Add longest-valid-parentheses

dynamic_programming/longest_valid_parentheses.cpp

@@ -0,0 +1,101 @@
+/**
+ * @file
+ * @details
+ * Given a string containing just the characters '(' and ')', 
+ * find the length of the longest valid (well-formed) parentheses substring.
+ *
+ * ### Approach
+ * This solution uses Dynamic Programming.
+ * We maintain an array `longest[]` where `longest[i]` represents 
+ * the length of the longest valid parentheses substring ending at index `i`.
+ *
+ * - If `s[i] == '('`, then `longest[i] = 0` (a valid substring cannot end with '(').
+ * - If `s[i] == ')'`:
+ *   - If `s[i - 1] == '('`, then `longest[i] = longest[i - 2] + 2`
+ *   - Else if `s[i - 1] == ')'` and `s[i - longest[i - 1] - 1] == '('`,
+ *     then `longest[i] = longest[i - 1] + 2 + longest[i - longest[i - 1] - 2]`
+ *
+ * Example:
+ * Input: "()(())"
+ * At i = 5, longest = [0, 2, 0, 0, 2, 0]
+ * longest[5] = longest[4] + 2 + longest[1] = 6
+ *
+ * @note
+ * Time Complexity: O(n)
+ * Space Complexity: O(n)
+ */
+
+#include <bits/stdc++.h>
+using namespace std;
+
+/**
+ * @brief Dynamic Programming based solution to find the longest valid parentheses substring.
+ * @param s Input string containing '(' and ')'
+ * @return Length of the longest valid parentheses substring
+ */
+int longestValidParentheses(const string &s) {
+    if (s.length() <= 1) return 0;
+
+    int curMax = 0;
+    vector<int> longest(s.size(), 0);
+
+    for (int i = 1; i < s.length(); i++) {
+        if (s[i] == ')') {
+            if (s[i - 1] == '(') {
+                longest[i] = (i - 2 >= 0 ? longest[i - 2] + 2 : 2);
+            } else if (i - longest[i - 1] - 1 >= 0 && s[i - longest[i - 1] - 1] == '(') {
+                longest[i] = longest[i - 1] + 2 +
+                             ((i - longest[i - 1] - 2 >= 0) ? longest[i - longest[i - 1] - 2] : 0);
+            }
+            curMax = max(curMax, longest[i]);
+        }
+    }
+    return curMax;
+}
+
+/**
+ * @brief Concise version 
+ * Same logic as above but written in a more compact form.
+ */
+int longestValidParenthesesConcise(const string &s) {
+    if (s.length() <= 1) return 0;
+
+    int curMax = 0;
+    vector<int> longest(s.size(), 0);
+
+    for (int i = 1; i < s.length(); i++) {
+        if (s[i] == ')' && i - longest[i - 1] - 1 >= 0 && s[i - longest[i - 1] - 1] == '(') {
+            longest[i] = longest[i - 1] + 2 +
+                         ((i - longest[i - 1] - 2 >= 0) ? longest[i - longest[i - 1] - 2] : 0);
+            curMax = max(curMax, longest[i]);
+        }
+    }
+    return curMax;
+}
+
+/**
+ * @brief Driver code for demonstration and basic testing.
+ */
+int main() {
+    vector<string> test_cases = {
+        "(()",          // expected 2
+        ")()())",       // expected 4
+        "()(())",       // expected 6
+        "((((((",       // expected 0
+        "()(()))))"     // expected 6
+    };
+
+    cout << "Testing Longest Valid Parentheses using DP:\n";
+    for (const auto &s : test_cases) {
+        cout << "Input: " << s 
+             << " | Output: " << longestValidParentheses(s) << endl;
+    }
+
+    cout << "\nTesting Concise Version:\n";
+    for (const auto &s : test_cases) {
+        cout << "Input: " << s 
+             << " | Output: " << longestValidParenthesesConcise(s) << endl;
+    }
+
+    return 0;
+}