    {
        for(let j = 2 * i; j <= m; j += i)
        {
            di[j].push(i);
            mu[i].push(j);
        }
        di[i].push(i);
    }
  
    // Initialising for size 1 array for
    // each i <= m.
    for(let i = 1; i <= m; i++)
        dp[1][i] = 1;
  
    // Calculating the number of array possible
    // of size i and starting with j.
    for(let i = 2; i <= n; i++)
    {
        for(let j = 1; j <= m; j++)
        {
            dp[i][j] = 0;
  
            // For all previous possible values.
            // Adding number of factors.
            for(let x = 0; x < di[j].length; x++)
                dp[i][j] += dp[i - 1][di[j][x]];
  
            // Adding number of multiple.
            for(let x = 0; x < mu[j].length; x++)
                dp[i][j] += dp[i - 1][mu[j][x]];
        }
    }
  
    // Calculating the total count of array
    // which start from [1...m].
    let ans = 0;
    for(let i = 1; i <= m; i++)
    {
        ans += dp[n][i];
        di[i] = [];
        mu[i] = [];
    }
    return ans;
}
 
// Driver Code
let n = 3, m = 3;
 
document.write(numofArray(n, m));
 
// This code is contributed by rag2127
 
</script>

Output: 

17

Time Complexity: O(N*M).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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