LeetCode - Circular Array Loop

Question Definition

You are given an array of positive and negative integers. If a number n at an index is positive, then move forward n steps. Conversely, if it’s negative (-n), move backward n steps. Assume the first element of the array is forward next to the last element, and the last element is backward next to the first element. Determine if there is a loop in this array. A loop starts and ends at a particular index with more than 1 element along the loop. The loop must be “forward” or “backward’.

Example 1: Given the array [2, -1, 1, 2, 2], there is a loop, from index 0 -> 2 -> 3 -> 0.

Example 2: Given the array [-1, 2], there is no loop.

Note: The given array is guaranteed to contain no element “0”.

Can you do it in O(n) time complexity and O(1) space complexity?

Java Solution

public boolean circularArrayLoop(int[] nums) {
    if(nums == null || nums.length == 0) return false;
    int n = nums.length;
    for(int i=0; i<n; i++){
        if(nums[i] == 0) continue;
        int slow = i, fast = i, count = 0;
        boolean forward = nums[slow] > 0;
        do{
            int tempSlow = slow;
            slow = (slow + nums[slow] + n) % n;

            if(forward && nums[fast] < 0 || !forward && nums[fast] > 0) return false;
            fast = (fast + nums[fast] + n) % n;

            if(forward && nums[fast] < 0 || !forward && nums[fast] > 0) return false;
            fast = (fast + nums[fast] + n) % n;

            nums[tempSlow] = 0;
            count++;

        } while(slow != fast);
        if(count > 1) return true;
    }
    return false;
}