How do you solve Stone Game IV on LeetCode?

Stone Game IV (LeetCode #1510) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n√n) time complexity and O(n) space complexity. Key insight: Alice can win if she can force Bob into a losing state.

What is the brute force solution for Stone Game IV?

The brute force approach for Stone Game IV is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we recursively explore all possible moves Alice can make by removing square numbers of stones. We check if there exists a move that forces Bob into a losing position. If such a move exists, Alice wins; otherwise, she loses. The optimal Optimal Solution approach improves this to O(n√n).

What algorithm or data structure is used to solve Stone Game IV?

Stone Game IV uses the following concepts: Math, Dynamic Programming, Game Theory. The recommended patterns to study are: Dynamic Programming, Game Theory.

What are the interview tips for Stone Game IV?

Explain your thought process clearly while coding. Discuss potential edge cases and how your solution handles them. Practice dynamic programming problems to become familiar with state transitions.

What are common mistakes when solving Stone Game IV?

Not considering all square numbers when making a move. Forgetting to check base cases in dynamic programming.

#1510

Stone Game IV

Hard

MathDynamic ProgrammingGame TheoryDynamic ProgrammingGame Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n√n)Space O(n)

We can use dynamic programming to store the winning states for each number of stones. By building up from smaller numbers, we can determine if Alice can win with n stones by checking if she can force Bob into a losing state.

⚙️

Algorithm

3 steps

1Step 1: Create a boolean array dp of size n+1, where dp[i] indicates if Alice can win with i stones.
2Step 2: Initialize dp[0] to false (no stones means Bob wins).
3Step 3: For each number of stones from 1 to n, check all square numbers and update dp[i] based on whether removing that square leads to a losing state for Bob.

solution.py10 lines

1def winnerSquareGame(n):
2    dp = [False] * (n + 1)
3    for i in range(1, n + 1):
4        j = 1
5        while j * j <= i:
6            if not dp[i - j * j]:
7                dp[i] = True
8                break
9            j += 1
10    return dp[n]

ℹ

Complexity note: The time complexity is O(n√n) because for each of the n states, we are iterating through the square numbers up to √n, leading to a linear growth with respect to n and a logarithmic growth with respect to the number of squares.

1Alice can win if she can force Bob into a losing state.
2Dynamic programming helps in breaking down the problem into manageable subproblems.

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