How do you solve Count Distinct Numbers on Board on LeetCode?

Count Distinct Numbers on Board (LeetCode #2549) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: After a few iterations, all numbers from 2 to n will be present on the board, except for 1.

What is the brute force solution for Count Distinct Numbers on Board?

The brute force approach for Count Distinct Numbers on Board is Brute Force, with O(n²) time complexity and O(n) space complexity. In the brute-force approach, we simulate the process of adding numbers to the board for a limited number of iterations. We check for each number on the board which numbers can be added based on the modulo condition. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Count Distinct Numbers on Board?

Count Distinct Numbers on Board uses the following concepts: Array, Hash Table, Math, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Distinct Numbers on Board?

Always consider if there's a pattern or mathematical property that can simplify the problem. Don't hesitate to start with a brute-force solution to understand the problem better before optimizing. Be clear about your thought process and explain your reasoning, especially when you identify a pattern.

What are common mistakes when solving Count Distinct Numbers on Board?

Not recognizing that the process stabilizes quickly and continuing to simulate unnecessarily. Overcomplicating the problem by trying to track every individual addition instead of looking for patterns.

#2549

Count Distinct Numbers on Board

Easy

ArrayHash TableMathSimulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(n)	O(1)

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Intuition

Time O(1)Space O(1)

In the optimal solution, we recognize that after a few iterations, all numbers from 2 to n will be added to the board, except for 1. This is due to the properties of the modulo operation.

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Algorithm

3 steps

1Step 1: If n == 1, return 1 (only 1 is on the board).
2Step 2: If n == 2, return 2 (both 1 and 2 are on the board).
3Step 3: For n > 2, return n - 1 (all numbers from 2 to n are added).

solution.py4 lines

1def countDistinctNumbers(n):
2    if n == 1:
3        return 1
4    return n - 1

ℹ

Complexity note: The time complexity is O(1) because we directly compute the result without any loops. The space complexity is also O(1) as we use a constant amount of space.

1After a few iterations, all numbers from 2 to n will be present on the board, except for 1.
2The modulo operation reveals a pattern that allows us to predict the final count without simulating each day.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.