LeetCode-in-Java
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
xpresent on the board, find all numbers1 <= i <= nsuch thatx % 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 % 3is2.
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
Solution
public class Solution {
public int distinctIntegers(int n) {
if (n == 1) {
return 1;
}
return n - 1;
}
}