Package g0401_0500.s0457_circular_array_loop

Class Solution

  • public class Solution
extends Object
    457 - Circular Array Loop.

    Medium

    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?

    • Constructor Detail

      • Solution

        public Solution()

    • Method Detail

      • circularArrayLoop

        public boolean circularArrayLoop​(int[] arr)