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author | jack <0x6A73@pm.me> | 2024-06-14 13:13:15 -0600 |
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committer | jack <0x6A73@pm.me> | 2024-06-14 13:13:15 -0600 |
commit | d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch) | |
tree | 96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /count_num_nice_subarr.py | |
download | leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.tar.gz leetcode-d343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa.zip |
initial
Diffstat (limited to 'count_num_nice_subarr.py')
-rw-r--r-- | count_num_nice_subarr.py | 34 |
1 files changed, 34 insertions, 0 deletions
diff --git a/count_num_nice_subarr.py b/count_num_nice_subarr.py new file mode 100644 index 0000000..de30423 --- /dev/null +++ b/count_num_nice_subarr.py @@ -0,0 +1,34 @@ +# 1248. Count Number of Nice Subarrays + +""" +given an array of integers 'nums' and an integers 'k'. a continuous subarray +is called nice if there are 'k' odd numbers on it. return the number of nice +subarrays. +""" + + +class Solution(object): + def numberOfSubarrays(self, nums, k): + """ + :type nums: List[int] + :type k: int + :rtype: int + """ + i, cnt, ans = 0, 0, 0 + for j in range(len(nums)): + if nums[j] & 1: + k -= 1 + cnt = 0 + while k == 0: + k += nums[i] & 1 + i += 1 + cnt += 1 + ans += cnt + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.numberOfSubarrays(nums=[1, 1, 2, 1, 1], k=3)) + print(obj.numberOfSubarrays(nums=[2, 4, 6], k=1)) + print(obj.numberOfSubarrays(nums=[2, 2, 2, 1, 2, 2, 1, 2, 2, 2], k=2)) |