# 0456. 132 Pattern

<https://leetcode.com/problems/132-pattern>

## 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`

## ac1: Monotonic sequence

<https://leetcode.com/problems/132-pattern/discuss/94071/Single-pass-C%2B%2B-O(n)-space-and-time-solution-(8-lines)-with-detailed-explanation>.

This is very tricky to be honest, hard to come up with.

```java
class Solution {
    public boolean find132pattern(int[] nums) {
        Deque<Integer> stack = new ArrayDeque<>();
        int num2 = Integer.MIN_VALUE;
        for (int i = nums.length - 1; i >= 0; i--) {
            int num1 = nums[i];
            if (num1 < num2) return true; // Found [num1, num3, num2]: "stack.peek()" -> num3, which always > num2.
            // Decreasing stack.
            while (!stack.isEmpty() && nums[i] > stack.peek()) {
                num2 = stack.pop();
            }
            stack.push(nums[i]);
        }
        
        return false;
    }
}
```


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