feat: solve valid parantheses

Author: GangBean

valid-parentheses/GangBean.java

@@ -0,0 +1,38 @@
+class Solution {
+    public boolean isValid(String s) {
+        /**
+        1. understanding
+        - given string contains character which has paired characters.
+        - validate all given characters in input string has each pair, and valid in order.
+        2. strategy
+        - use stack, to save left brackets, and if you encounter right bracket, then pop from stack, and compare two characters are paired.
+        - if not paired on any right character, or stack is empty then return false.
+        - all characters are iterated and stack is not empty, then return false.
+        3. complexity
+        - time: O(N), N is the length of input string s.
+        - space: O(N), for stack variable memory.
+        */
+        Stack<Character> leftBracket = new Stack<>();
+        for (char c : s.toCharArray()) {
+            if (c == '(' || c == '{' || c == '[') {
+                // when character is left bracket, then push to stack.
+                leftBracket.push(c);
+            } else if (c == ')' || c == '}' || c == ']') {
+                if (leftBracket.isEmpty()) return false;
+                char left = leftBracket.pop();
+                if (isPair(left, c)) continue;
+                return false;
+            } else {
+                throw new RuntimeException(String.format("Not valid input character: %c in input %s", c, s));
+            }
+        }
+        return leftBracket.isEmpty();
+    }
+
+    private boolean isPair(char left, char right) {
+        return (left == '(' && right == ')') 
+            || (left == '{' && right == '}')
+            || (left == '[' && right == ']');
+    }
+}
+