LeetCode 486: Predict the Winner Solution in Python – A Step-by-Step Guide

What Is LeetCode 486: Predict the Winner?

In LeetCode 486: Predict the Winner, you’re given an array of non-negative integers nums representing scores, and two players take turns picking a score from either end of the array, adding it to their total. Player 1 goes first, and both play optimally to maximize their own scores. Your task is to determine if Player 1 can achieve a score at least as high as Player 2’s (i.e., win or tie). For example, with nums = [1, 5, 2], Player 1 gets at most 3 vs. Player 2’s 5 (lose), but with [1, 5, 233, 7], Player 1 can secure 234 vs. 12 (win). It’s like a strategic tug-of-war over treasures, where foresight decides the victor.

Problem Statement

• Input: nums (List[int])—array of non-negative integers (scores).
• Output: bool—True if Player 1 can win or tie, False otherwise.
• Rules:
• Players alternate, picking from left or right end.
• Player 1 starts, both play optimally.
• Win if Player 1’s score ≥ Player 2’s.

Constraints

• \( 1 \leq nums.length \leq 20 \).
• \( 0 \leq nums[i] \leq 10^7 \).

Examples to Get Us Started

• Input: nums = [1,5,2]
• Output: False (Player 1: 3, Player 2: 5).
• Input: nums = [1,5,233,7]
• Output: True (Player 1: 234, Player 2: 12).
• Input: nums = [1]
• Output: True (Player 1: 1, Player 2: 0).

Understanding the Problem: Planning the Heist

To solve LeetCode 486: Predict the Winner in Python, we need to determine if Player 1 can secure a score at least as high as Player 2’s by optimally picking from either end of nums, assuming Player 2 also plays optimally. A naive approach—trying all possible move sequences—could be O(2^n), infeasible for n = 20. The key? Use dynamic programming with a min-max strategy to compute the best possible score difference efficiently, leveraging the fact that each move reduces the problem size. We’ll explore:

• Best Solution (Dynamic Programming with Min-Max Strategy): O(n²) time, O(n²) space—fast and optimal.
• Alternative Solution (Recursive Brute Force): O(2^n) time, O(n) space—simple but slow.

Let’s dive into the DP solution—it’s the strategist’s master plan we need.

Best Solution: Dynamic Programming with Min-Max Strategy

Why This Is the Best Solution

The dynamic programming with min-max strategy is the top pick because it’s O(n²) time (n = array length) and O(n²) space, efficiently solving the game by computing the maximum score difference Player 1 can achieve over Player 2 for every subarray, using a 2D DP table to store results. It accounts for optimal play by both players, avoiding exponential recursion. It’s like mapping out every possible heist scenario in a playbook, ensuring Player 1’s best outcome—clever and swift!

How It Works

Here’s the strategy:

• Step 1: Define DP table:
• \( dp[i][j] \) = max net score Player 1 can get over Player 2 for nums[i:j+1].
• Step 2: Base cases:
• \( i = j \): \( dp[i][i] = nums[i] \) (Player 1 takes all, Player 2 gets 0).
• Step 3: Recurrence:
• For subarray i to j, Player 1 picks:
• Left: \( nums[i] - dp[i+1][j] \) (Player 2’s turn on i+1 to j).
• Right: \( nums[j] - dp[i][j-1] \) (Player 2’s turn on i to j-1).
• \( dp[i][j] = \max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1]) \).
• Step 4: Return \( dp[0][n-1] \geq 0 \) (Player 1 wins if net score ≥ 0).
• Why It Works:
• Min-max: Player 1 maximizes, Player 2 minimizes the difference.
• DP builds solution bottom-up, avoiding redundant calculations.

Step-by-Step Example

Example: nums = [1,5,2]

• n = 3, init \( dp[3][3] \).
• Base:
• \( dp[0][0] = 1 \), \( dp[1][1] = 5 \), \( dp[2][2] = 2 \).
• Len 2:
• \( dp[0][1] = \max(1 - dp[1][1], 5 - dp[0][0]) = \max(1-5, 5-1) = \max(-4, 4) = 4 \).
• \( dp[1][2] = \max(5-2, 2-5) = \max(3, -3) = 3 \).
• Len 3:
• \( dp[0][2] = \max(1 - dp[1][2], 2 - dp[0][1]) = \max(1-3, 2-4) = \max(-2, -2) = -2 \).
• Result: \( dp[0][2] = -2 < 0 \), False (Player 1 loses).

Example: nums = [1,5]

• **dp[0][1] = \max(1-5, 5-1) = \max(-4, 4) = 4\).
• Result: 4 ≥ 0, True.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

Copy codeclass Solution:
    def predictTheWinner(self, nums: List[int]) -> bool:
        n = len(nums)
        # Step 1: Initialize DP table
        dp = [[0] * n for _ in range(n)]

        # Step 2: Base cases (single elements)
        for i in range(n):
            dp[i][i] = nums[i]

        # Step 3: Fill DP table for increasing lengths
        for length in range(2, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                # Player 1 picks left or right
                left_pick = nums[i] - dp[i + 1][j]
                right_pick = nums[j] - dp[i][j - 1]
                dp[i][j] = max(left_pick, right_pick)

        # Step 4: Check if Player 1 wins
        return dp[0][n - 1] >= 0

• Line 4-5: Init \( n \) and \( dp[n][n] \) with zeros.
• Line 8-9: Base case: \( dp[i][i] = nums[i] \) (Player 1 takes all).
• Line 12-17: Fill DP:
• For each length and i,j pair.
• Compute left pick (\( nums[i] - dp[i+1][j] \)) and right pick.
• \( dp[i][j] = \max(left, right) \).
• Line 20: Return True if \( dp[0][n-1] \geq 0 \).
• Time Complexity: O(n²)—n² subproblems, O(1) per fill.
• Space Complexity: O(n²)—DP table.

It’s like a strategic game planner!

Alternative Solution: Recursive Brute Force Approach

Why an Alternative Approach?

The recursive brute force approach—O(2^n) time, O(n) space—explores all possible move sequences by recursively simulating Player 1 and Player 2’s optimal picks, computing the score difference without memoization. It’s slow but intuitive, like playing out every heist scenario by hand. Good for small arrays or learning!

How It Works

• Step 1: Define recursive function:
• \( solve(i, j) \) = max net score for nums[i:j+1], Player 1’s turn.
• Step 2: Base case:
• \( i = j \): return nums[i].
• Step 3: Recurrence:
• Left: \( nums[i] - solve(i+1, j) \).
• Right: \( nums[j] - solve(i, j-1) \).
• Max of two.
• Step 4: Return \( solve(0, n-1) \geq 0 \).

Step-by-Step Example

Example: nums = [1,5]

• solve(0,1):
• Left: 1 - solve(1,1) = 1 - 5 = -4.
• Right: 5 - solve(0,0) = 5 - 1 = 4.
• Max = 4.
• Result: 4 ≥ 0, True.

Code for Brute Force

Copy codeclass Solution:
    def predictTheWinner(self, nums: List[int]) -> bool:
        def solve(i: int, j: int) -> int:
            if i == j:
                return nums[i]
            # Player 1 picks left or right
            left = nums[i] - solve(i + 1, j)
            right = nums[j] - solve(i, j - 1)
            return max(left, right)

        n = len(nums)
        score_diff = solve(0, n - 1)
        return score_diff >= 0

• Line 3-9: Recursive solve:
• Base: single element.
• Max of left or right pick, subtract opponent’s score.
• Line 12-13: Compute diff, check if ≥ 0.
• Time Complexity: O(2^n)—exponential branches.
• Space Complexity: O(n)—recursion stack.

It’s a slow game simulator!

Comparing the Two Solutions

• DP with Min-Max (Best):
• Pros: O(n²), fast, scalable.
• Cons: O(n²) space.
• Brute Force (Alternative):
• Pros: O(2^n), simple.
• Cons: Impractical for n > 15.

DP wins for efficiency.

Edge Cases and Examples

• Input: [0] → True.
• Input: [1,1] → True.
• Input: [0,0,0] → True.

DP handles all perfectly.

Complexity Recap

• DP: Time O(n²), Space O(n²).
• Brute Force: Time O(2^n), Space O(n).

DP’s the champ.

Key Takeaways

• DP: Plan with min-max.
• Brute Force: Simulate all moves.
• Python Tip: DP optimizes—see [Python Basics](/python/basics).

Final Thoughts: Claim That Victory

LeetCode 486: Predict the Winner in Python is a strategic treasure duel. DP with min-max is your fast playbook, while brute force is a slow rehearsal. Want more game theory? Try LeetCode 464: Can I Win or LeetCode 877: Stone Game. Ready to predict some wins? Head to Solve LeetCode 486 on LeetCode and strategize today—your coding skills are victory-ready!