Problem Description

In this problem, you are working with a circular array of non-zero integers. The array is called "circular" because if you move past the last element, you wrap around to the first element, and vice versa. Each value in the array tells you how many steps to move from your current position. A positive value means you move that number of steps forward, while a negative value means you move backward.

The challenge is to determine if there exists a "cycle" in the array. A cycle means that if you start at some index and follow the steps, you eventually return to the starting index after k moves, where k is greater than 1. Furthermore, all the steps taken during this process should be exclusively positive or exclusively negative, enforcing that the loop goes in a single direction.

Your task is to return true if there is such a cycle in the array, otherwise return false.

Intuition

To address this problem, we need to consider that a cycle can only exist if we're moving consistently in one direction and eventually end up where we started. This naturally brings the "fast and slow pointers" technique to mind, which is often used for cycle detection in linked lists.

The fast and slow pointers method involves two pointers moving at different speeds, and if there is a cycle, they will eventually meet. We apply the same principle here:

• The slow pointer moves one step at a time.
• The fast pointer moves two steps at a time.

If slow and fast meet at the same index, and this index is not the same as the next step (to prevent single-element loops, which aren't considered valid cycles), we have found a cycle.

At each step, we also verify that the direction does not change. If the product of nums[slow] and nums[fast] is positive, they are either both positive or both negative, thus maintaining a consistent direction. If this product is negative or if we reach an element that is already marked as visited (a value of 0), we do not have a valid cycle from that start point.

For each element, if it does not lead to a cycle, we mark the visited elements as 0 to avoid re-checking them in the future, thereby optimizing our algorithm. This marking also helps to invalidate any non-cycle paths swiftly.

Overall, the algorithm is to iterate over each element and use the fast and slow pointer method to detect a cycle. If any cycle is found, return true. After checking all possibilities, if no cycle is found, return false.

Learn more about Two Pointers patterns.

Solution Approach

The implementation of the solution for detecting a cycle in the circular array follows these main steps:

1. Array Length Extraction: We start by obtaining the length n of the input array nums. This is crucial since we need to calculate the next index correctly within the circular context:

   n = len(nums)

2. Helper Function for Index Calculation: Since the array is circular, we define a function named next() that takes an index i and returns the next index we should move to, according to nums[i], and wraps around the array if necessary:

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

   We ensure that the result of the movement remains within the bounds of the array indices by taking the modulo with n.

3. Main Loop to Check for Cycles: We iterate through each element in the array to check for cycles starting from that index:

   for i in range(n):
       if nums[i] == 0:  # Skip already marked elements (no cycle from this point)
           continue

4. Fast and Slow Pointers Initialization: For each starting index, we initiate slow and fast pointers, which represent the current position of each pointer:

   slow, fast = i, next(i)

5. Cycle Detection Loop: Next, we loop to detect cycles using the following conditions:

   • The product of nums[slow] and nums[fast] must be positive, indicating they move in the same direction.
   • The product of nums[slow] and nums[next(fast)] must also be positive, ensuring that the fast pointer also continues in the same direction after two moves.

   while nums[slow] * nums[fast] > 0 and nums[slow] * nums[next(fast)] > 0:
       if slow == fast:  # Pointers meet, indicating a potential cycle
           if slow != next(slow):  # Check to avoid single-length cycle
               return True
           break

6. Marking Elements: If not a valid cycle, we need to mark elements associated with the failed attempt to find a cycle to prevent re-processing them in future iterations. This is achieved by setting each involved element to 0:

   j = i
   while nums[j] * nums[next(j)] > 0:
       nums[j] = 0  # Marking the element
       j = next(j)

7. Final Return: After exhaustively checking all indices, if no cycle is found, the function returns false.

This solution leverages the cyclical two-pointer technique to identify cycles and uses in-place marking to improve efficiency by reducing redundant checks. The use of the modulo operator ensures proper index wrapping within the circular array boundaries, and thorough condition checks maintain consistency in direction for cycle validation.

Example Walkthrough

To illustrate the solution approach using an example, let's consider the circular array nums = [2, -1, 1, 2, 2].

Now let’s walk through the algorithm step by step:

1. Array Length Extraction: The length n of the array nums is 5.

2. Helper Function for Index Calculation: We use the next() function to determine the subsequent index after taking a step in the array. For instance, next(0) would calculate (0 + nums[0]) % 5, which equals 2 % 5, resulting in the next index as 2.

3. Main Loop to Check for Cycles: We start with index i = 0.

4. Fast and Slow Pointers Initialization: At index 0, slow is initiated at 0 and fast is initiated at next(0), which is 2.

5. Cycle Detection Loop:

   • On the first iteration, slow = 0 and fast = 2. We calculate nums[slow] * nums[fast] which is 2 * 1 = 2 (positive, moving in the same direction).
   • slow then moves to next(slow), which is 2, and fast moves to next(next(fast)), first to 4 then wrapped to 1. Both moves are in the forward direction, and nums[slow] is still positive. We check nums[2] * nums[1] which is 1 * (-1) = -1, this indicates a change in direction, so we break out of this loop.

6. Marking Elements: Elements associated with index 0 are not leading to a valid cycle, so they should be marked. However, since the product of nums[j] and nums[next(j)] was not positive, we do not proceed with marking in this iteration.

7. Continuing with the Loop: We now increment i to 1 and continue the process. The array element at index 1 is -1, a backward step, so both slow and fast pointers move backward. If at any time the pointers meet and they have traversed more than a single-element loop (by ensuring slow != next(slow)), it indicates there's a cycle.

   However, in this example, no cycle will be found, and each element that has been verified not to contribute to a cycle will ultimately be marked as 0 to prevent redundant future checks.

8. Final Return: After iterating through the whole array, if no cycle is found, the function returns false. For the given example, since a cycle exists when starting at index 0 where slow pointer would eventually catch up to the fast pointer while maintaining a consistent direction, we would return true.

Please note that this example assumes we did not encounter a valid cycle in the first iteration for illustrative purposes. In reality, once a cycle is detected, we would return true immediately.

Solution Implementation

Time and Space Complexity

The given Python code defines a method for detecting a cycle in a circular array. The cycle must follow certain rules – it cannot consist of a single element looping to itself, and it must maintain a consistent direction (all elements in the cycle are either all positive or all negative).

Time Complexity:

The time complexity of this method is O(n), where n is the length of the array. This results from the fact that each element is visited at most twice – once by the slow pointer and once by the fast pointer. Even though there are nested loops, the inner loop executes a maximum of two times for each element: once by the slow pointer and once by the fast pointer (since we nullify elements once visited to avoid revisiting). Thus, the while loops do not multiply the complexity, but rather, each contributes to the linear visit of each element.

Space Complexity:

The space complexity is O(1) since the algorithm only uses a fixed amount of extra space. Additional variables such as slow, fast, i, and j are used for indexing, and these do not scale with the input size. The computation is done in place, and the input list is modified directly without using any extra space proportional to the input size.

Learn more about how to find time and space complexity quickly using problem constraints.