457. Circular Array Loop

Problem

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 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:
There is a cycle from index 0 -> 2 -> 3 -> 0 -> ...
The cycle's length is 3.

Example 2:

Input: nums = [-1,2]
Output: false
Explanation:
The sequence from index 1 -> 1 -> 1 -> ... is not a cycle because the sequence's length is 1.
By definition the sequence's length must be strictly greater than 1 to be a cycle.

Example 3:

Input: nums = [-2,1,-1,-2,-2]
Output: false
Explanation:
The sequence from index 1 -> 2 -> 1 -> ... is not a cycle because nums[1] is positive, but nums[2] is negative.
Every nums[seq[j]] must be either all positive or all negative.

  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?

Solution (Java)

class Solution {
    public boolean circularArrayLoop(int[] arr) {
        int n = arr.length;
        Map<Integer, Integer> map = new HashMap<>();
        // arr[i]%n==0 (cycle)
        for (int start = 0; start < n; start++) {
            if (map.containsKey(start)) {
                // check if already visited
                continue;
            }
            int curr = start;
            // Check if the current index is valid
            while (isValidCycle(start, curr, arr)) {
                // marking current index visited and set value as start of loop
                map.put(curr, start);
                // steps to jump;
                int jump = arr[curr] % n;
                // Jumping x steps backwards is same as jumping (n-x) steps forward
                // going to next index;
                curr = (curr + jump + n) % n;
                if (map.containsKey(curr)) {
                    // value already processed
                    if (map.get(curr) == start) {
                        // If equal to start then we have found a loop
                        return true;
                    }
                    // Else we can break since this has already been visited hence we will get the
                    // same result as before
                    break;
                }
            }
        }
        return false;
    }

    private boolean isValidCycle(int start, int curr, int[] arr) {
        return (arr[start] <= 0 || arr[curr] >= 0)
                && (arr[start] >= 0 || arr[curr] <= 0)
                && (arr[curr] % arr.length != 0);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).