LeetCode 456. 132 Pattern

最新推荐文章于 2022-11-30 22:05:20 发布

junchen1992

最新推荐文章于 2022-11-30 22:05:20 发布

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Problem Statement

(Source) Given a sequence of n integers $a_1, a_2, ..., a_n$ , a 132 pattern is a subsequence $a_i, a_j, a_k$ such that i < j < k and $a_i < a_k < a_j$ . Design an algorithm that takes a list of n numbers as input and checks whether there is a 132 pattern in the list.

Note: n will be less than 15,000.

Example 1:

Input: [1, 2, 3, 4]

Output: False

Explanation: There is no 132 pattern in the sequence.

Example 2:

Input: [3, 1, 4, 2]

Output: True

Explanation: There is a 132 pattern in the sequence: [1, 4, 2].

Example 3:

Input: [-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].

Solution - java.TreeSet

The idea is to first construct a prefix-min array (as I name it): left[i] stores the minimal value from position 0 to i (inclusive). Then scan the original array from end to start and check if all scanned elements contains some element x that is larger than left[i] but smaller than nums[i]. If there exists such an element, return true. Otherwise, it will return false at the end.

Java TreeSet comes very handy for this question :)

public class Solution {
    public boolean find132pattern(int[] nums) {
        int n = nums.length;
        if (n < 3) {
            return false;
        }

        int[] left = new int[n];
        left[0] = nums[0];
        for (int i = 1; i < n; i++) {
            left[i] = left[i-1] > nums[i] ? nums[i] : left[i-1];
        }

        TreeSet<Integer> set = new TreeSet<>();
        for (int i = n - 1; i >= 0; i--) {
            set.add(nums[i]);
            if (!set.headSet(nums[i], false).tailSet(left[i], false).isEmpty()) {
                return true;
            }
        }
        return false;
    }
}

If using Python with the same idea, will get an TLE:

class Solution(object):
    def find132pattern(self, nums):
        """
        :type nums: List[int]
        :rtype: bool
        """
        n = len(nums)
        if n < 3:
            return False

        left = [nums[0]] * n
        for i in xrange(1, n):
            left[i] = min(left[i-1], nums[i])
        st = set()
        for i in xrange(n - 1, -1, -1):
            st.add(nums[i])
            if any(left[i] < x < nums[i] for x in st):
                return True
        return False