How do you solve Counter II on LeetCode?

Counter II (LeetCode #2665) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Functions can maintain state using closures.

What is the brute force solution for Counter II?

The brute force approach for Counter II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating a counter that simply modifies a value based on the operations called. Each function directly manipulates the current count without any optimization. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Counter II?

Explain your thought process clearly as you code. Test edge cases like negative and zero values for init. Be prepared to discuss the implications of state management.

What are common mistakes when solving Counter II?

Not using closures correctly to maintain state. Confusing the return values of the functions.

#2665

Counter II

Easy

ClosureState Management

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution uses the same approach as brute force but emphasizes that each function operates in constant time, making it efficient for multiple calls.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable `currentCount` with the value of `init`.
2Step 2: Define the `increment` function to increase `currentCount` by 1 and return it.
3Step 3: Define the `decrement` function to decrease `currentCount` by 1 and return it.
4Step 4: Define the `reset` function to set `currentCount` back to `init` and return it.
5Step 5: Return an object containing the three functions.

solution.py15 lines

1def createCounter(init):
2    currentCount = init
3    def increment():
4        nonlocal currentCount
5        currentCount += 1
6        return currentCount
7    def decrement():
8        nonlocal currentCount
9        currentCount -= 1
10        return currentCount
11    def reset():
12        nonlocal currentCount
13        currentCount = init
14        return currentCount
15    return {'increment': increment, 'decrement': decrement, 'reset': reset}

ℹ

Complexity note: The time complexity is O(n) where n is the number of calls made to the functions. Each function operates in constant time, so the overall complexity is linear with respect to the number of operations.

1Functions can maintain state using closures.
2Understanding scope and variable lifetime is crucial.

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