. Day 7 - Challenge 2 - Valid Parentheses: A Guide to Checking Bracket Validity in JavaScript Skip to main content

Day 7 - Challenge 2 - Valid Parentheses: A Guide to Checking Bracket Validity in JavaScript

Valid Parentheses: A Guide to Checking Bracket Validity in JavaScript

Brackets are a fundamental part of programming languages and are used to group and structure code. In this blog post, we will explore the problem of determining whether a given string of brackets is valid or not using JavaScript. This problem is commonly referred to as the "Valid Parentheses" problem.

Understanding the Problem:

The problem statement provides a string that consists of three types of brackets: (, ), {, }, [, and ]. The task is to determine whether the brackets in the string are arranged in a valid manner. For a string to be valid, each opening bracket must have a corresponding closing bracket of the same type, and they must be arranged in the correct order.

For example, the strings "(){}[]" and "{[()]}" are valid, while "([)]" and "{{]}" are not.

Approach:

To solve this problem, we can use a stack data structure. The stack will help us keep track of the opening brackets encountered in the string. Whenever we encounter a closing bracket, we can check if the top of the stack contains the corresponding opening bracket. If it does, we pop the opening bracket from the stack; otherwise, the string is not valid.

Here's how the algorithm works:

  1. Initialize an empty stack.
  2. Iterate through each character in the string.
    • If the character is an opening bracket ((, {, or [), push it onto the stack.
    • If the character is a closing bracket (), }, or ]), check if the stack is empty. If it is, the string is not valid. Otherwise, pop the top element from the stack and compare it to the current closing bracket. If they do not match, the string is not valid.
  3. After iterating through the entire string, if the stack is empty, the string is valid; otherwise, it is not.

JavaScript Implementation:


function isValid(str) {
    const stack = [];
    const bracketPairs = {
        ')': '(',
        '}': '{',
        ']': '['
    };

    for (let char of str) {
        if (char === '(' || char === '{' || char === '[') {
            stack.push(char);
        } else {
            if (stack.length === 0 || stack.pop() !== bracketPairs[char]) {
                return false;
            }
        }
    }

    return stack.length === 0;
}
console.log(isValid("()"));        // true
console.log(isValid("()[]{}"));    // true
console.log(isValid("{[]}"));      // true
console.log(isValid("(]"));        // false
console.log(isValid("([)]"));      // false
console.log(isValid("{{]}"));      // false
console.log(isValid("["));         // false
console.log(isValid(")"));         // false
console.log(isValid(""));          // true (an empty string is considered valid)

console.log(isValid("[({})]"));    // true
console.log(isValid("([{}])"));    // true
console.log(isValid("{[()]}"));    // true
console.log(isValid("({[)}"));    // false
console.log(isValid("(([]){})"));  // true
console.log(isValid("["));         // false
console.log(isValid("{[("));       // false

Demo:

Valid Parentheses Checker

Conclusion

The "Valid Parentheses" problem can be effectively solved using a stack data structure in JavaScript. By keeping track of the opening brackets encountered and comparing them to the corresponding closing brackets, we can determine whether a given string of brackets is valid or not. This problem showcases how fundamental data structures and algorithmic techniques can be applied to real-world scenarios in programming.

Other Challenges:

  1. Day 4 Challenges
  2. Day 5 Challenges
  3. Day 6 Challenges

 

Labels:

Comments

Post a Comment

Popular posts from this blog

Day 9 - Challenge 1 - Product of Array Except Self

Solving the "Product of Array Except Self" Problem in JavaScript Are you ready to dive into a common coding challenge that not only tests your programming skills but also sharpens your problem-solving mindset? If you're up for the challenge, let's tackle the "Product of Array Except Self" problem together using JavaScript. This problem requires us to return an array where each element at index i is the product of all the elements in the original array except the one at index i . Understanding the Problem:   Imagine you're given an array of integers, let's call it nums . Your task is to create a new array where the value at index i in this new array is the product of all the elements in nums , except the one at index i . In other words, you're calculating the product of all the elements to the left of nums[i] and the product of all the elements to the right of nums[i] , and then multiplying these two products to get the final value at index i ...
Post a Comment
Read more

Day 9 - Challenge 2 - Reverse Linked List

Reversing a Singly Linked List in JavaScript: An In-Place Approach Introduction:   Singly linked lists are fundamental data structures in computer science that consist of a sequence of nodes, each containing data and a reference to the next node in the list. Reversing a singly linked list is a classic problem that challenges programmers to manipulate pointers effectively to achieve the desired outcome. In this blog post, we'll explore the problem of reversing a singly linked list using an in-place approach and provide a step-by-step solution in JavaScript. Problem Statement:   Given the head of a singly linked list, our task is to reverse the list in-place and return its new head. In other words, we need to modify the pointers of the nodes in such a way that the direction of the linked list is reversed. Solution Approach:   To solve this problem, we will iterate through the linked list while maintaining three pointers: previous , current , and next . The previous pointer...
Post a Comment
Read more

Day 8 - Challenge 2 - Move Zeroes to the End

Moving Zeroes to the End: A JavaScript Solution Introduction:   When working with arrays, there are often times when we need to manipulate their elements to achieve a specific goal. One common problem is moving all zeroes to the end of an array while keeping the order of non-zero elements unchanged. In this blog post, we will explore an elegant solution to this problem using JavaScript. We'll discuss the problem statement, the approach we'll take, and provide a step-by-step guide to implementing the solution. The Problem:   Given an array of integers, the task is to move all zeroes to the end of the array while maintaining the relative order of the non-zero elements. This means that after rearranging the array, all the zeroes should be at the end, and the order of the non-zero elements should remain the same. The Approach:   To solve this problem, we can utilize a two-pointer approach. We'll maintain two pointers, one for iterating through the array and another for keepin...
Post a Comment
Read more