Given a quantity N, the duty is to rely the variety of methods to create an N digit quantity from digits 1 to 9 such that each digit is divisible by its earlier digit that’s if the quantity is represented by an array of digits A then A[i + 1] % A[i] == 0. print the reply modulo 109 + 7.
Examples:
Enter: N =Â 2
Output: 23
Rationalization: For N = 2 potential solutions are 11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, and 99.Enter: N =Â 3
Output: Â 44Â
Naive method: The fundamental technique to clear up the issue is as follows:
The fundamental technique to clear up this drawback is to generate all potential combos by utilizing a recursive method.
Time Complexity: O(9N)
Auxiliary Area: O(1)
Environment friendly Method: Â The above method might be optimized based mostly on the next thought:
Dynamic programming can be utilized to resolve this drawback
- dp[i][j] represents the variety of methods of making numerous dimension i and j is its earlier digit taken.
- It may be noticed that the recursive perform known as exponential instances. That implies that some states are known as repeatedly.Â
- So the concept is to retailer the worth of every state. This may be carried out by storing the worth of a state and every time the perform known as, returning the saved worth with out computing once more.
Observe the steps under to resolve the issue:
- Create a recursive perform that takes two parameters representing ith place to be stuffed and j representing the final digit taken.
- Name the recursive perform for selecting all digits from 1 to 9.
- Base case if all positions stuffed return 1.
- Create a second array of dp[N][10] initially full of -1.
- If the reply for a specific state is computed then put it aside in dp[i][j].
- If the reply for a specific state is already computed then simply return dp[i][j].
Under is the implementation of the above method:
C++
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Time Complexity: O(N) Â
Auxiliary Area: O(N)
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