動機

atMost或是在回收時計算長度

Problem

Given an array of integers nums and an integer k. A continuous subarray is called nice if there are k odd numbers on it.

Return the number of nice sub-arrays.

 

Example 1:

Input: nums = [1,1,2,1,1], k = 3Output: 2Explanation: The only sub-arrays with 3 odd numbers are [1,1,2,1] and [1,2,1,1].

Example 2:

Input: nums = [2,4,6], k = 1Output: 0Explanation: There is no odd numbers in the array.

Example 3:

Input: nums = [2,2,2,1,2,2,1,2,2,2], k = 2Output: 16

 

Constraints:

  • 1 <= nums.length <= 50000
  • 1 <= nums[i] <= 10^5
  • 1 <= k <= nums.length

Sol

class Solution:
    def numberOfSubarrays(self, nums: List[int], k: int) -> int:
        def atmost(limit):
            i = cnt = ret = 0
            for j,n in enumerate(nums):
                if n % 2 == 1:
                    cnt += 1
                while i <= j and cnt > limit:
                    if nums[i] % 2 == 1:
                        cnt -= 1
                    i += 1
                ret += j-i+1
            return ret
        return atmost(k)-atmost(k-1)

另一個是在回收時計算長度

def numberOfSubarrays(self, A, k):
    i = count = res = 0
    for j in range(len(A)):
        if A[j] & 1:
            k -= 1
            count = 0 # 等累積到合法時才計算長度
        while k == 0:
            k += A[i] & 1
            i += 1
            count += 1
        res += count
    return res