What is the time complexity of 24 Game?

The optimal solution has a time complexity of O(1).

What data structures are used to solve 24 Game?

This problem uses Array, Math, Backtracking. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving 24 Game?

Insight 1: Backtracking is the core approach. Since there are a limited number of cards (4), we can explore all possible combinations of operations and numbers recursively. Insight 2: Floating-point precision is important when dealing with division. Use a small epsilon value to compare floating-point numbers for equality.

Which companies ask LeetCode 679. 24 Game in interviews?

This problem is frequently asked at Huawei, Roku.

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24 Game - LeetCode 679 Solution

24 Game - Complete Solution Guide

24 Game is LeetCode problem 679, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer array cards of length 4 . You have four cards, each containing a number in the range [1, 9] . You should arrange the numbers on these cards in a mathematical expression using the operators ['+', '-', '*', '/'] and the parentheses '(' and ')' to get the value 24. You are restricted with the following rules: The division operator '/' represents real division, not integer division. For example, 4 / (1 - 2 / 3) = 4 / (1 / 3) = 12 . Every operation done is between two numbers

Detailed Explanation

The 24 Game problem challenges you to use four given numbers (each between 1 and 9) and the basic arithmetic operations (+, -, *, /) along with parentheses to achieve a final result of 24. The division operator represents real division. Each operation must be between two numbers. Number concatenation is not allowed. The task is to determine if it's possible to reach 24 using the given numbers and operations.

Solution Approach

The solution uses a recursive backtracking approach to explore all possible combinations of numbers and operations. It starts with the four given numbers. In each recursive step, it picks two numbers, applies all possible operations to them, and then recursively calls itself with the remaining numbers (including the result of the operation). When only one number is left, it checks if that number is approximately equal to 24 (within a small epsilon).

Step-by-Step Algorithm

Step 1: Convert the input integer array to a double array for floating-point operations.
Step 2: Define a recursive function `solve` that takes a list of numbers as input.
Step 3: Base case for `solve`: if the list contains only one number, check if its absolute difference with 24 is less than a small epsilon value. Return `true` if it is, `false` otherwise.
Step 4: Iterate through all possible pairs of numbers in the list.
Step 5: For each pair, apply all six possible operations (+, -, (a-b), *, /, (a/b), (b/a)) to them, if the divisor is not too close to zero. Calculate the result of each operation.
Step 6: Create a new list containing the remaining numbers and the result of the current operation.
Step 7: Recursively call `solve` with the new list.
Step 8: If the recursive call returns `true`, return `true` immediately. If the base case is not reached or if no valid combination of operations leads to 24, return `false`.

Key Insights

Insight 1: Backtracking is the core approach. Since there are a limited number of cards (4), we can explore all possible combinations of operations and numbers recursively.
Insight 2: Floating-point precision is important when dealing with division. Use a small epsilon value to compare floating-point numbers for equality.
Insight 3: There are only a limited number of permutations and operator combinations to explore, meaning the actual time complexity is acceptable even if it looks high at first glance.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Backtracking.

Companies

Asked at: Huawei, Roku.

24 Game | LeetCode 679

Hard

Array Math Backtracking

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

Solution Code

Available in:

class Solution {
    public boolean judgePoint24(int[] cards) {
        double[] nums = new double[cards.length];
        for (int i = 0; i < cards.length; i++) {
            nums[i] = (double) cards[i];
        }
        return solve(nums);
    }

    private boolean solve(double[] nums) {
        double EPSILON = 1e-6;
        int n = nums.length;
        if (n == 1) {
            return Math.abs(nums[0] - 24) < EPSILON;
        }

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                double a = nums[i];
                double b = nums[j];

                double[] next_nums = new double[n - 2];
                int index = 0;
                for (int k = 0; k < n; k++) {
                    if (k != i && k != j) {
                        next_nums[index++] = nums[k];
                    }
                }

                double[] possible_results = {a + b, a - b, b - a, a * b};
                
                if (Math.abs(b) > EPSILON) {
                  double[] temp = new double[possible_results.length + 1];
                  System.arraycopy(possible_results, 0, temp, 0, possible_results.length);
                  temp[possible_results.length] = a / b;
                  possible_results = temp;
                }
                if (Math.abs(a) > EPSILON) {
                  double[] temp = new double[possible_results.length + 1];
                  System.arraycopy(possible_results, 0, temp, 0, possible_results.length);
                  temp[possible_results.length] = b / a;
                  possible_results = temp;
                }
                
                for (double res : possible_results) {
                    double[] new_nums = new double[next_nums.length + 1];
                    System.arraycopy(next_nums, 0, new_nums, 0, next_nums.length);
                    new_nums[next_nums.length] = res;
                    if (solve(new_nums)) {
                        return true;
                    }
                }
            }
        }

        return false;
    }
}

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Problem Info

Description

Problem 679: 24 Game - Detailed explanation and solution approaches.

Topics

Array Math Backtracking

Companies

HuaweiRoku

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:679

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: Convert the input integer array to a double array for floating-point operations.
2
Step 2: Define a recursive function `solve` that takes a list of numbers as input.
3
Step 3: Base case for `solve`: if the list contains only one number, check if its absolute difference with 24 is less than a small epsilon value. Return `true` if it is, `false` otherwise.
4
Step 4: Iterate through all possible pairs of numbers in the list.
5
Step 5: For each pair, apply all six possible operations (+, -, (a-b), *, /, (a/b), (b/a)) to them, if the divisor is not too close to zero. Calculate the result of each operation.
6
Step 6: Create a new list containing the remaining numbers and the result of the current operation.
7
Step 7: Recursively call `solve` with the new list.
8
Step 8: If the recursive call returns `true`, return `true` immediately. If the base case is not reached or if no valid combination of operations leads to 24, return `false`.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is constant because the number of possibilities we need to check is very limited. In each recursive call, we reduce the number of elements by 1. With 4 numbers initially, we will have at most 3 levels of recursion. In each level, we iterate through the remaining numbers and perform different arithmetic operations. The number of such operations is also limited, making the time complexity effectively constant.

📖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

Brute Force with Permutations

Generate all possible permutations of the input numbers and then try all possible combinations of operators and parentheses. This approach would also be O(1) because we are dealing with only 4 numbers. The advantage to the backtracking approach is that we can stop execution once we find one correct solution.

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

Trade-offs: This approach is possible but might be less elegant than the backtracking solution, as it requires explicitly generating permutations. The backtracking approach implicitly generates and tests permutations during the recursive calls.

✨

Key Insights

1Insight 1: Backtracking is the core approach. Since there are a limited number of cards (4), we can explore all possible combinations of operations and numbers recursively.
2Insight 2: Floating-point precision is important when dealing with division. Use a small epsilon value to compare floating-point numbers for equality.
3Insight 3: There are only a limited number of permutations and operator combinations to explore, meaning the actual time complexity is acceptable even if it looks high at first glance.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Division by zero. The solution checks if the divisor is close to zero (using `abs(b) > EPSILON`) before performing division, preventing runtime errors and inaccurate results.
Edge case 2: Floating-point precision. The solution uses an epsilon value to compare floating-point numbers for equality, addressing the inherent imprecision of floating-point arithmetic. Without epsilon, the comparison might incorrectly return false even when the numbers are very close to each other.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer division. Failing to cast the input integers to doubles before performing division can lead to incorrect results due to integer division. Ensure that all numbers are treated as floating-point numbers throughout the calculation.

✗Common mistake 2: Ignoring floating-point precision. Directly comparing floating-point numbers for equality using `==` can be unreliable. Always use an epsilon value to account for potential imprecision.

✗Common mistake 3: Not considering all possible operations. For each pair of numbers, remember to consider both `a - b` and `b - a`, as well as `a / b` and `b / a` when appropriate.

✅

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: Game development. This type of logic can be used in puzzle games or educational games that involve mathematical calculations and problem-solving.

🏦Real-world scenario 2: Mathematical expression evaluation. The core concepts of backtracking and operator precedence can be applied to more complex expression evaluation scenarios in compilers or scientific software.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

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: October 18, 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 13

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

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "24 Game"?

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

🤖 Need help with this problem?