LC-2762 Medium LeetCode

2762. Continuous Subarrays

Array Queue Sliding Window Heap (Priority Queue) Ordered Set Monotonic Queue

Read the full problem statement on LeetCode.

Difficulty: medium Acceptance: 58% Topics: Array, Queue, Sliding Window, Heap (Priority Queue), Ordered Set, Monotonic Queue

View full problem on LeetCode

Reading material

Arrays & Hashing 12 min Stacks & Queues 11 min Sliding Window 11 min

Reference solution (spoiler · python)

# Time:  O(n)
# Space: O(1)

import collections


# two pointers
class Solution(object):
    def continuousSubarrays(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        result = left = 0
        mn, mx = float("inf"), float("-inf")
        for right in xrange(len(nums)):
            if mn <= nums[right] <= mx:
                mn, mx = max(mn, nums[right]-2), min(mx, nums[right]+2)
            else:
                mn, mx = nums[right]-2, nums[right]+2
                for left in reversed(xrange(right)):
                    if not mn <= nums[left] <= mx:
                        break
                    mn, mx = max(mn, nums[left]-2), min(mx, nums[left]+2)
                else:
                    left = -1
                left += 1
            result += right-left+1
        return result
    

# Time:  O(n)
# Space: O(n)
import collections


# mono deque, two pointers
class Solution2(object):
    def continuousSubarrays(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        mn, mx = collections.deque(), collections.deque()
        result = left = 0
        for right in xrange(len(nums)):
            while mn and nums[mn[-1]] > nums[right]:
                mn.pop()
            mn.append(right)
            while mx and nums[mx[-1]] < nums[right]:
                mx.pop()
            mx.append(right)
            while not nums[right]-nums[mn[0]] <= 2:
                left = max(left, mn.popleft()+1)
            while not nums[mx[0]]-nums[right] <= 2:
                left = max(left, mx.popleft()+1)
            result += right-left+1
        return result


# Time:  O(nlogn)
# Space: O(n)
from sortedcontainers import SortedDict


# ordered dict, two pointers
class Solution3(object):
    def continuousSubarrays(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        result = left = 0
        lookup = SortedDict()
        for right in xrange(len(nums)):
            lookup[nums[right]] = right
            to_del = []
            for x, i in lookup.items():
                if nums[right]-x <= 2:
                    break
                left = max(left, i+1)
                to_del.append(x)
            for x, i in reversed(lookup.items()):
                if x-nums[right] <= 2:
                    break
                left = max(left, i+1)
                to_del.append(x) 
            for x in to_del:
                del lookup[x]
            result += right-left+1
        return result

Solution from kamyu104/LeetCode-Solutions · MIT