easy Valid Parentheses

Problem Statement

Determine if an input string containing only the characters '(', ')', '{', '}', '[', and ']' is valid. A string is considered valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Each close bracket has a corresponding open bracket of the same type.

Examples

- Input: "(]"
- Expected Output: false
- Justification: The opening parenthesis '(' is not closed by its corresponding closing parenthesis.
- Input: "{[]}"
- Expected Output: true
- Justification: The string contains pairs of opening and closing brackets in the correct order.
- Input: "[{]}"
- Expected Output: false
- Justification: The opening square bracket '[' is closed by a curly brace '}', which is incorrect.

Constraints:

1 <= s.length <= 10⁴
s consists of parentheses only '()[]{}'.

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✅ Solution Valid Parentheses

Problem Statement

Determine if an input string containing only the characters '(', ')', '{', '}', '[', and ']' is valid. A string is considered valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Each close bracket has a corresponding open bracket of the same type.

Examples

- Input: "(]"
- Expected Output: false
- Justification: The opening parenthesis '(' is not closed by its corresponding closing parenthesis.
- Input: "{[]}"
- Expected Output: true
- Justification: The string contains pairs of opening and closing brackets in the correct order.
- Input: "[{]}"
- Expected Output: false
- Justification: The opening square bracket '[' is closed by a curly brace '}', which is incorrect.

Constraints:

1 <= s.length <= 10⁴
s consists of parentheses only '()[]{}'.

Solution

Initialization:
- Start with an empty stack.
- For every bracket in the input string:
  - If the bracket is an opening bracket ('(', '{', '['), push it onto the stack.
  - If the bracket is a closing bracket (')', '}', ']'):
    - If the stack is empty, return false.
    - Pop the top of the stack and check if it matches the corresponding opening bracket.
Validation:
- If the current closing bracket doesn't match the top of the stack, return false.
- If the stack is not empty after processing all brackets in the string, return false.
Final Output:
- If the stack is empty at the end, return true, indicating that the string has valid parentheses.

Algorithm Walkthrough

Using the input "{[]}":

Start with an empty stack.
Encounter '{', push onto the stack.
Encounter '[', push onto the stack.
Encounter ']', pop from the stack and check if the popped character is '['. It matches, so continue.
Encounter '}', pop from the stack and check if the popped character is '{'. It matches.
The string is processed and the stack is empty, so return true.

Code

java

import java.util.Stack;

public class Solution {

  public boolean isValid(String s) {
    Stack<Character> stack = new Stack<>();
    // Iterate through each character in the string
    for (char c : s.toCharArray()) {
      // If it's an opening bracket, push to stack
      if (c == '(' || c == '{' || c == '[') {
        stack.push(c);
      } else {
        // If stack is empty, it means no matching opening bracket
        if (stack.isEmpty()) return false;
        char top = stack.pop(); // Get the top element of the stack
        // Check if the current closing bracket matches the top of the stack
        if (c == ')' && top != '(') return false;
        if (c == '}' && top != '{') return false;
        if (c == ']' && top != '[') return false;
      }
    }
    // If stack is empty, all brackets were matched
    return stack.isEmpty();
  }

  public static void main(String[] args) {
    Solution solution = new Solution();
    System.out.println(solution.isValid("(]")); // false
    System.out.println(solution.isValid("{[]}")); // true
    System.out.println(solution.isValid("[{]}")); // false
  }
}

Complexity Analysis

Time Complexity: O(n) where n is the length of the string. Each character is processed once.
Space Complexity: O(n) in the worst case when all characters are opening brackets.

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