> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Predict the Winner Python Solution with Tests

> Tested Python solution for LeetCode 486 with 24 pytest cases. Generate a practice environment with lcpy.

LeetCode 486, [Medium](/catalog/medium). Topics: [Array](/catalog/topics/array), [Math](/catalog/topics/math), [Dynamic Programming](/catalog/topics/dynamic-programming), [Recursion](/catalog/topics/recursion), [Game Theory](/catalog/topics/game-theory). [View on LeetCode](https://leetcode.com/problems/predict-the-winner/description/).

Generate this problem as a practice environment: tested reference solution, 24 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 486   # by problem number
lcpy gen -s predict_the_winner   # by problem name
```

## Problem

You are given an integer array `nums`. Two players are playing a game with this array: player 1 and player 2.

Player 1 and player 2 take turns, with player 1 starting first. Both players start the game with a score of `0`. At each turn, the player takes one of the numbers from either end of the array (i.e., `nums[0]` or `nums[nums.length - 1]`) which reduces the size of the array by `1`. The player adds the chosen number to their score. The game ends when there are no more elements in the array.

Return `true` if Player 1 can win the game. If the scores of both players are equal, then player 1 is still the winner, and you should also return `true`. You may assume that both players are playing optimally.

### Examples

```
Input: nums = [1,5,2]
Output: false
Explanation: 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.
```

```
Input: nums = [1,5,233,7]
Output: true
Explanation: 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] \<= 10^7

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/predict_the_winner/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/predict_the_winner/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n^2)
    # Space: O(n^2)
    def predict_the_winner(self, nums: list[int]) -> bool:
        n = len(nums)
        # dp[l][r] is the best score difference (current player minus opponent)
        # achievable on the subarray nums[l:r + 1].
        dp = [[0] * n for _ in range(n)]
        for i in range(n):
            dp[i][i] = nums[i]
        for length in range(2, n + 1):
            for left in range(n - length + 1):
                right = left + length - 1
                take_left = nums[left] - dp[left + 1][right]
                take_right = nums[right] - dp[left][right - 1]
                dp[left][right] = max(take_left, take_right)
        return dp[0][n - 1] >= 0
```

## Complexity

| Time | Space |
| - | - |
| O(n^2) | O(n^2) |

## Tags


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