In this guide, you will get 457. Circular Array Loop LeetCode Solution with the best time and space complexity. The solution to Circular Array Loop problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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: 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).
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:
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).
Follow up: Could you solve it in O(n) time complexity and O(1) extra space complexity?
Complexity Analysis
Time Complexity: O(n)
Space Complexity: O(1)
457. Circular Array Loop LeetCode Solution in C++
class Solution {
public:
bool circularArrayLoop(vector<int>& nums) {
const int n = nums.size();
if (n < 2)
return false;
auto advance = [&](int i) {
const int val = (i + nums[i]) % n;
return i + nums[i] >= 0 ? val : n + val;
};
for (int i = 0; i < n; ++i) {
if (nums[i] == 0)
continue;
int slow = i;
int fast = advance(slow);
while (nums[i] * nums[fast] > 0 && nums[i] * nums[advance(fast)] > 0) {
if (slow == fast) {
if (slow == advance(slow))
break;
return true;
}
slow = advance(slow);
fast = advance(advance(fast));
}
slow = i;
const int sign = nums[i];
while (sign * nums[slow] > 0) {
const int next = advance(slow);
nums[slow] = 0;
slow = next;
}
}
return false;
}
}; /* code provided by PROGIEZ */
457. Circular Array Loop LeetCode Solution in Java
class Solution {
public boolean circularArrayLoop(int[] nums) {
if (nums.length < 2)
return false;
for (int i = 0; i < nums.length; ++i) {
if (nums[i] == 0)
continue;
int slow = i;
int fast = advance(nums, slow);
while (nums[i] * nums[fast] > 0 && nums[i] * nums[advance(nums, fast)] > 0) {
if (slow == fast) {
if (slow == advance(nums, slow))
break;
return true;
}
slow = advance(nums, slow);
fast = advance(nums, advance(nums, fast));
}
slow = i;
final int sign = nums[i];
while (sign * nums[slow] > 0) {
final int next = advance(nums, slow);
nums[slow] = 0;
slow = next;
}
}
return false;
}
private int advance(int[] nums, int i) {
final int n = nums.length;
final int val = (i + nums[i]) % n;
return i + nums[i] >= 0 ? val : n + val;
}
} // code provided by PROGIEZ
457. Circular Array Loop LeetCode Solution in Python
class Solution:
def circularArrayLoop(self, nums: list[int]) -> bool:
def advance(i: int) -> int:
return (i + nums[i]) % len(nums)
if len(nums) < 2:
return False
for i, num in enumerate(nums):
if num == 0:
continue
slow = i
fast = advance(slow)
while num * nums[fast] > 0 and num * nums[advance(fast)] > 0:
if slow == fast:
if slow == advance(slow):
break
return True
slow = advance(slow)
fast = advance(advance(fast))
slow = i
sign = num
while sign * nums[slow] > 0:
next = advance(slow)
nums[slow] = 0
slow = next
return False # code by PROGIEZ
