#486

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Predict the Winner

Predict the Winner involves two players taking turns to maximize their score by picking from either end of an array, optimizing with dynamic programming.

array math dynamic programming recursion game theory

Problem Statement

In the game 'Predict the Winner', two players, Player 1 and Player 2, take turns selecting a number from either end of an integer array 'nums'. Each player adds the selected number to their score. The game ends when the array is empty, and the player with the highest score wins. If both players have equal scores, Player 1 is still considered the winner. The challenge is to determine if Player 1 can win, assuming both players play optimally.

You are tasked with returning true if Player 1 can win. You can assume that both players make the best possible decisions at every step, and the game follows the given rules of taking turns from either end of the array. The optimal strategy for each player needs to be evaluated using dynamic programming and state transitions.

Examples

Example 1

Input: nums = [1,5,2]

Output: false

Initially, player 1 can choose between 1 and 2. If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. If player 2 chooses 5, then player 1 will be left with 1 (or 2). So, final score of player 1 is 1 + 2 = 3, and player 2 is 5. Hence, player 1 will never be the winner and you need to return false.

Example 2

Input: nums = [1,5,233,7]

Output: true

Player 1 first chooses 1. Then player 2 has to choose between 5 and 7. No matter which number player 2 choose, player 1 can choose 233. Finally, player 1 has more score (234) than player 2 (12), so you need to return True representing player1 can win.

Constraints

1 <= nums.length <= 20
0 <= nums[i] <= 107

Solution Approach

State Transition Dynamic Programming

The problem can be approached with dynamic programming by storing the optimal outcomes for subarrays. Define a recursive function that evaluates the best score difference for Player 1 and Player 2 for each subarray, ensuring that the current player always makes the optimal choice based on future states.

Memoization to Avoid Redundant Calculations

To optimize the solution, use memoization to store the results of previously computed subarrays. This helps avoid recalculating the results for the same subarray multiple times, reducing time complexity.

Base Case and Recursive Transition

The base case of the recursion is when only one element remains in the subarray. The current player selects that element. From there, recursively calculate the optimal choices by evaluating the two possible end values and the results of future subarrays.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this solution is O(n^2) due to the two nested loops for each subarray. The space complexity is O(n^2) for storing the results of subarrays during the memoization process.

What Interviewers Usually Probe

Can the candidate explain the optimal substructure and state transition of this problem?
Does the candidate properly implement dynamic programming and memoization techniques?
Can the candidate recognize the need for recursion and optimize with memoization?

Common Pitfalls or Variants

Common pitfalls

Failing to implement the correct state transition for subarrays, leading to incorrect predictions of the winner.
Not using memoization properly, causing unnecessary recomputation and inefficient performance.
Overlooking the need to handle both player choices optimally, potentially assuming one player will always make a bad choice.

Follow-up variants

What if the array contains negative numbers? How would the strategy change?
Can the solution be optimized further in terms of space complexity?
What happens if Player 1 and Player 2 can only pick the same number on each turn?

FAQ

What is the optimal strategy for Player 1 in Predict the Winner?

Player 1 needs to always pick the number that maximizes their score while minimizing Player 2’s future score by considering both ends of the current subarray.

How does dynamic programming help in Predict the Winner?

Dynamic programming stores the results of subarrays, avoiding recomputation and allowing for efficient evaluation of optimal choices for both players.

What is the time complexity of the solution to Predict the Winner?

The time complexity is O(n^2), where n is the length of the array, due to the two nested loops for each subarray calculation.

Can Player 1 ever lose if they play optimally?

Yes, Player 1 can lose if the numbers on the array are distributed in such a way that Player 2 can always outscore them, even when Player 1 plays optimally.

How can I implement memoization in Predict the Winner?

Memoization can be implemented by storing the results of computed subarrays in a cache to avoid recalculating the same result multiple times, thereby improving performance.

terminal

Solution

Solution 1: Memoization Search

We design a function $\textit{dfs}(i, j)$, which represents the maximum difference in scores between the current player and the other player from the $i$-th number to the $j$-th number. The answer is $\textit{dfs}(0, n - 1) \geq 0$.

class Solution:
    def predictTheWinner(self, nums: List[int]) -> bool:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            return max(nums[i] - dfs(i + 1, j), nums[j] - dfs(i, j - 1))

        return dfs(0, len(nums) - 1) >= 0

Solution 2: Dynamic Programming

We can also use dynamic programming. Define $f[i][j]$ to represent the maximum score difference the current player can achieve in the range $\textit{nums}[i..j]$. The final answer is $f[0][n - 1] \geq 0$.

class Solution:
    def predictTheWinner(self, nums: List[int]) -> bool:
        @cache
        def dfs(i: int, j: int) -> int:
            if i > j:
                return 0
            return max(nums[i] - dfs(i + 1, j), nums[j] - dfs(i, j - 1))

        return dfs(0, len(nums) - 1) >= 0

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