Sync LeetCode submission Runtime - 26 ms (64.72%), Memory - 30.2 MB (24.30%)

Author: jimit105

0287-find-the-duplicate-number/README.md

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+<p>Given an array of integers <code>nums</code> containing&nbsp;<code>n + 1</code> integers where each integer is in the range <code>[1, n]</code> inclusive.</p>
+
+<p>There is only <strong>one repeated number</strong> in <code>nums</code>, return <em>this&nbsp;repeated&nbsp;number</em>.</p>
+
+<p>You must solve the problem <strong>without</strong> modifying the array <code>nums</code>&nbsp;and using only constant extra space.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [1,3,4,2,2]
+<strong>Output:</strong> 2
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,1,3,4,2]
+<strong>Output:</strong> 3
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums = [3,3,3,3,3]
+<strong>Output:</strong> 3</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
+	<li><code>nums.length == n + 1</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= n</code></li>
+	<li>All the integers in <code>nums</code> appear only <strong>once</strong> except for <strong>precisely one integer</strong> which appears <strong>two or more</strong> times.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><b>Follow up:</b></p>
+
+<ul>
+	<li>How can we prove that at least one duplicate number must exist in <code>nums</code>?</li>
+	<li>Can you solve the problem in linear runtime complexity?</li>
+</ul>

0287-find-the-duplicate-number/solution.py

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+# Approach 7: Floyd's Tortoise and Hare (Cycle Detection)
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def findDuplicate(self, nums: List[int]) -> int:
+        # Find the intersection point of the two runners.
+        slow, fast = 0, 0
+        while True:
+            slow = nums[slow]
+            fast = nums[nums[fast]]
+            if slow == fast:
+                break
+
+        # Find the "entrance" to the cycle.
+        slow2 = 0
+        while slow != slow2:
+            slow = nums[slow]
+            slow2 = nums[slow2]
+
+        return slow
+
+            
+