From a8893c41331e6d787c189ff698d73925966b80ca Mon Sep 17 00:00:00 2001
From: Maheshwari Love <l.maheshwari2008@gmail.com>
Date: Tue, 25 Jun 2024 16:33:16 +0530
Subject: [PATCH] added solution for lc-2550

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 ...ount Collisions of Monkeys on a Polygon.md | 106 ++++++++++++++++++
 1 file changed, 106 insertions(+)
 create mode 100644 dsa-solutions/lc-solutions/2500 - 3000/2550. Count Collisions of Monkeys on a Polygon.md

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+---
+id: alternating-digit-sum
+title: Count Collisions of Monkeys on a Polygonaximum
+sidebar_label: 2550 Count Collisions of Monkeys on a Polygonaximum
+  - Array
+  - Recursion
+  - LeetCode
+  - C++
+description: "This is a solution to the Count Collisions of Monkeys on a Polygonaximum problem on LeetCode."
+---
+
+## Problem Description
+
+There is a regular convex polygon with n vertices. The vertices are labeled from 0 to n - 1 in a clockwise direction, and each vertex has exactly one monkey. The following figure shows a convex polygon of 6 vertices.
+
+Simultaneously, each monkey moves to a neighboring vertex. A collision happens if at least two monkeys reside on the same vertex after the movement or intersect on an edge.
+
+Return the number of ways the monkeys can move so that at least one collision happens. Since the answer may be very large, return it modulo 10^9 + 7.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: n = 3
+Output: 6
+Explanation:
+There are 8 total possible movements.
+
+```
+
+**Example 2:**
+
+```
+Input: n = 4
+Output: 14
+
+```
+
+
+### Constraints
+
+- `3 <= n <= 10^9`
+ 
+
+### Approach 
+
+According to the problem description, each monkey has two ways of moving, either clockwise or counterclockwise. Therefore, there are a total of $2^n$ ways to move. The non-collision ways of moving are only two, that is, all monkeys move clockwise or all monkeys move counterclockwise. Therefore, the number of collision ways of moving is $2^n - 2$.
+
+We can use fast power to calculate the value of $2^n$, then use $2^n - 2$ to calculate the number of collision ways of moving, and finally take the remainder of $10^9 + 7$.
+
+The time complexity is $O(\log n)$, where $n$ is the number of monkeys. The space complexity is $O(1)$.
+
+#### Python3
+
+```python
+class Solution:
+    def monkeyMove(self, n: int) -> int:
+        mod = 10**9 + 7
+        return (pow(2, n, mod) - 2) % mod
+```
+
+#### Java
+
+```java
+class Solution {
+    public int monkeyMove(int n) {
+        final int mod = (int) 1e9 + 7;
+        return (qpow(2, n, mod) - 2 + mod) % mod;
+    }
+
+    private int qpow(long a, int n, int mod) {
+        long ans = 1;
+        for (; n > 0; n >>= 1) {
+            if ((n & 1) == 1) {
+                ans = ans * a % mod;
+            }
+            a = a * a % mod;
+        }
+        return (int) ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int monkeyMove(int n) {
+        const int mod = 1e9 + 7;
+        using ll = long long;
+        auto qpow = [&](ll a, int n) {
+            ll ans = 1;
+            for (; n; n >>= 1) {
+                if (n & 1) {
+                    ans = ans * a % mod;
+                }
+                a = a * a % mod;
+            }
+            return ans;
+        };
+        return (qpow(2, n) - 2 + mod) % mod;
+    }
+};
+```