Introduction

In this blog post, we will explore how to solve the LeetCode problem "457" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are playing a game involving a circular array of non-zero integers nums. Each nums[i] denotes the number of indices forward/backward you must move if you are located at index i: If nums[i] is positive, move nums[i] steps forward, and If nums[i] is negative, move nums[i] steps backward. Since the array is circular, 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. A cycle in the array consists of a sequence of indices seq of length k where: Following the movement rules above results in the repeating index sequence seq[0] -> seq[1] -> ... -> seq[k - 1] -> seq[0] -> ... Every nums[seq[j]] is either all positive or all negative. k > 1 Return true if there is a cycle in nums, or false otherwise. Example 1: Input: nums = [2,-1,1,2,2] Output: true Explanation: The graph shows how the indices are connected. White nodes are jumping forward, while red is jumping backward. We can see the cycle 0 --> 2 --> 3 --> 0 --> ..., and all of its nodes are white (jumping in the same direction). Example 2: Input: nums = [-1,-2,-3,-4,-5,6] Output: false Explanation: 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. Example 3: Input: nums = [1,-1,5,1,4] Output: true Explanation: The graph shows how the indices are connected. White nodes are jumping forward, while red is jumping backward. We can see the cycle 0 --> 1 --> 0 --> ..., and while it is of size > 1, it has a node jumping forward and a node jumping backward, so it is not a cycle. We can see the cycle 3 --> 4 --> 3 --> ..., and all of its nodes are white (jumping in the same direction). Constraints: 1 <= nums.length <= 5000 -1000 <= nums[i] <= 1000 nums[i] != 0 Follow up: Could you solve it in O(n) time complexity and O(1) extra space complexity?

Explanation

• Cycle Detection: Iterate through each element of the array. For each element, treat it as a potential starting point of a cycle and trace the path it creates.
  • Direction Check: While tracing the path, verify that all elements encountered along the path have the same sign (either all positive or all negative). If a change in sign is detected, break the path tracing and mark the elements visited as invalid (by changing their value to 0).
  • Cycle Length: If a cycle is detected (i.e., the path returns to the starting point), ensure that the cycle's length is greater than 1.

• Time Complexity: O(n)
• Space Complexity: O(1)

Code

    class Solution:
    def circularArrayLoop(self, nums: list[int]) -> bool:
        n = len(nums)

        def get_next_index(i: int) -> int:
            return (i + nums[i]) % n

        for start in range(n):
            if nums[start] == 0:
                continue

            slow, fast = start, start
            direction = nums[start] > 0

            while True:
                slow = get_next_index(slow)

                if nums[slow] == 0 or (nums[slow] > 0) != direction:
                    break

                fast = get_next_index(fast)
                if nums[fast] == 0 or (nums[fast] > 0) != direction:
                    break
                fast = get_next_index(fast)
                if nums[fast] == 0 or (nums[fast] > 0) != direction:
                    break

                if slow == fast:
                    if slow == get_next_index(slow):
                        break
                    return True

            add = start
            direction_add = nums[start] > 0
            while nums[add] != 0 and (nums[add] > 0) == direction_add :
                temp = add
                add = get_next_index(add)
                nums[temp] = 0

        return False