Added 'Valid Parentheses'

Author: anthonyzhub

Diff for: Valid Parentheses.java

@@ -0,0 +1,51 @@
+// https://leetcode.com/problems/valid-parentheses/
+// https://www.youtube.com/watch?v=WTzjTskDFMg&t=316
+
+class Solution {
+    public boolean isValid(String s) {
+
+        /*
+         * Time Complexity: O(n) where n = length of input string. String will be iterated
+         *  and worse case is that it's long and has a valid parentheses.
+         * 
+         * Space Complexity: O(n) where n = length of stack. Stack height is based of string's
+         *  length. Also, although hash map was created, it size isn't dynamic, which is why
+         *  it didn't influence the space complexity.
+         */
+        
+        // Create a hash map
+        Map<Character, Character> closingCharacters = new HashMap<>(Map.of(
+            '}', '{',
+            ']', '[',
+            ')', '('
+        ));
+
+        // Create a stack
+        Stack<Character> mainStack = new Stack<>();
+
+        // Iterate string
+        for (int i = 0; i < s.length(); i++) {
+
+            Character c = Character.valueOf(s.charAt(i));
+            
+            // Check if c is a closing character
+            if (closingCharacters.containsKey(c)) {
+
+                // If stack isn't empty AND stack head is a matching pair with c, pop stack head.
+                // If not, return false because the string doesn't have a valid parentheses
+                if (!mainStack.empty() && mainStack.peek() == closingCharacters.get(c)) {
+                    mainStack.pop();
+                }
+                else {
+                    return false;
+                }
+            }
+            else { // If c is another opening character, add it to stack
+                mainStack.push(c);
+            }
+        }
+
+        // If stack is empty, then string had valid parentheses. If not, return false
+        return mainStack.empty();
+    }
+}