How do you solve Circular Array Loop on LeetCode?

Circular Array Loop (LeetCode #457) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The direction of movement must remain consistent for a valid cycle.

What is the brute force solution for Circular Array Loop?

The brute force approach for Circular Array Loop is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible starting index to see if it can form a cycle. We simulate the movement based on the values in the array and track visited indices until we either find a cycle or exhaust all possibilities. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Circular Array Loop?

Circular Array Loop uses the following concepts: Array, Hash Table, Two Pointers. The recommended patterns to study are: Two Pointers, Cycle Detection.

What are the interview tips for Circular Array Loop?

Clarify the problem statement and edge cases before jumping into coding. Explain your thought process while coding to demonstrate your understanding. Practice similar problems to get comfortable with cycle detection techniques.

What are common mistakes when solving Circular Array Loop?

Not considering the direction of movement when checking for cycles. Assuming a cycle exists if two pointers meet without checking the length of the cycle.

#457

Circular Array Loop

Medium

ArrayHash TableTwo PointersTwo PointersCycle Detection

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a modified version of Floyd's Tortoise and Hare algorithm to detect cycles. It efficiently checks for cycles without needing extra space for tracking visited indices, leveraging the properties of the circular array.

⚙️

Algorithm

3 steps

1Step 1: Iterate through each index in the array as a potential starting point.
2Step 2: For each starting index, use two pointers (slow and fast) to traverse the array according to the movement rules.
3Step 3: If slow and fast meet at the same index and the cycle length is greater than 1, return true. If the direction of movement changes, break and start from the next index.

solution.py16 lines

1def circularArrayLoop(nums):
2    n = len(nums)
3    for i in range(n):
4        slow, fast = i, i
5        direction = nums[i] > 0
6        while True:
7            slow = (slow + nums[slow]) % n
8            fast = (fast + nums[fast]) % n
9            fast = (fast + nums[fast]) % n
10            if slow == fast:
11                if slow != (slow + nums[slow]) % n:
12                    return True
13                break
14            if (nums[slow] > 0) != direction or (nums[fast] > 0) != direction:
15                break
16    return False

ℹ

Complexity note: This complexity is achieved because each index is processed at most twice (once by slow and once by fast), leading to a linear time complexity.

1The direction of movement must remain consistent for a valid cycle.
2Using two pointers can significantly reduce the time complexity compared to tracking visited indices.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.