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1. Introduction
In this blog post, we'll explore a common problem in string processing and stack usage - the "Valid Parentheses" problem. This problem is a typical example in coding interviews to assess understanding of basic data structures in Java.
Problem
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', the task is to determine if the input string is valid. An input string is considered valid if:
1. Open brackets are closed by the same type of brackets.
2. Open brackets are closed in the correct order.
Note: An empty string is also considered valid.
2. Solution Steps
1. Initialize a stack to keep track of opening brackets.
2. Iterate through each character in the string.
3. If an opening bracket is encountered, push it onto the stack.
4. If a closing bracket is encountered, check if it matches the top element of the stack.
5. If it matches, pop the top element from the stack; otherwise, return false.
6. After the iteration, if the stack is empty, return true; otherwise, return false.
3. Code Program
import java.util.Stack;
public class Solution {
// Main method for testing
public static void main(String[] args) {
String input1 = "({[]})";
String input2 = "([)]";
System.out.println("Input: " + input1 + " - Valid: " + isValid(input1));
System.out.println("Input: " + input2 + " - Valid: " + isValid(input2));
}
// Method to check if the string has valid parentheses
public static boolean isValid(String s) {
Stack<Character> stack = new Stack<>(); // Stack to keep track of opening brackets
for (char c : s.toCharArray()) {
// Push corresponding opening bracket onto stack if closing bracket is found
if (c == ')' && !stack.isEmpty() && stack.peek() == '(') {
stack.pop();
} else if (c == '}' && !stack.isEmpty() && stack.peek() == '{') {
stack.pop();
} else if (c == ']' && !stack.isEmpty() && stack.peek() == '[') {
stack.pop();
} else if (c == '(' || c == '{' || c == '[') {
stack.push(c); // Push opening bracket onto stack
} else {
return false; // Invalid case
}
}
return stack.isEmpty(); // Return true if stack is empty, false otherwise
}
}
Output:
Input: ({[]}) - Valid: true
Input: ([)] - Valid: false
Explanation:
The input "({[]})" is valid as each type of opening bracket is matched and closed by the same type of closing bracket in the correct order.
The method isValid uses a stack to keep track of the opening brackets. When a closing bracket is encountered, it checks whether it matches the last opened bracket.
Conversely, the input "([)]" is invalid as the brackets are not closed in the correct order, demonstrated by the round bracket closing before the square bracket.
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