solution(python): 456. 132 Pattern

456. 132 Pattern
- Python

Medium/456. 132 Pattern/README.md

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+# LeetCode 456.
+
+## Problem Statement
+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.
+
+### Test Cases
+
+#### 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].
+
+## Solution
+We iterate through the 'nums' list in reverse order. We maintain a stack to keep track of potential '2' candidates.
+We also keep track of the '3' candidate as 'third'. If we encounter a number smaller than 'third', we return True as we found a '132' pattern.
+If the current number is greater than the top of the stack, it becomes a potential '3' candidate, and we update 'third'.
+Finally, we push the current number onto the stack for potential future '2' candidates.
+
+### Complexity
+- Time complexity: O(n), where n is the length of the 'nums' list.
+- Space complexity: O(n), as the stack can potentially hold all elements of the 'nums' list.

Medium/456. 132 Pattern/sol.py

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+'''
+ * LeetCode 456.
+ * Problem Statement- 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.
+ *
+ * Test Cases:
+
+  * 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].
+'''
+
+# Solution
+class Solution:
+    def find132pattern(self, nums: List[int]) -> bool:
+        # Initialize an empty stack and set 'third' to negative infinity.
+        stack, third = [], float('-inf')
+
+        # Iterate through the 'nums' list in reverse order.
+        for num in reversed(nums):
+            # If the current number is smaller than 'third', we found the '132' pattern.
+            if num < third:
+                return True
+
+            # While the stack is not empty and the top of the stack is smaller than the current number,
+            # pop elements from the stack and update 'third' if necessary.
+            while stack and stack[-1] < num:
+                third = stack.pop()
+
+            # Push the current number onto the stack for potential future comparisons.
+            stack.append(num)
+
+        # If no '132' pattern is found after iterating through the entire list, return False.
+        return False
+