Formatted question description: https://leetcode.ca/all/456.html

456. 132 Pattern (Medium)

Given a sequence of n integers a1, a2, ..., an, a 132 pattern is a subsequence ai, aj, ak such that i < j < k and ai < ak < aj. 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].

Related Topics:
Stack

Solution 1. Multiset

For a given middle element A[i], we’d want to use the minimum element to the left of A[i] (denote as a), and then find the greatest elemnt to the right of A[i] which is smaller than A[i] (denote as b).

For a it’s easy: we can just use a single variable to keep track of the minimum elemenent we’ve seen thus far.

For b, we can use a multiset to hold all the values to the right of A[i] in descending order, and use binary search to find the greatest element that is smaller than A[i].

// OJ: https://leetcode.com/problems/132-pattern/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    bool find132pattern(vector<int>& A) {
        multiset<int, greater<int>> s(begin(A), end(A));
        int mn = INT_MAX;
        for (int n : A) {
            mn = min(mn, n);
            s.erase(s.find(n));
            auto it = s.upper_bound(n);
            if (it != s.end() && mn < *it) return true;
        }
        return false;
    }
};

Solution 2. Monotonic Stack

Assume the sequence is a, b, c (a < c < b).

We scan from the right to the left. Keep a monotonically decreasing stack s and keep track of the best candidate of c in a variable right.

The best candidate of c must be monotonically increasing because the moment we find that the current element is < c, we’ve found a valid sequence.

As for the stack s, we pop all the elements that are smaller than A[i], store the last popped element in right, and then push A[i] into the stack. Thus s.top() and right forms a best candidate b, c sequence.

Once we find A[i] < right, we can return true.

// OJ: https://leetcode.com/problems/132-pattern/

// Time: O(N)
// Space: O(N)
// Ref: https://leetcode.com/problems/132-pattern/discuss/94071/Single-pass-C%2B%2B-O(n)-space-and-time-solution-(8-lines)-with-detailed-explanation.
class Solution {
public:
    bool find132pattern(vector<int>& A) {
        stack<int> s;
        int right = INT_MIN;
        for (int i = A.size() - 1; i >= 0; --i) {
            if (A[i] < right) return true;
            while (s.size() && s.top() < A[i]) {
                right = s.top();
                s.pop();
            }
            s.push(A[i]);
        }
        return false;
    }
};

Java

class Solution {
    public boolean find132pattern(int[] nums) {
        if (nums == null || nums.length < 3)
            return false;
        int length = nums.length;
        int[] minArray = new int[length];
        minArray[0] = nums[0];
        for (int i = 1; i < length; i++)
            minArray[i] = Math.min(minArray[i - 1], nums[i]);
        Stack<Integer> stack = new Stack<Integer>();
        for (int i = length - 1; i >= 0; i--) {
            int num = nums[i];
            while (!stack.isEmpty() && stack.peek() <= minArray[i])
                stack.pop();
            if (!stack.isEmpty() && stack.peek() < num)
                return true;
            else
                stack.push(num);
        }
        return false;
    }
}

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