Increasing triplet subsequence

Time: O(N); Space: O(1); medium

Given an unsorted array return whether an increasing subsequence of length 3 exists or not in the array.

Formally the function should: * Return true if there exists i, j, k such that arr[i] < arr[j] < arr[k] given 0 <= i < j < k <= n-1 else return false.

Your algorithm should run in O(N) time complexity and O(1) space complexity.

Example 1:

Input: nums = [1, 2, 3, 4, 5]

Output: True

Example 2:

Input: nums = [5, 4, 3, 2, 1]

Output: False

class Solution1(object):
    def increasingTriplet(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        min_num, a, b = float("inf"), float("inf"), float("inf")
        for c in nums:
            if min_num >= c:
                min_num = c
            elif b >= c:
                a, b = min_num, c
            else:  # a < b < c
                return True
        return False

s = Solution1()
nums = [1, 2, 3, 4, 5]
assert s.increasingTriplet(nums) == True
nums = [5, 4, 3, 2, 1]
assert s.increasingTriplet(nums) == False

import bisect

class Solution_Generalization(object):
    """
    Time: O(N * logk); Space: O(k)
    Generalization of k-uplet.
    """
    def increasingTriplet(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        def increasingKUplet(nums, k):
            inc = [float('inf')] * (k - 1)
            for num in nums:
                i = bisect.bisect_left(inc, num)
                if i >= k - 1:
                    return True
                inc[i] = num
            return k == 0

        return increasingKUplet(nums, 3)

s = Solution_Generalization()
nums = [1, 2, 3, 4, 5]
assert s.increasingTriplet(nums) == True
nums = [5, 4, 3, 2, 1]
assert s.increasingTriplet(nums) == False

See also:

https://leetcode.com/problems/increasing-triplet-subsequence

https://www.lintcode.com/problem/increasing-triplet-subsequence