Guess Number Higher Or Lower Leetcode Problem 374 [Python Solution]

Have you ever played a guessing game as a child? Well, we’re about to apply a similar strategy to solve the Guess Number Higher Or Lower Leetcode problem.

In this LeetCode challenge, we’re given a range from 1 to n, and our goal is to guess a number chosen by someone else.

To help us make informed guesses, we have access to a predefined API function, guess(int num), which returns one of three possible results:

• -1: Our guess is higher than the chosen number (i.e., num > pick).
• 1: Our guess is lower than the chosen number (i.e., num < pick).
• 0: Our guess is equal to the chosen number (i.e., num == pick).

Our task is to efficiently determine the number that has been picked.

Let’s explore this problem in detail.

Problem Overview

Example:

Let’s illustrate the problem with an example.

Suppose we’re given n = 10, and the picked number is 6.

Our goal is to find the chosen number efficiently.

Understanding the Constraints

Before diving into the solution, let’s analyze the constraints:

• n ranges from 1 to 2^31 – 1.
• The chosen number, “pick,” also lies within the range from 1 to n.

Binary Search Approach

Now, how can we efficiently guess the chosen number within this given range?

The key to an efficient solution lies in employing a strategy that narrows down the possibilities with each guess.

Instead of guessing one number at a time, we can significantly reduce the search space by using a binary search algorithm.

Binary search works by dividing the search space in half with each guess.

It’s a classic algorithm for finding an element in a sorted list.

In our case, we’re essentially looking for the chosen number within a range, which is a sorted list from 1 to n.

The binary search algorithm starts with two pointers: left and right.

In our problem, left begins at 1, and right starts at n because these values define the range we are searching within.

Binary Search Algorithm:

1. Initialize left to 1 and right to n.
2. In a loop, calculate the midpoint mid between left and right.
3. Call the guess(mid) function, which returns one of three possible results.
   • If guess(mid) is 1, it means our guess is lower than the chosen number. We update left to mid + 1.
   • If guess(mid) is -1, it means our guess is higher than the chosen number. We update right to mid - 1.
   • If guess(mid) is 0, our guess is correct, and we return mid as the chosen number.

This process continues until we find the correct number, at which point the loop exits.

The time complexity of binary search is O(log n), and it requires only O(1) extra space.

Python Code Solution

Let’s implement the binary search approach in Python:

def guessNumber(self, n: int) -> int:
    # Initialize the left and right boundaries
    left = 1
    right = n

    while True:
        # Calculate the midpoint
        mid = left + (right - left) // 2

        # Use the guess function to determine the next step
        result = guess(mid)

        if result == 1:
            # Our guess is lower than the chosen number
            left = mid + 1
        elif result == -1:
            # Our guess is higher than the chosen number
            right = mid - 1
        else:
            # We found the correct number
            return mid

This Python solution efficiently determines the chosen number by repeatedly halving the search space.

It ensures that we’re making the most informed guesses, eliminating half of the remaining possibilities with each iteration.

Time and Space Complexity

The binary search approach to solving the Guess Number Higher Or Lower problem offers an efficient solution with the following complexities:

• Time Complexity: O(log n)
• Binary search reduces the search space by half with each guess, resulting in a logarithmic time complexity.
• Space Complexity: O(1)
• We only need a constant amount of extra space to store the left and right boundaries.

Reasoning Behind Our Approach

Binary search is a powerful algorithm for finding an element in a sorted list or efficiently solving problems like Guess Number Higher Or Lower The key insight behind this approach is that, by consistently narrowing down the search space, we can reach the correct solution much faster than guessing one number at a time.

In our Python implementation, we maintain two pointers, left and right, to represent the current search boundaries.

By calculating the midpoint mid and using the guess function to determine the next step, we are guided towards the correct answer.

This method ensures that, in the worst case, we find the answer in O(log n) time.

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Conclusion

In this blog post, we explored the Guess Number Higher Or Lower problem from LeetCode.

By applying the binary search algorithm, we efficiently determine the chosen number within the given range of 1 to n.

This approach ensures that we make informed guesses and systematically eliminate possibilities with each iteration.

Binary search, a classic algorithm, is a valuable tool for solving a wide range of problems.

In this particular case, it provides a solution with a time complexity of O(log n), making it a highly efficient approach.

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For the original problem statement and to explore the Guess Number Higher Or Lower challenge on LeetCode, visit the following link:
Guess Number Higher Or Lower LeetCode Problem

Thank you for reading, and we look forward to seeing you in future posts as we continue to tackle exciting coding challenges and explore efficient algorithms!