Day 1 - Parenthesis Checker.md

Difficulty	Source	Tags
Easy	160 Days of Problem Solving	Strings Stack STL

🚀 Day 1. Parenthesis Checker 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

Given a string s, composed of different combinations of (, ), {, }, [, ], verify the validity of the arrangement.

A string is valid if:

1. Open brackets must be closed by the same type of brackets.
2. Open brackets must be closed in the correct order.

🔍 Example Walkthrough:

Example 1:

Input:

s = "[{()}]"

Output:

true

Explanation:

All brackets are correctly paired and well-formed.

Example 2:

Input:

s = "[()()]{}"

Output:

true

Explanation:

All brackets are well-formed.

Example 3:

Input:

s = "([]"

Output:

false

Explanation:

The expression is not balanced as there is a missing ')' at the end.

Example 4:

Input:

s = "([{]})"

Output:

false

Explanation:

The expression is not balanced as there is a closing ] before the closing }.

Constraints:

• $1 \leq |s| \leq 10^6$
• s[i] ∈ { '(', ')', '{', '}', '[', ']' }

🎯 My Approach:

Stack-Based Validation (O(N) Time, O(N) Space)

1. Iterate over the string and use a stack to track opening brackets.
2. When encountering a closing bracket:
   • If the stack is empty, return false.
   • Check whether the top of the stack matches the expected opening bracket.
   • If matched, pop the stack. Otherwise, return false.
3. After iterating the string, the stack should be empty for a valid expression.

Algorithm Steps:

1. Use a stack to store open brackets {[(.
2. Iterate through the string:
   • If s[i] is an opening bracket, push it onto the stack.
   • If s[i] is a closing bracket:
     • Check if the stack is empty (invalid case).
     • Compare the top element of the stack with s[i].
     • If it matches, pop the stack. Otherwise, return false.
3. After the loop, return true if the stack is empty, else false.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N), since each character is processed once in the loop.
• Expected Auxiliary Space Complexity: O(N), in the worst case, all characters are pushed onto the stack.

📝 Solution Code

Code (C++)

class Solution {
  public:
    bool isBalanced(string& s) {
        stack<char> st;
        for (char c : s) {
            if (c == '(' || c == '{' || c == '[') st.push(c);
            else if (st.empty() || abs(st.top() - c) > 2) return false;
            else st.pop();
        }
        return st.empty();
    }
};

⚡ Alternative Approaches

2️⃣ Hash Map Lookup (O(N) Time, O(N) Space)

1. Store matching pairs in a hash map.
2. Push opening brackets onto a stack.
3. On encountering a closing bracket:
   • Check stack is empty.
   • Compare top of stack with map lookup.
   • If matched, pop.
4. Return true if stack is empty.

class Solution {
  public:
    bool isBalanced(string& s) {
        unordered_map<char, char> m = {{')', '('}, {'}', '{'}, {']', '['}};
        stack<char> st;
        for (char c : s) {
            if (m.count(c)) {
                if (st.empty() || st.top() != m[c]) return false;
                st.pop();
            } else st.push(c);
        }
        return st.empty();
    }
};

🔹 Pros: Explicit matching is better than abs(top - c) > 2.
🔹 Cons: Uses extra hash map (though small overhead).

📊 Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Stack-Based Matching	🟢 O(N)	🟡 O(N)	Simple and effective	Uses extra stack space
Hash Map Lookup	🟢 O(N)	🟡 O(N)	Explicit and easy-to-read matching	Slightly more memory

💡 Best Choice?

✅ For general use: Stack-Based Matching (O(N)).
✅ For minimal space usage: Two-Pointer (O(1)), but fails for mixed brackets.
✅ For explicit matching: Hash Map (O(N)), great for readability.

Code (Java)

class Solution {
    static boolean isBalanced(String s) {
        Stack<Character> st = new Stack<>();
        for (char c : s.toCharArray()) 
            if ("({[".indexOf(c) >= 0) st.push(c);
            else if (st.isEmpty() || Math.abs(st.peek() - c) > 2) return false;
            else st.pop();
        return st.isEmpty();
    }
}

Code (Python)

class Solution:
    def isBalanced(self, s):
        st = []
        for c in s:
            if c in "({[": st.append(c)
            elif not st or abs(ord(st[-1]) - ord(c)) > 2: return False
            else: st.pop()
        return not st

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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