LeetCode 2481. Minimum Cuts to Divide a Circle

Author: raul-sauco

README.md

@@ -350,6 +350,7 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 | [2115. Find All Possible Recipes from Given Supplies][lc2115]                  | 🟠 Medium  | [![python](res/py.png)][lc2115py]                                   |
 | [2131. Longest Palindrome by Concatenating Two Letter Words][lc2131]           | 🟠 Medium  | [![python](res/py.png)][lc2131py]                                   |
 | [2136. Earliest Possible Day of Full Bloom][lc2136]                            | 🔴 Hard    | [![python](res/py.png)][lc2136py]                                   |
+| [2481. Minimum Cuts to Divide a Circle][lc2481]                                | 🟢 Easy    | [![python](res/py.png)][lc2481py]                                   |
 
 [🔝 Back to Top 🔝](#coding-challenges)
 
@@ -1024,6 +1025,8 @@ Solutions to LeetCode problems. The first column links to the problem in LeetCod
 [lc2131py]: leetcode/longest-palindrome-by-concatenating-two-letter-words.py
 [lc2136]: https://leetcode.com/problems/earliest-possible-day-of-full-bloom/
 [lc2136py]: leetcode/earliest-possible-day-of-full-bloom.py
+[lc2481]: https://leetcode.com/problems/minimum-cuts-to-divide-a-circle/
+[lc2481py]: leetcode/minimum-cuts-to-divide-a-circle.py
 
 ## CodeWars problems
 

leetcode/minimum-cuts-to-divide-a-circle.py

@@ -0,0 +1,52 @@
+# 2481. Minimum Cuts to Divide a Circle
+# 🟢 Easy
+#
+# https://leetcode.com/problems/minimum-cuts-to-divide-a-circle/
+#
+# Tags: Math - Geometry
+
+import timeit
+
+
+# If n == 1, we do not need to cut, otherwise, if the number is uneven
+# we need to make n radial cuts, but if n is uneven, we can make edge
+# to edge cuts, diametral, and we only need n // 2 of them.
+#
+# Time complexity: O(1)
+# Space complexity: O(1)
+#
+# Runtime: 59 ms, faster than 40.00%
+# Memory Usage: 13.9 MB, less than 20.00%
+class Solution:
+    def numberOfCuts(self, n: int) -> int:
+        if n % 2:
+            return n if n != 1 else 0
+        return n // 2
+
+
+def test():
+    executors = [Solution]
+    tests = [
+        [1, 0],
+        [3, 3],
+        [4, 2],
+    ]
+    for executor in executors:
+        start = timeit.default_timer()
+        for _ in range(1):
+            for col, t in enumerate(tests):
+                sol = executor()
+                result = sol.numberOfCuts(t[0])
+                exp = t[1]
+                assert result == exp, (
+                    f"\033[93m» {result} <> {exp}\033[91m for"
+                    + f" test {col} using \033[1m{executor.__name__}"
+                )
+        stop = timeit.default_timer()
+        used = str(round(stop - start, 5))
+        cols = "{0:20}{1:10}{2:10}"
+        res = cols.format(executor.__name__, used, "seconds")
+        print(f"\033[92m» {res}\033[0m")
+
+
+test()