LeetCode #457 — MEDIUM

Circular Array Loop

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

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?

Step 01

Brute Force Baseline

Problem summary: 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. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Two Pointers

Example 1

[2,-1,1,2,2]

Example 2

[-1,-2,-3,-4,-5,6]

Example 3

[1,-1,5,1,4]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #457: Circular Array Loop
class Solution {
    private int n;
    private int[] nums;

    public boolean circularArrayLoop(int[] nums) {
        n = nums.length;
        this.nums = nums;
        for (int i = 0; i < n; ++i) {
            if (nums[i] == 0) {
                continue;
            }
            int slow = i, fast = next(i);
            while (nums[slow] * nums[fast] > 0 && nums[slow] * nums[next(fast)] > 0) {
                if (slow == fast) {
                    if (slow != next(slow)) {
                        return true;
                    }
                    break;
                }
                slow = next(slow);
                fast = next(next(fast));
            }
            int j = i;
            while (nums[j] * nums[next(j)] > 0) {
                nums[j] = 0;
                j = next(j);
            }
        }
        return false;
    }

    private int next(int i) {
        return (i + nums[i] % n + n) % n;
    }
}

// Accepted solution for LeetCode #457: Circular Array Loop
func circularArrayLoop(nums []int) bool {
	for i, num := range nums {
		if num == 0 {
			continue
		}
		slow, fast := i, next(nums, i)
		for nums[slow]*nums[fast] > 0 && nums[slow]*nums[next(nums, fast)] > 0 {
			if slow == fast {
				if slow != next(nums, slow) {
					return true
				}
				break
			}
			slow, fast = next(nums, slow), next(nums, next(nums, fast))
		}
		j := i
		for nums[j]*nums[next(nums, j)] > 0 {
			nums[j] = 0
			j = next(nums, j)
		}
	}
	return false
}

func next(nums []int, i int) int {
	n := len(nums)
	return (i + nums[i]%n + n) % n
}

# Accepted solution for LeetCode #457: Circular Array Loop
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

// Accepted solution for LeetCode #457: Circular Array Loop
struct Solution;

impl Solution {
    fn circular_array_loop(mut nums: Vec<i32>) -> bool {
        let n = nums.len();
        for i in 0..n {
            if Self::next(&nums, i) == i {
                nums[i] = 0;
            }
        }
        for i in 0..n {
            if nums[i] == 0 {
                continue;
            }
            let mut slow = i;
            let mut fast = i;
            while nums[slow] * nums[Self::next(&nums, fast)] > 0
                && nums[slow] * nums[Self::next(&nums, Self::next(&nums, fast))] > 0
            {
                slow = Self::next(&nums, slow);
                fast = Self::next(&nums, Self::next(&nums, fast));
                if slow == fast {
                    return true;
                }
            }
            let mut j = i;
            let val = nums[i];
            while nums[j] * val > 0 {
                let next = Self::next(&nums, j);
                nums[j] = 0;
                j = next;
            }
        }
        false
    }

    fn next(nums: &[i32], index: usize) -> usize {
        let n = nums.len();
        let index = index as i32 + nums[index];
        let index = if index < 0 {
            n as i32 + (index % n as i32)
        } else {
            index % n as i32
        };
        (index as usize) % n
    }
}

#[test]
fn test() {
    let nums = vec![2, -1, 1, 2, 2];
    let res = true;
    assert_eq!(Solution::circular_array_loop(nums), res);
    let nums = vec![-1, 2];
    let res = false;
    assert_eq!(Solution::circular_array_loop(nums), res);
    let nums = vec![-2, 1, -1, -2, -2];
    let res = false;
    assert_eq!(Solution::circular_array_loop(nums), res);
    let nums = vec![-1];
    let res = false;
    assert_eq!(Solution::circular_array_loop(nums), res);
}

// Accepted solution for LeetCode #457: Circular Array Loop
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #457: Circular Array Loop
// class Solution {
//     private int n;
//     private int[] nums;
// 
//     public boolean circularArrayLoop(int[] nums) {
//         n = nums.length;
//         this.nums = nums;
//         for (int i = 0; i < n; ++i) {
//             if (nums[i] == 0) {
//                 continue;
//             }
//             int slow = i, fast = next(i);
//             while (nums[slow] * nums[fast] > 0 && nums[slow] * nums[next(fast)] > 0) {
//                 if (slow == fast) {
//                     if (slow != next(slow)) {
//                         return true;
//                     }
//                     break;
//                 }
//                 slow = next(slow);
//                 fast = next(next(fast));
//             }
//             int j = i;
//             while (nums[j] * nums[next(j)] > 0) {
//                 nums[j] = 0;
//                 j = next(j);
//             }
//         }
//         return false;
//     }
// 
//     private int next(int i) {
//         return (i + nums[i] % n + n) % n;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.