Solution Explanation

This problem asks whether all occurrences of the digit 1 in a binary array are separated by at least k zeros. The solution uses a single pass through the array to efficiently determine this.

Approach:

The core idea is to track the index j of the last encountered 1. For each element, we check:

1. Is it a 1? If so, we calculate the distance between the current index i and the previously seen 1's index j. If this distance (minus 1 to exclude the 1s themselves) is less than k, it means the 1s are too close, and we return false. Otherwise, we update j to the current index i.

2. Is it a 0? If it's a 0, we simply continue iterating.

If the loop completes without finding any 1s that are too close, we return true.

Time Complexity Analysis:

The solution iterates through the input array nums exactly once. Therefore, the time complexity is O(n), where n is the length of the array.

Space Complexity Analysis:

The solution uses a constant amount of extra space to store the variables i, j, and k. Therefore, the space complexity is O(1), which is constant space.

Code Explanation (Python):

class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        j = -float('inf')  # Initialize j to negative infinity. This handles the case of the first 1.
        for i, x in enumerate(nums):
            if x: #if x ==1
                if i - j - 1 < k: #check if the distance between the current 1 and the previous 1 is less than k
                    return False #if yes return false
                j = i #update j to current index if the distance is greater than k
        return True #return true if all 1s are at least k places away

The other languages (Java, C++, Go, TypeScript) follow the same logic, with minor syntactic variations. The key idea of tracking the last seen 1 and efficiently checking the distance remains consistent across all implementations.