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This approach involves iterating through the array and keeping track of the most recent index where a '1' has been encountered. For every '1' encountered, check if the distance from the last '1' is at least 'k'. If any distance is found to be less than 'k', return false. If the loop ends without returning false, return true.
Time Complexity: O(n), where n is the number of elements in nums.
Space Complexity: O(1), as only a few variables are used.
1def kLengthApart(nums, k):
2 lastIndex = -1
3 for i in range(len(nums)):
4 if nums[i] == 1:
5 if lastIndex != -1 and i - lastIndex - 1 < k:
6 return False
7 lastIndex = i
8 return True
9The Python solution employs a similar logic of iterating the entire array and checking the distances between the indices of the '1's.
This method includes scanning the array to compute gaps between '1's using a counter. If the gap calculated is ever less than k while traversing, return false. This process continues until the scan ends, signifying all gaps are adequate, thus returning true.
Time Complexity: O(n), with n being the size of nums array.
Space Complexity: O(1), utilizing constant space to track count.
The solution checks each element, increasing count when a '0' is found and resetting or checking it when a '1' appears. If count is ever less than k upon seeing '1', false is returned instantly.