LeetCode 1437. Check If All 1's Are at Least Length K Places Away (Medium)

Given an array nums of 0s and 1s and an integer k, return True if all 1's are at least k places away from each other, otherwise return False.

 

Example 1:

Input: nums = [1,0,0,0,1,0,0,1], k = 2
Output: true
Explanation: Each of the 1s are at least 2 places away from each other.

Example 2:

Input: nums = [1,0,0,1,0,1], k = 2
Output: false
Explanation: The second 1 and third 1 are only one apart from each other.

Example 3:

Input: nums = [1,1,1,1,1], k = 0
Output: true

Example 4:

Input: nums = [0,1,0,1], k = 1
Output: true

 

Constraints:

• 1 <= nums.length <= 10^5
• 0 <= k <= nums.length
• nums[i] is 0 or 1

方法：指针

思路：用一根指针去记录两个相邻的1之间的距离。如果这个距离小于k，那么return False。要注意的是，距离等于两个1的index差值再减1

time complexity: O(n) space complexity: O(1) 

class Solution:
    def kLengthApart(self, nums: List[int], k: int) -> bool:
        if k == 0: return True 
        
        pre = None 
        for i, num in enumerate(nums):
            if num == 1:
                if pre is None:
                    pre = i 
                else:
                    if i - pre -1 < k:
                        return False 
                    pre = i 
                    
        return True