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 :
- Each element is in range [1, m]
- 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 adjacent are// divisible#include <bits/stdc++.h>#define MAX 1000using 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 Programint 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 adjacent are// divisibleimport 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 Codepublic 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 = 1000def 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 adjacent are// divisibleusing 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 Codepublic 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 adjacent are// divisiblelet 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 Codelet n = 3, m = 3;document.write(numofArray(n, m));// This code is contributed by rag2127</script> |
Output
17
Time Complexity: O(N*M).
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