## Algorithm

Problem Name: 456. 132 Pattern

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`

## Code Examples

### #1 Code Example with Java Programming

```Code - Java Programming```

``````
class Solution {
public boolean find132pattern(int[] nums) {
if (nums.length < 3) {
return false;
}
int[] minTillIndex = new int[nums.length];
minTillIndex[0] = nums[0];
for (int idx = 1; idx  <  nums.length; idx++) {
minTillIndex[idx] = Math.min(minTillIndex[idx - 1], nums[idx]);
}
Stack < Integer> stack = new Stack<>();
for (int idx = nums.length - 1; idx >= 0; idx--) {
if (nums[idx] > minTillIndex[idx]) {
while (!stack.isEmpty() && stack.peek()  < = minTillIndex[idx]) {
stack.pop();
}
if (!stack.isEmpty() && stack.peek() < nums[idx]) {
return true;
}
stack.push(nums[idx]);
}
}
return false;
}
}
``````
Copy The Code &

Input

cmd
nums = [1,2,3,4]

Output

cmd
false

### #2 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const find132pattern = function(nums) {
let [stack, s3] = [[], -Infinity]
for (let i = nums.length - 1; i >= 0; i--) {
if (nums[i] < s3) {
return true
}
while (stack[stack.length - 1] < nums[i]) {
s3 = stack.pop()
}
stack.push(nums[i])
}
return false
}
``````
Copy The Code &

Input

cmd
nums = [1,2,3,4]

Output

cmd
false

### #3 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def find132pattern(self, nums):
stack, s3 = [], -float("inf")
for n in nums[::-1]:
if n < s3: return True
while stack and stack[-1] < n: s3 = stack.pop()
stack.append(n)
return False
``````
Copy The Code &

Input

cmd
nums = [3,1,4,2]

Output

cmd
true