Teaching Kids Programming – Count Distinct Numbers on Board (Recursion, Simulation, Math, Hash Set) | ninjasquad

**Teaching Kids Programming**: Videos on **Data Structures and Algorithms**

You are given a positive integer n, that is initially placed on a board. Every day, for 109 days, you perform the following procedure:

For each number x present on the board, find all numbers 1 <= i <= n such that x % i == 1.

Then, place those numbers on the board.

Return the number of distinct integers present on the board after 109 days have elapsed.Note:

Once a number is placed on the board, it will remain on it until the end.

% stands for the modulo operation. For example, 14 % 3 is 2.Example 1:

Input: n = 5

Output: 4

Explanation: Initially, 5 is present on the board.

The next day, 2 and 4 will be added since 5 % 2 == 1 and 5 % 4 == 1.

After that day, 3 will be added to the board because 4 % 3 == 1.

At the end of a billion days, the distinct numbers on the board will be 2, 3, 4, and 5.Example 2:

Input: n = 3

Output: 2

Explanation:

Since 3 % 2 == 1, 2 will be added to the board.

After a billion days, the only two distinct numbers on the board are 2 and 3.Constraints:

1 <= n <= 100Hints:

For n > 2, n % (n – 1) == 1 thus n – 1 will be added on the board the next day.

As the operations are performed for so long time, all the numbers lesser than n except 1 will be added to the board.

What will happen if n == 1?

### Count Distinct Numbers on Board

The given input N is small, thus we can simulate the process. We use a hash set to store the unique/distinct numbers, and if n % i is 1, we add it to the hash set, and repeat the process for i. We need to check numbers from 2 to N-1 only. The time/space complexity is O(N) as at most there are N-1 numbers put on the whiteboard.

The following is the Recursive algorithm for simulating the process.

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class Solution: def distinctIntegers(self, n: int) -> int: nums = set([n]) def f(n): for i in range(2, n): if n % i == 1 and i not in nums: nums.add(i) f(i) f(n) return len(nums) |

class Solution: def distinctIntegers(self, n: int) -> int: nums = set([n]) def f(n): for i in range(2, n): if n % i == 1 and i not in nums: nums.add(i) f(i) f(n) return len(nums)

Since we will repeat enough number of times, all the numbers between 2 to N-1 will be eventually on the whiteboard. Consider N which is greater than 2, N%(N-1)==1, and next round, N-2 will be qualified, and then N-3 etc until 2. Special case is when N=1 and in this case, only 1 will be on the board. Thus, if N is 1, the answer is one, otherwise, the answer is N-1. The time/space complexity is O(1) constant since this is just pure math.

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class Solution: def distinctIntegers(self, n: int) -> int: return max(n - 1, 1) |

class Solution: def distinctIntegers(self, n: int) -> int: return max(n - 1, 1)

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