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 trueif there is a cycle innums, orfalseotherwise.
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:
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2
3
4
5
6
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:
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2
3
4
5
6
7
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).
We want to detect a cycle in a circular array where all elements in the cycle move in the same direction (all positive or all negative) and the cycle length is greater than 1. We use the two pointers (Floyd’s Tortoise and Hare) technique to detect cycles efficiently. To avoid revisiting the same paths, we mark visited elements as 0.
classSolution {
funcircularArrayLoop(nums: IntArray): Boolean {
val n = nums.size
funnext(i: Int): Int = (i + nums[i] % n + n) % n
for (i in0 until n) {
if (nums[i] ==0) continuevar slow = i
var fast = next(i)
while (nums[slow] * nums[fast] > 0&& nums[slow] * nums[next(fast)] > 0) {
if (slow == fast) {
if (slow != next(slow)) returntruebreak }
slow = next(slow)
fast = next(next(fast))
}
var j = i
while (nums[j] * nums[next(j)] > 0) {
nums[j] = 0 j = next(j)
}
}
returnfalse }
}
from typing import List
classSolution:
defcircularArrayLoop(self, nums: List[int]) -> bool:
n = len(nums)
defnext(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] >0and nums[slow] * nums[next(fast)] >0:
if slow == fast:
if slow != next(slow):
returnTruebreak slow, fast = next(slow), next(next(fast))
j = i
while nums[j] * nums[next(j)] >0:
nums[j] =0 j = next(j)
returnFalse