Merge pull request #1873 from Maheshwari-Love/add/solution-lc-2549

added solution to leetcode-2549

Author: ajay-dhangar

dsa-solutions/lc-solutions/2500 - 3000/2549-Count Distinct Numbers on Board.md

@@ -0,0 +1,97 @@
+---
+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."
+---
+
+## Problem Description
+
+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.
+
+### Examples
+
+**Example 1:**
+
+```
+
+Input: n = 5
+Output: 4
+Explanation: Initially, 5 is present on the board. 
+
+```
+
+**Example 2:**
+
+```
+Input: n = 3
+Output: 2
+Explanation: 
+Since 3 % 2 == 1, 2 will be added to the board. 
+```
+
+
+### Constraints
+
+- `1 <= n <= 100`
+ 
+
+### Approach 
+
+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.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the given number.
+
+#### Python3
+
+```python
+class Solution:
+    def distinctIntegers(self, n: int) -> int:
+        return max(1, n - 1)
+```
+
+#### Java
+
+```java
+class Solution {
+    public int distinctIntegers(int n) {
+        return Math.max(1, n - 1);
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int distinctIntegers(int n) {
+        return max(1, n - 1);
+    }
+};
+```
+
+#### Go
+
+```go
+func distinctIntegers(n int) int {
+	return max(1, n-1)
+}
+```
+
+#### TypeScript
+
+```ts
+function distinctIntegers(n: number): number {
+    return Math.max(1, n - 1);
+}
+```