## The challenge

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`.

**Follow up: **The `O(n^2)` is trivial, could you come up with the `O(n logn)` or the `O(n)` solution?

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 <= 10<sup>4</sup>`
• `-10<sup>9</sup>&nbsp;<= nums[i] <= 10<sup>9</sup>`

## The solution in Java

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 `````` ``````class Solution { public boolean find132pattern(int[] nums) { int n = nums.length, top = n, third = Integer.MIN_VALUE; for (int i = n - 1; i >= 0; i--) { if (nums[i] < third) return true; while (top < n && nums[i] > nums[top]) third = nums[top++]; nums[--top] = nums[i]; } return false; } } ``````

You can also solve this problem by means of using a `Stack`.

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 `````` ``````class Solution { public boolean find132pattern(int[] nums) { if(nums == null || nums.length < 3){ return false; } int [] min = new int[nums.length]; min = nums; for(int i = 1; i < min.length; i++){ min[i] = Math.min(nums[i], min[i - 1]); } Stack stack = new Stack<>(); for(int i = nums.length - 1; i >= 0; i--){ if(nums[i] > min[i]){ while(!stack.empty() && stack.peek() <= min[i]){ stack.pop(); } if(!stack.empty() && stack.peek() < nums[i]){ return true; } stack.push(nums[i]); } } return false; } } ``````