Leetcode #457: Circular Array Loop
In this guide, we solve Leetcode #457 Circular Array Loop in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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 abs(nums[i]) steps backward.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Two Pointers
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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).
Python Solution
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
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.