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, movenums[i]
steps forward, and - If
nums[i]
is negative, movenums[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?
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
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;
}
}
class Solution {
public:
bool circularArrayLoop(vector<int>& nums) {
int n = nums.size();
for (int i = 0; i < n; ++i) {
if (!nums[i]) continue;
int slow = i, fast = next(nums, i);
while (nums[slow] * nums[fast] > 0 && nums[slow] * nums[next(nums, fast)] > 0) {
if (slow == fast) {
if (slow != next(nums, slow)) return true;
break;
}
slow = next(nums, slow);
fast = next(nums, next(nums, fast));
}
int j = i;
while (nums[j] * nums[next(nums, j)] > 0) {
nums[j] = 0;
j = next(nums, j);
}
}
return false;
}
int next(vector<int>& nums, int i) {
int n = nums.size();
return (i + nums[i] % n + n) % n;
}
};
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
}