Merge pull request #191 from BrianLusina/feat/algorithms-stack-longest-valid-parentheses

feat(algorithms, stack): longest valid parentheses

Author: BrianLusina

DIRECTORY.md

@@ -386,6 +386,8 @@
       * [Test Decimal To Binary](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/stack/decimal_to_binary/test_decimal_to_binary.py)
     * Decode String
       * [Test Decode String](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/stack/decode_string/test_decode_string.py)
+    * Longest Valid Parentheses
+      * [Test Longest Valid Parentheses](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/stack/longest_valid_parentheses/test_longest_valid_parentheses.py)
     * Minimum String Length After Removing Substrings
       * [Test Min Str Length After Removing Substrings](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/stack/minimum_string_length_after_removing_substrings/test_min_str_length_after_removing_substrings.py)
     * Nextgreater

algorithms/stack/longest_valid_parentheses/README.md

@@ -0,0 +1,85 @@
+# Longest Valid Parentheses
+
+You are given a string composed entirely of ‘(’ and ‘)’ characters. Your goal is to identify the longest contiguous segment (substring) within this string that represents a “well-formed” or “valid” sequence of parentheses.
+
+A substring is considered valid if:
+
+1. Every opening parenthesis ‘(’ has a corresponding closing parenthesis ‘)’. 
+2. The pairs of parentheses are correctly nested.
+
+Return the length of this longest valid substring.
+
+## Constraints
+
+- 0 <= s.length <= 3 * 10^4
+- s[i] is '(', or ')'.
+
+## Examples
+
+Example 1:
+```text
+Input: s = "(()"
+Output: 2
+Explanation: The longest valid parentheses substring is "()".
+```
+
+Example 2:
+```text
+Input: s = ")()())"
+Output: 4
+Explanation: The longest valid parentheses substring is "()()".
+```
+
+Example 3:
+```text
+Input: s = ""
+Output: 0
+```
+
+## Topics
+
+- String
+- Dynamic Programming
+- Stack
+
+## Solution
+
+The problem asks for the length of the longest valid (well-formed) parenthesis substring. This type of matching problem,
+where you need to pair opening brackets with their corresponding closing brackets, is a classic example of a problem that
+lends itself to a stack pattern. The stack’s last in, first out (LIFO) property naturally models the nesting structure of
+parentheses: the last opening parenthesis must be the first one to be closed.
+
+The essence of this algorithm is to use the stack to keep track of the indexes of parentheses that have not yet been
+matched. Instead of just pushing the characters themselves, pushing their indexes allows us to calculate lengths. The
+stack maintains a “base” index at the bottom, which marks the start of a potential valid substring. When a closing
+parenthesis, ‘)’, finds a matching opening parenthesis, ‘(’, (by popping its index from the stack), the length of the
+newly formed valid substring is calculated as the difference between the current index and the new top of the stack.
+
+The following steps can be performed to implement the algorithm above:
+
+1. Initialize a stack, indexStack, and push an initial index of -1. This value acts as a sentinel or a “base” for the
+   first potential valid substring.
+2. Next, iterate over the input string from the current index to the right (length of the input string), character by
+   character.
+   - **Opening parenthesis (**: If we encounter an opening parenthesis, we push its index onto the stack.
+   - **Closing parenthesis )**: If we encounter a closing parenthesis:
+     - We pop the top element from the indexStack.
+     - If the stack becomes empty after popping, it means the current ‘)’ does not have a matching ‘(’. In this case, we
+       push the current ‘)’s index onto the stack to serve as the new “base” for any future valid substrings.
+     - Otherwise, if the stack is not empty, this means a valid pair has been formed. The length of this valid substring
+       is the difference between the currentIndex and the index at the new top of the stack (which is the index right
+       before the start of this valid pair). Compare this length with the maximum length, currentLength, found so far,
+       and update it if necessary.
+3. After the iteration ends, the maximum length, as recorded using max(maxLength, currentLength), is the answer.
+
+### Time Complexity
+
+The solution involves a single pass through the input string s. The loop runs exactly n times, where n is the length of
+the string, and inside the loop, all operations (push, pop, top on the stack) take constant time, i.e., O(1). Therefore,
+the total time complexity is dominated by the loop, resulting in O(n).
+
+### Space Complexity
+
+In the worst-case scenario, the input string, s, may consist entirely of opening parenthesis (e.g., “((((...”). In this
+case, the index of every character will be pushed onto the stack, where the size of the stack can grow up to n, where n
+is the length of the string. Therefore, the space complexity is O(n).

algorithms/stack/longest_valid_parentheses/__init__.py

@@ -0,0 +1,29 @@
+def longest_valid_parentheses(s: str) -> int:
+    # Stack to store indexes. We initialize it with -1 to serve as a
+    # sentinel value. This handles the edge case of a valid substring
+    # starting from index 0, allowing us to calculate its length correctly.
+    stack = [-1]
+    # Variable to store the maximum length of valid parentheses found so far.
+    max_length = 0
+
+    for index, char in enumerate(s):
+        # If the character is an opening parenthesis, push its index onto the stack. It represents a potential start of
+        # a valid sequence
+        if char == "(":
+            stack.append(index)
+        elif char == ")":
+            stack.pop()
+            # iF the stack is now empty, it means the current ')' has no matching '('. We push the current index to act
+            # as a new base or starting point for the next potential valid substring
+            if len(stack) == 0:
+                stack.append(index)
+            else:
+                # If the stack is not empty, a valid pair was just formed
+                # The new top of the stack holds the index of the character just 'before' the start of the current valid
+                # substring. The length is the difference between the current index and that 'base' index
+                current_length = index - stack[len(stack) - 1]
+
+                # Update the overall maximum length if the current one is greater
+                max_length = max(max_length, current_length)
+
+    return max_length

algorithms/stack/longest_valid_parentheses/test_longest_valid_parentheses.py

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+import unittest
+from parameterized import parameterized
+from algorithms.stack.longest_valid_parentheses import longest_valid_parentheses
+
+LONGEST_VALID_PARENTHESES_TEST_CASES = [
+    ("(()", 2),
+    (")()())", 4),
+    ("", 0),
+    ("())()", 2),
+    ("(())", 4),
+    (")(", 0),
+    ("(", 0),
+    ("()", 2),
+    ("))(((", 0),
+    ("(()())", 6),
+    (")()((()))((((", 8),
+    ("()(()))", 6),
+    ("(()))())(", 4),
+]
+
+
+class LongestValidParenthesesTestCase(unittest.TestCase):
+    @parameterized.expand(LONGEST_VALID_PARENTHESES_TEST_CASES)
+    def test_longest_valid_parentheses(self, s: str, expected: int):
+        actual = longest_valid_parentheses(s)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()