# 456. 132 Pattern

Given a sequence of n integers a1, a2, ..., an, 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].
```

```cpp
// O(n) space
bool find132pattern(vector<int>& nums) { // time: O(n); space: O(n)
    int s3 = numeric_limits<int>::min();
    stack<int> st; // monotonous descreasing stack
    for (int i = nums.size() - 1; i >= 0; --i) {
        if (nums[i] < s3) return true;
        while (!st.empty() && nums[i] > st.top()) {
            s3 = st.top(); st.pop();
        }
        st.push(nums[i]);
    }
    return false;
}
```

```cpp
// O(1) Space
bool find132pattern(vector<int>& nums) { // time: O(n); space: O(1)
    int n = nums.size(), s3 = numeric_limits<int>::min(), idx = nums.size();
    for (int i = n - 1; i >= 0; --i) {
        if (nums[i] < s3) return true;
        while (idx < n && nums[i] > nums[idx]) {
            s3 = nums[idx++];
        }
        nums[--idx] = nums[i];
    }
    return false;
}
```
