How do you solve Guess Number Higher or Lower on LeetCode?

Guess Number Higher or Lower (LeetCode #374) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Binary search drastically reduces the number of guesses needed.

What is the brute force solution for Guess Number Higher or Lower?

The brute force approach for Guess Number Higher or Lower is Brute Force, with O(n) time complexity and O(1) space complexity. In the brute force approach, we simply guess every number from 1 to n until we find the correct one. This is straightforward but inefficient as it can take up to n guesses in the worst case. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Guess Number Higher or Lower?

Guess Number Higher or Lower uses the following concepts: Binary Search, Interactive. The recommended patterns to study are: Binary Search, Interactive Problems.

What are the interview tips for Guess Number Higher or Lower?

Always consider the range of values and how to narrow it down. Explain your thought process clearly while coding. Practice binary search problems to become comfortable with the approach.

What are common mistakes when solving Guess Number Higher or Lower?

Not using binary search when applicable, leading to inefficient solutions. Failing to handle edge cases, such as when n = 1.

#374

Guess Number Higher or Lower

Easy

Binary SearchInteractiveBinary SearchInteractive Problems

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(log n)Space O(1)

The optimal solution uses a binary search approach, which significantly reduces the number of guesses needed. By halving the search space with each guess, we can quickly hone in on the correct number.

⚙️

Algorithm

3 steps

1Step 1: Initialize two pointers, left = 1 and right = n.
2Step 2: While left is less than or equal to right, calculate mid = (left + right) // 2.
3Step 3: Call the guess API with mid. If the result is 0, return mid. If it's -1, set right = mid - 1. If it's 1, set left = mid + 1.

solution.py10 lines

1def guessNumber(n):
2    left, right = 1, n
3    while left <= right:
4        mid = (left + right) // 2
5        if guess(mid) == 0:
6            return mid
7        elif guess(mid) == -1:
8            right = mid - 1
9        else:
10            left = mid + 1

ℹ

Complexity note: The time complexity is O(log n) because we halve the search space with each guess. The space complexity is O(1) as we are using a constant amount of space.

1Binary search drastically reduces the number of guesses needed.
2Understanding how to adjust search boundaries is crucial.

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