24 Game
Time O(1) · Space O(1) · Official statement on LeetCode
Solutions
// Time: O(n^3 * 4^n) = O(1), n = 4
// Space: O(n^2) = O(1)
class Solution {
public:
bool judgePoint24(vector<int>& nums) {
vector<double> doubles;
std::transform(nums.begin(), nums.end(), std::back_inserter(doubles),
[](const int num) { return double(num); });
return dfs(doubles);
}
private:
bool dfs(const vector<double>& nums) {
if (nums.size() == 1) {
return fabs(nums[0] - 24) < 1e-6;
}
static unordered_map<char, std::function<double(double, double)>> ops =
{
{'+', std::plus<double>()},
{'-', std::minus<double>()},
{'*', std::multiplies<double>()},
{'/', std::divides<double>()},
};
for (int i = 0; i < nums.size(); ++i) {
for (int j = 0; j < nums.size(); ++j) {
if (i == j) {
continue;
}
vector<double> next_nums;
for (int k = 0; k < nums.size(); ++k) {
if (k == i || k == j) {
continue;
}
next_nums.emplace_back(nums[k]);
}
for (const auto& op : ops) {
if (((op.first == '+' || op.first == '*') && i > j) ||
(op.first == '/' && nums[j] == 0)) {
continue;
}
next_nums.emplace_back(op.second(nums[i], nums[j]));
if (dfs(next_nums)) {
return true;
}
next_nums.pop_back();
}
}
}
return false;
}
};
class Fraction {
public:
Fraction() = default;
Fraction(int n)
: Fraction(n, 1)
{
}
Fraction(int n, int d)
: numerator_(n)
, denominator_(d)
{
}
~Fraction() = default;
void set_num(int value) { numerator_ = value; }
void set_den(int value) { denominator_ = value; }
int get_num() const { return numerator_; }
int get_den() const { return denominator_; }
void reduce();
int calculate_gcd(int, int) const;
private:
int numerator_, denominator_;
};
void Fraction::reduce()
{
const auto gcd = calculate_gcd(numerator_, denominator_);
numerator_ = numerator_ / gcd;
denominator_ = denominator_ / gcd;
}
int Fraction::calculate_gcd(int a, int b) const
{
a = std::abs(a);
b = std::abs(b);
while (b != 0) {
int tmp = b;
b = a % b;
a = tmp;
}
return a;
}
Fraction operator+(const Fraction& lhs, const Fraction& rhs)
{
Fraction result{};
result.set_num((lhs.get_num() * rhs.get_den()) + (lhs.get_den() * rhs.get_num()));
result.set_den(lhs.get_den() * rhs.get_den());
result.reduce();
return result;
}
Fraction operator-(const Fraction& lhs, const Fraction& rhs)
{
Fraction result{};
result.set_num((lhs.get_num() * rhs.get_den()) - (lhs.get_den() * rhs.get_num()));
result.set_den(lhs.get_den() * rhs.get_den());
result.reduce();
return result;
}
Fraction operator*(const Fraction& lhs, const Fraction& rhs)
{
Fraction result{};
result.set_num(lhs.get_num() * rhs.get_num());
result.set_den(lhs.get_den() * rhs.get_den());
result.reduce();
return result;
}
Fraction operator/(const Fraction& lhs, const Fraction& rhs)
{
Fraction result{};
result.set_num(lhs.get_num() * rhs.get_den());
result.set_den(lhs.get_den() * rhs.get_num());
result.reduce();
return result;
}
bool operator==(const Fraction &lhs, const Fraction &rhs) {
return (((lhs.get_num() * rhs.get_den()) - (rhs.get_num() * lhs.get_den())) == 0);
}
std::ostream &operator<<(std::ostream &os, const Fraction &value) {
os << value.get_num() << "/" << value.get_den();
return os;
}
// Time: O(n^3 * 4^n) = O(1), n = 4
// Space: O(n^2) = O(1)
class Solution2 {
public:
bool judgePoint24(vector<int>& nums) {
vector<Fraction> fraction_nums;
std::transform(nums.begin(), nums.end(), std::back_inserter(fraction_nums),
[](const int num) { return Fraction(num); });
return dfs(fraction_nums);
}
private:
bool dfs(const vector<Fraction>& nums) {
if (nums.size() == 1) {
return nums[0] == 24;
}
static unordered_map<char, std::function<Fraction(Fraction, Fraction)>> ops =
{
{'+', std::plus<Fraction>()},
{'-', std::minus<Fraction>()},
{'*', std::multiplies<Fraction>()},
{'/', std::divides<Fraction>()},
};
for (int i = 0; i < nums.size(); ++i) {
for (int j = 0; j < nums.size(); ++j) {
if (i == j) {
continue;
}
vector<Fraction> next_nums;
for (int k = 0; k < nums.size(); ++k) {
if (k == i || k == j) {
continue;
}
next_nums.emplace_back(nums[k]);
}
for (const auto& op : ops) {
if (((op.first == '+' || op.first == '*') && i > j) ||
(op.first == '/' && nums[j] == 0)) {
continue;
}
next_nums.emplace_back(op.second(nums[i], nums[j]));
if (dfs(next_nums)) {
return true;
}
next_nums.pop_back();
}
}
}
return false;
}
};
Beginner Explanation
What is 24 Game?
24 Game (LeetCode #679) is a Hard problem that primarily trains backtracking.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with dfs backtracking.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: DFS.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for 24 Game
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to dfs backtracking.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(1)) and space (O(1)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(1) time and O(1) space.
Pattern focus: dfs backtracking
Use the pattern as a checklist:
- dfs backtracking — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(1) |
| Space | O(1) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on 24 Game
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dfs backtracking — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to dfs backtracking:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: backtracking.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve 24 Game in a second language (cpp, python).
- Drill 3–5 more problems tagged backtracking.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the dfs backtracking approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
24 Game (#679) — Hard. Pattern: dfs backtracking. Complexity: O(1) time / O(1) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of 24 Game?+
The reference solutions aim for O(1) time and O(1) space. Always re-derive complexity from the code you write in the interview.
What pattern does 24 Game use?+
It primarily maps to dfs backtracking, within the broader topic of backtracking.
Is 24 Game good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/24-game/