2419. Longest Subarray With Maximum Bitwise AND

Medium


You are given an integer array nums of size n.

Consider a non-empty subarray from nums that has the maximum possible bitwise AND.

  • In other words, let k be the maximum value of the bitwise AND of any subarray of nums. Then, only subarrays with a bitwise AND equal to k should be considered.

Return the length of the longest such subarray.

The bitwise AND of an array is the bitwise AND of all the numbers in it.

A subarray is a contiguous sequence of elements within an array.

 

Example 1:

Input: nums = [1,2,3,3,2,2]
Output: 2
Explanation:
The maximum possible bitwise AND of a subarray is 3.
The longest subarray with that value is [3,3], so we return 2.

Example 2:

Input: nums = [1,2,3,4]
Output: 1
Explanation:
The maximum possible bitwise AND of a subarray is 4.
The longest subarray with that value is [4], so we return 1.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 106




 class Solution:
    def longestSubarray(self, nums: List[int]) -> int:

#         count = 1
#         max_num = nums[0] 
#         mm = 1
#         sub_array = True

#         for i, num in enumerate(nums[1:]):

#             if num == max_num and sub_array:
#                 count+=1
#                 mm = max(count, mm)
#                 #sub_array = False
#                 #print(count)

#             elif num >= max_num:
#                 if num > max_num:
#                     mm = 1
#                 max_num = num
#                 sub_array = True
#                 count = 1

#             else:
#                 sub_array = False

#         return mm

        max_num = max(nums)
        max_count = 0
        for x, y in groupby(nums):
            if x == max_num:
                max_count = max(max_count, len(list(y)))
        return max_count

Random Note


From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict