How do you solve To Be Or Not To Be on LeetCode?

To Be Or Not To Be (LeetCode #2704) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Understanding strict equality in JavaScript and other languages is crucial.

What is the brute force solution for To Be Or Not To Be?

The brute force approach for To Be Or Not To Be is Brute Force, with O(1) time complexity and O(1) space complexity. The brute force approach involves creating a simple function that checks for equality and inequality using basic conditional statements. This method is straightforward but not the most efficient. The optimal Optimal Solution approach improves this to O(1).

What are the interview tips for To Be Or Not To Be?

Explain your thought process clearly while coding. Test edge cases, such as null or undefined values. Be ready to discuss why you chose a particular approach.

What are common mistakes when solving To Be Or Not To Be?

Confusing == with === in JavaScript. Not handling exceptions properly.

#2704

To Be Or Not To Be

Easy

Error HandlingFunction Closures

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(1)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

The optimal solution is similar to the brute force approach but emphasizes clarity and error handling. It ensures that the functions are encapsulated and can be reused without redundancy.

⚙️

Algorithm

3 steps

1Step 1: Define the expect function that takes a value val.
2Step 2: Inside expect, define toBe and notToBe functions that handle equality and inequality checks respectively.
3Step 3: Return an object containing both functions.

solution.py12 lines

1def expect(val):
2    def toBe(other):
3        if val == other:
4            return {'value': True}
5        raise ValueError('Not Equal')
6
7    def notToBe(other):
8        if val != other:
9            return {'value': True}
10        raise ValueError('Equal')
11
12    return {'toBe': toBe, 'notToBe': notToBe}

ℹ

Complexity note: Both time and space complexity remain O(1) as we are performing constant-time checks and not using additional data structures.

1Understanding strict equality in JavaScript and other languages is crucial.
2Error handling is important to provide clear feedback to developers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.