# How to Solve the 132 Pattern in Java

## 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

``````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`.

``````class Solution {
public boolean find132pattern(int[] nums) {
if(nums == null || nums.length < 3){
return false;
}
int [] min = new int[nums.length];
min[0] = nums[0];
for(int i = 1; i < min.length; i++){
min[i] = Math.min(nums[i], min[i - 1]);
}
Stack<Integer> 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;
}
}
``````