Circular Array Loop
<p>You are playing a game involving a <strong>circular</strong> array of non-zero integers <code>nums</code>. Each <code>nums[i]</code> denotes the number of indices forward/backward you must move if you are located at index <code>i</code>:</p>

<ul>
	<li>If <code>nums[i]</code> is positive, move <code>nums[i]</code> steps <strong>forward</strong>, and</li>
	<li>If <code>nums[i]</code> is negative, move <code>nums[i]</code> steps <strong>backward</strong>.</li>
</ul>

<p>Since the array is <strong>circular</strong>, you may assume that moving forward from the last element puts you on the first element, and moving backwards from the first element puts you on the last element.</p>

<p>A <strong>cycle</strong> in the array consists of a sequence of indices <code>seq</code> of length <code>k</code> where:</p>

<ul>
	<li>Following the movement rules above results in the repeating index sequence <code>seq[0] -&gt; seq[1] -&gt; ... -&gt; seq[k - 1] -&gt; seq[0] -&gt; ...</code></li>
	<li>Every <code>nums[seq[j]]</code> is either <strong>all positive</strong> or <strong>all negative</strong>.</li>
	<li><code>k &gt; 1</code></li>
</ul>

<p>Return <code>true</code><em> if there is a <strong>cycle</strong> in </em><code>nums</code><em>, or </em><code>false</code><em> otherwise</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0457.Circular%20Array%20Loop/images/img1.jpg" style="width: 402px; height: 289px;" />
<pre>
<strong>Input:</strong> nums = [2,-1,1,2,2]
<strong>Output:</strong> true
<strong>Explanation:</strong> The graph shows how the indices are connected. White nodes are jumping forward, while red is jumping backward.
We can see the cycle 0 --&gt; 2 --&gt; 3 --&gt; 0 --&gt; ..., and all of its nodes are white (jumping in the same direction).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0457.Circular%20Array%20Loop/images/img2.jpg" style="width: 402px; height: 390px;" />
<pre>
<strong>Input:</strong> nums = [-1,-2,-3,-4,-5,6]
<strong>Output:</strong> false
<strong>Explanation:</strong> The graph shows how the indices are connected. White nodes are jumping forward, while red is jumping backward.
The only cycle is of size 1, so we return false.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0400-0499/0457.Circular%20Array%20Loop/images/img3.jpg" style="width: 497px; height: 242px;" />
<pre>
<strong>Input:</strong> nums = [1,-1,5,1,4]
<strong>Output:</strong> true
<strong>Explanation:</strong> The graph shows how the indices are connected. White nodes are jumping forward, while red is jumping backward.
We can see the cycle 0 --&gt; 1 --&gt; 0 --&gt; ..., and while it is of size &gt; 1, it has a node jumping forward and a node jumping backward, so <strong>it is not a cycle</strong>.
We can see the cycle 3 --&gt; 4 --&gt; 3 --&gt; ..., and all of its nodes are white (jumping in the same direction).
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
	<li><code>-1000 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>nums[i] != 0</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve it in <code>O(n)</code> time complexity and <code>O(1)</code> extra space complexity?</p>
Array; Hash Table; Two Pointers
class Solution:
    def circularArrayLoop(self, nums: List[int]) -> bool:
        n = len(nums)

        def next(i):
            return (i + nums[i] % n + n) % n

        for i in range(n):
            if nums[i] == 0:
                continue
            slow, fast = i, next(i)
            while nums[slow] * nums[fast] > 0 and nums[slow] * nums[next(fast)] > 0:
                if slow == fast:
                    if slow != next(slow):
                        return True
                    break
                slow, fast = next(slow), next(next(fast))
            j = i
            while nums[j] * nums[next(j)] > 0:
                nums[j] = 0
                j = next(j)
        return False
