456-132-pattern

Try it on leetcode

Description

Given an array of n integers nums, a 132 pattern is a subsequence of three integers nums[i], nums[j] and nums[k] such that i < j < k and nums[i] < nums[k] < nums[j].

Return true if there is a 132 pattern in nums, otherwise, return false.

 

Example 1:

Input: nums = [1,2,3,4]
Output: false
Explanation: There is no 132 pattern in the sequence.

Example 2:

Input: nums = [3,1,4,2]
Output: true
Explanation: There is a 132 pattern in the sequence: [1, 4, 2].

Example 3:

Input: nums = [-1,3,2,0]
Output: true
Explanation: There are three 132 patterns in the sequence: [-1, 3, 2], [-1, 3, 0] and [-1, 2, 0].

 

Constraints:

• n == nums.length
• 1 <= n <= 2 * 105
• -109 <= nums[i] <= 109

Solution(Python)

class Solution:
    def find132pattern(self, nums: List[int]) -> bool:
        return self.arrayasstack(nums)

    # Time Compleity :O(n^3)
    # space Complexity: O(1)
    def bruteforce(self, nums):
        n = len(nums)

        for i in range(n):
            for j in range(i + 1, n):
                for k in range(j + 1, n):
                    if nums[i] < nums[k] < nums[j]:
                        return True

        return False

    # Time Compleity :O(n^2)
    # space Complexity: O(1)
    def betterbruteforce(self, nums):
        n = len(nums)
        min_i = inf

        for j in range(n):
            if nums[j] < min_i:
                min_i = nums[j]
            for k in range(j + 1, n):
                if min_i < nums[k] < nums[j]:
                    return True
        return False

    # Time Compleity :O(n^2)
    # space Complexity: O(n)
    def interval(self, nums):
        n = len(nums)
        intervals = []
        min_point_after_last_peak_index = 0
        for i in range(1, n):
            if nums[i] < nums[i - 1]:
                if min_point_after_last_peak_index < i - 1:
                    intervals.append(
                        nums[min_point_after_last_peak_index], nums[i - 1])
            min_point_after_last_peak_index = i
            for i_num, j_num in intervals:
                if i_num < nums[i] < j_num:
                    return True
        return False

    # Time Compleity :O(n)
    # space Complexity: O(n)
    def stack(self, nums):
        n = len(nums)

        if n < 3:
            return False

        stack = []
        min_array = [-1] * n

        min_array[0] = nums[0]

        for i in range(1, n):
            min_array[i] = min(min_array[i - 1], nums[i])

        for j in range(n - 1, -1, -1):
            if nums[j] <= min_array[j]:
                continue
            while stack and stack[-1] <= min_array[j]:
                stack.pop()

            if stack and min_array[j] < stack[-1] < nums[j]:
                return True
            stack.append(nums[j])

        return False

    # Time Compleity :O(nlogn)
    # space Complexity: O(n)
    def binarysearch(self, nums):
        n = len(nums)

        if n < 3:
            return False

        min_array = [-1] * n
        min_array[0] = nums[0]
        min_array[0] = nums[0]

        for i in range(1, n):
            min_array[i] = min(min_array[i - 1], nums[i])

        k = n

        for j in range(n - 1, -1, -1):
            if nums[j] <= min_array[j]:
                continue
            k = bisect_left(nums, min_array[j] + 1, k, n)
            if k < n and min_array[j] < nums[k] < nums[j]:
                return True

            k -= 1
            nums[k] = nums[j]
        return False

    # Time Compleity :O(n)
    # space Complexity: O(n)
    def arrayasstack(self, nums):
        if len(nums) < 3:
            return False
        min_array = [-1] * len(nums)
        min_array[0] = nums[0]
        for i in range(1, len(nums)):
            min_array[i] = min(min_array[i - 1], nums[i])

        k = len(nums)
        for j in range(len(nums) - 1, -1, -1):
            if nums[j] <= min_array[j]:
                continue
            while k < len(nums) and nums[k] <= min_array[j]:
                k += 1
            if k < len(nums) and nums[k] < nums[j]:
                return True
            k -= 1
            nums[k] = nums[j]
        return False