How do you solve 24 Game on LeetCode?

24 Game (LeetCode #679) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n!) time complexity and O(n) space complexity. Key insight: Understanding operator precedence is crucial.

What is the brute force solution for 24 Game?

The brute force approach for 24 Game is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible combination of the four numbers and the operators to see if we can get to 24. This involves generating all permutations of the numbers and all combinations of the operators. The optimal Optimal Solution approach improves this to O(n!).

What algorithm or data structure is used to solve 24 Game?

24 Game uses the following concepts: Array, Math, Backtracking. The recommended patterns to study are: Backtracking, Permutations.

What are the interview tips for 24 Game?

Explain your thought process clearly. Discuss edge cases, like zeros or negative results. Practice writing clean, efficient code.

What are common mistakes when solving 24 Game?

Forgetting to handle division by zero. Not considering all permutations and combinations.

#679

24 Game

Hard

ArrayMathBacktrackingBacktrackingPermutations

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n!)
Space	O(1)	O(n)

💡

Intuition

Time O(n!)Space O(n)

We can use a backtracking approach to recursively try different combinations of numbers and operations, reducing unnecessary calculations by pruning paths that cannot lead to 24.

⚙️

Algorithm

3 steps

1Step 1: Create a recursive function that takes the current numbers and checks if we can reach 24.
2Step 2: For each pair of numbers, apply each operator and recursively call the function with the new number and the remaining numbers.
3Step 3: If at any point we reach 24, return true.

solution.py25 lines

1# Full working Python code
2from itertools import permutations
3
4def canGet24(cards):
5    def backtrack(nums):
6        if len(nums) == 1:
7            return abs(nums[0] - 24) < 1e-6
8        for i in range(len(nums)):
9            for j in range(len(nums)):
10                if i != j:
11                    next_nums = [nums[k] for k in range(len(nums)) if k != i and k != j]
12                    for op in ['+', '-', '*', '/']:
13                        if op == '+':
14                            next_nums.append(nums[i] + nums[j])
15                        elif op == '-':
16                            next_nums.append(nums[i] - nums[j])
17                        elif op == '*':
18                            next_nums.append(nums[i] * nums[j])
19                        elif nums[j] != 0:
20                            next_nums.append(nums[i] / nums[j])
21                        if backtrack(next_nums):
22                            return True
23        return False
24
25    return backtrack(cards)

ℹ

Complexity note: The time complexity is O(n!) due to the permutations of numbers and combinations of operations, but we significantly reduce the number of evaluations through backtracking.

1Understanding operator precedence is crucial.
2Recognizing that not all combinations will yield valid results helps in pruning.

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