algoadvance

Given a circular array nums of integers, determine if there is a loop in nums. A loop must start and end at the same index, and the loop’s length is at least 2. The movement for a loop is defined in the following way:

• A move from index i to index j is determined by j = (i + nums[i]) % n, where n is the length of the array.
• A loop must be either entirely positive or entirely negative, meaning you can only move forwards or backwards through the array depending on the sign of nums[i].

Return True if there is a loop in nums, or False otherwise.

Clarifying Questions

1. Are there any constraints on the values in the nums array?
   • There are no explicit constraints on values in the nums array from the problem statement. They can be positive or negative.
2. Can the values within nums be zero?
   • We will assume values can be zero, but a zero value at any index means we cannot proceed forward from that index.
3. How large can the input array nums be?
   • Although not specified, we will assume typical constraints for competitive programming, potentially up to 10^5 elements.
4. Is the array guaranteed to be non-null?
   • Yes, the input will be a properly initialized array.

Let’s proceed with the solution now.

Strategy

1. Mark Visitation: We need to keep track of visited elements to avoid infinite loops and ensure we are not retracing our steps.
2. Two Pointers: Use a slow and fast pointer (Floyd’s Cycle Detection Algorithm) to detect cycles:
   • Slow pointer moves one step at a time.
   • Fast pointer moves two steps at a time.
   • If both pointers meet, a cycle is detected.
3. Direction Consistency: Ensure that the cycle maintains consistent direction (all positive or all negative values).

Code

def circularArrayLoop(nums):
    def next_index(i):
        n = len(nums)
        return (i + nums[i]) % n  # ensures circular indexing

    n = len(nums)
    if n < 2:
        return False

    for i in range(n):
        if nums[i] == 0:
            continue

        slow, fast = i, next_index(i)
        
        # Detect loop using two pointers.
        while (nums[fast] * nums[i] > 0) and (nums[next_index(fast)] * nums[i] > 0):
            if slow == fast:
                if slow == next_index(slow):  # Check for one-element loop
                    break
                return True
                
            slow = next_index(slow)
            fast = next_index(next_index(fast))

        # If no loop found, set all elements along the path to 0 to mark them.
        slow = i
        sign = nums[i]
        while nums[slow] * sign > 0:
            next_i = next_index(slow)
            nums[slow] = 0
            slow = next_i

    return False

Time Complexity

• Time Complexity: O(n)
  • Each element is visited at most twice.
  • Initial traversal to find loops and marking non-loop elements.
• Space Complexity: O(1)
  • No extra space used except for pointers.

This solution uses Floyd’s Cycle Detection Algorithm in an array context with boundary conditions for circular indexing and ensuring direction consistency.

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