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dsa-solutions/lc-solutions/2500 - 3000/2549-Count Distinct Numbers on Board.md
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--- | ||
id: count-distinct-numbers-on-board | ||
title: Count Distinct Numbers on Board | ||
sidebar_label: 2544 Count Distinct Numbers on Board | ||
tags: | ||
- Array | ||
- Hash Table | ||
- LeetCode | ||
- C++ | ||
description: "This is a solution to the Count Distinct Numbers on Board problem on LeetCode." | ||
--- | ||
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## Problem Description | ||
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You are given a positive integer n, that is initially placed on a board. Every day, for 109 days, you perform the following procedure: | ||
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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. | ||
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### Examples | ||
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**Example 1:** | ||
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``` | ||
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Input: n = 5 | ||
Output: 4 | ||
Explanation: Initially, 5 is present on the board. | ||
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``` | ||
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**Example 2:** | ||
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``` | ||
Input: n = 3 | ||
Output: 2 | ||
Explanation: | ||
Since 3 % 2 == 1, 2 will be added to the board. | ||
``` | ||
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### Constraints | ||
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- `1 <= n <= 100` | ||
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### Approach | ||
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Since every operation on the number n on the desktop will also cause the number n-1 to appear on the desktop, the final numbers on the desktop are [2,...n], that is, n-1 numbers. | ||
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the given number. | ||
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#### Python3 | ||
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```python | ||
class Solution: | ||
def distinctIntegers(self, n: int) -> int: | ||
return max(1, n - 1) | ||
``` | ||
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#### Java | ||
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```java | ||
class Solution { | ||
public int distinctIntegers(int n) { | ||
return Math.max(1, n - 1); | ||
} | ||
} | ||
``` | ||
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#### C++ | ||
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```cpp | ||
class Solution { | ||
public: | ||
int distinctIntegers(int n) { | ||
return max(1, n - 1); | ||
} | ||
}; | ||
``` | ||
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#### Go | ||
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```go | ||
func distinctIntegers(n int) int { | ||
return max(1, n-1) | ||
} | ||
``` | ||
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#### TypeScript | ||
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```ts | ||
function distinctIntegers(n: number): number { | ||
return Math.max(1, n - 1); | ||
} | ||
``` |
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