Medium
Number of Subarrays with Bounded Maximum — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(1)
class Solution(object):
def numSubarrayBoundedMax(self, A, L, R):
"""
:type A: List[int]
:type L: int
:type R: int
:rtype: int
"""
def count(A, bound):
result, curr = 0, 0
for i in A :
curr = curr + 1 if i <= bound else 0
result += curr
return result
return count(A, R) - count(A, L-1)