2549

2549. Count Distinct Numbers on Board (Easy)

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 <= 100

Companies: Oracle

Related Topics:
Array, Hash Table, Math, Simulation

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Solution 1.

// OJ: https://leetcode.com/problems/count-distinct-numbers-on-board
// Author: github.com/lzl124631x
// Time: O(1)
// Space: O(1)
class Solution {
public:
    int distinctIntegers(int n) {
        return max(1, n - 1);
    }
};