Technical Interview Question

Problem: Circular Array Loop

You are given an array of integers called nums where each element represents the maximum number of steps that can be jumped going forward from that element. Write a function circularArrayLoop(nums) that determines if there is a loop (or cycle) in the array where the sequence of jumps forms a circular loop. A circular loop is a sequence of elements in the array that starts and ends at the same index and follows the direction of the elements' values. The length of the loop must be greater than 1.

Implement the circularArrayLoop function and return True if there is a circular loop in nums, or False otherwise.

Example:

Input: nums = [2,-1,1,2,2]

Output: True

Explanation: There is a circular loop in the array. The loop starts at index 0 -> 2 -> 3 -> 0 -> ...

Note:

• The given array nums will not be empty and will not contain elements that are 0.

Answer

To solve this problem, we can use the "two pointers" algorithm. We'll start with two pointers: one slow pointer that moves one step at a time, and one fast pointer that moves two steps at a time. We'll keep track of whether the loop is clockwise or counterclockwise.

Here's the step-by-step algorithm for the circularArrayLoop function:

1. Create a function getNextIndex(arr, currentIndex) that takes the array arr and the current index currentIndex as parameters and returns the next index based on the element at the current index.

2. Create a function isSameDirection(a, b) that takes two integers a and b as parameters and returns True if both a and b have the same sign, indicating that they are moving in the same direction.

3. Iterate through each element in the array using a for loop:

   a. Initialize the slow pointer (slow) and fast pointer (fast) at the current index i.

   b. Check if the loop is clockwise or counterclockwise. Set a boolean variable isForward to True if the current element is positive, indicating clockwise direction, and False otherwise.

   c. While the slow and fast pointers are not equal and both indices are not pointing to the same value:

   i. Move the slow pointer to the next index by calling the getNextIndex function.

   ii. Move the fast pointer to the next index twice by calling the getNextIndex function twice.

   iii. If the loop is not clockwise, update the isForward variable to False if the next element is negative.

   iv. Check if the slow and fast pointers have the same value, indicating the presence of a loop.

     - If they have the same value, return `True` since we have found a circular loop.
   
     - If they have different values, but the direction of the loop is different than the `isForward` variable, break out of the loop since it's not a valid circular loop.
   

   d. Reset the visited elements in this attempt to zero by iterating through all the elements in the loop again, starting from the current index i, and changing their values to 0.

4. If no circular loop is found after iterating through every element, return False.

Here's the implementation of the circularArrayLoop function in Python:

def circularArrayLoop(nums):
    n = len(nums)
    
    def getNextIndex(arr, currentIndex):
        nextIndex = (currentIndex + arr[currentIndex]) % n
        if nextIndex < 0:
            nextIndex += n
        return nextIndex
    
    def isSameDirection(a, b):
        return (a >= 0 and b >= 0) or (a < 0 and b < 0)
    
    for i, num in enumerate(nums):
        if num == 0:
            continue
        
        slow = fast = i
        isForward = num > 0
        
        while True:
            slow = getNextIndex(nums, slow)
            fast = getNextIndex(nums, fast)
            if isForward != isSameDirection(nums[fast], nums[i]):
                break
            fast = getNextIndex(nums, fast)
            if isForward != isSameDirection(nums[fast], nums[i]):
                break
            if slow == fast:
                return True
        
        j = i
        while isSameDirection(nums[j], nums[i]):
            nextIndex = getNextIndex(nums, j)
            nums[j] = 0
            j = nextIndex
    
    return False

The time complexity of this solution is O(n), where n is the size of the input array nums. This is because we iterate through every element of the array once. The space complexity is O(1) since we use a constant amount of extra space to store variables.