What is the time complexity of Guess Number Higher or Lower?

The optimal solution has a time complexity of O(log n).

What data structures are used to solve Guess Number Higher or Lower?

This problem uses Binary Search, Interactive. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Guess Number Higher or Lower?

Insight 1: Binary search is the most efficient approach because it halves the search space with each guess. Insight 2: Understanding the 'guess' API and how to use its return values to adjust the search range is crucial.

Which companies ask LeetCode 374. Guess Number Higher or Lower in interviews?

This problem is frequently asked at Samsung.

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Guess Number Higher or Lower - LeetCode 374 Solution

Q: How to solve Guess Number Higher or Lower?

The provided code implements a binary search algorithm. It initializes a 'low' pointer to 1 and a 'high' pointer to 'n'. In each iteration, it calculates the middle value 'mid' as (low + high) / 2. It then calls the 'guess(mid)' API to get feedback. Based on the feedback, it either returns 'mid' if the guess is correct, updates 'high' to 'mid - 1' if the guess is too high, or updates 'low' to 'mid + 1' if the guess is too low. This process continues until the correct number is found.

Guess Number Higher or Lower - Complete Solution Guide

Guess Number Higher or Lower is LeetCode problem 374, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

We are playing the Guess Game. The game is as follows: I pick a number from 1 to n . You have to guess which number I picked. Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess. You call a pre-defined API int guess(int num) , which returns three possible results: -1 : Your guess is higher than the number I picked (i.e. num > pick ). 1 : Your guess is lower than the number I picked (i.e. num < pick ). 0 : your guess is equal to the number I

Detailed Explanation

The 'Guess Number Higher or Lower' problem is a classic guessing game. The system picks a secret number between 1 and n (inclusive). Your task is to write code to guess this number. After each guess, you receive feedback from a pre-defined API, 'guess(num)', which returns -1 if your guess is too high, 1 if your guess is too low, and 0 if your guess is correct. The goal is to minimize the number of guesses and find the correct number efficiently.

Solution Approach

The provided code implements a binary search algorithm. It initializes a 'low' pointer to 1 and a 'high' pointer to 'n'. In each iteration, it calculates the middle value 'mid' as (low + high) / 2. It then calls the 'guess(mid)' API to get feedback. Based on the feedback, it either returns 'mid' if the guess is correct, updates 'high' to 'mid - 1' if the guess is too high, or updates 'low' to 'mid + 1' if the guess is too low. This process continues until the correct number is found.

Step-by-Step Algorithm

Step 1: Initialize 'low' to 1 and 'high' to 'n'.
Step 2: While 'low' is less than or equal to 'high':
Step 3: Calculate 'mid' as the average of 'low' and 'high'. To prevent overflow use `low + (high - low) / 2`.
Step 4: Call the 'guess(mid)' API.
Step 5: If 'guess(mid)' returns 0, return 'mid' (the number is found).
Step 6: If 'guess(mid)' returns -1, set 'high' to 'mid - 1' (the number is lower).
Step 7: If 'guess(mid)' returns 1, set 'low' to 'mid + 1' (the number is higher).
Step 8: If the loop finishes without finding the number (which shouldn't happen given the problem constraints), you can return -1 or throw an exception since the number *must* exist.

Key Insights

Insight 1: Binary search is the most efficient approach because it halves the search space with each guess.
Insight 2: Understanding the 'guess' API and how to use its return values to adjust the search range is crucial.
Insight 3: Handling integer overflow when calculating the middle value 'mid' in binary search is important for large 'n'.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Binary Search, Interactive.

Companies

Asked at: Samsung.

Guess Number Higher or Lower | LeetCode 374

Easy

Binary Search Interactive

Time: O(log n)Space: O(1)

Solution Code

Available in:

/** 
 * Forward declaration of guess API.
 * @param  num   your guess
 * @return 	     -1 if num is higher than the picked number
 *			      1 if num is lower than the picked number
 *               otherwise return 0
 * int guess(int num);
 */

public class Solution extends GuessGame {
    public int guessNumber(int n) {
        int low = 1;
        int high = n;
        while (low <= high) {
            int mid = low + (high - low) / 2;
            int result = guess(mid);
            if (result == 0) {
                return mid;
            } else if (result == -1) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return -1; 
    }
}

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Problem Info

Description

Problem 374: Guess Number Higher or Lower - Detailed explanation and solution approaches.

Topics

Binary Search Interactive

Companies

Samsung

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:374

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize 'low' to 1 and 'high' to 'n'.
2
Step 2: While 'low' is less than or equal to 'high':
3
Step 3: Calculate 'mid' as the average of 'low' and 'high'. To prevent overflow use `low + (high - low) / 2`.
4
Step 4: Call the 'guess(mid)' API.
5
Step 5: If 'guess(mid)' returns 0, return 'mid' (the number is found).
6
Step 6: If 'guess(mid)' returns -1, set 'high' to 'mid - 1' (the number is lower).
7
Step 7: If 'guess(mid)' returns 1, set 'low' to 'mid + 1' (the number is higher).
8
Step 8: If the loop finishes without finding the number (which shouldn't happen given the problem constraints), you can return -1 or throw an exception since the number *must* exist.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The binary search algorithm efficiently narrows down the search range. Each step reduces the possible values by half, resulting in a logarithmic time complexity. The algorithm's memory usage remains constant, making it space-efficient.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Linear Search

Iterate through the numbers from 1 to n, guessing each number until the correct one is found.

Time: O(n)Space: O(1)

Trade-offs: Linear search is much less efficient than binary search for large 'n'. Use binary search unless 'n' is extremely small or if you have a reason to believe the hidden number is likely at the start.

✨

Key Insights

1Insight 1: Binary search is the most efficient approach because it halves the search space with each guess.
2Insight 2: Understanding the 'guess' API and how to use its return values to adjust the search range is crucial.
3Insight 3: Handling integer overflow when calculating the middle value 'mid' in binary search is important for large 'n'.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: n = 1. The code handles this case correctly. low and high are both 1, and mid is also 1. If pick is 1, the code immediately returns 1.
Edge case 2: pick = 1 (the number is the lowest possible value). The code correctly searches from 1 to n and finds 1 as the result.
Edge case 3: pick = n (the number is the highest possible value). The code correctly searches and finds n as the result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer Overflow. Calculating 'mid' as (low + high) / 2 can lead to integer overflow if low + high exceeds the maximum integer value. The solution is to calculate mid as low + (high - low) / 2, which avoids the overflow.

✗Common mistake 2: Incorrectly updating 'low' and 'high'. It's important to update 'low' to 'mid + 1' and 'high' to 'mid - 1', *not* just 'mid', because we've already checked 'mid' and it's not the correct answer.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Searching for a value in a sorted database index. Databases use binary search to quickly locate specific records based on indexed fields.

🏦Real-world scenario 2: Finding a specific version of software in a sorted list of available versions. Version control systems can leverage binary search to identify versions efficiently.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Binary search isn't just for sorted arrays. It can be applied to any monotonic function. Look for the "search space" in the problem.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Integer overflow in (left + right) / 2. Use left + (right - left) / 2 instead. Also watch for infinite loops.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 15, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Guess Number Higher or Lower"?

For this easy-level Binary Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether interactive techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?