LeetCodee

486. Predict the Winner

Jump to Solution: Python Java C++ JavaScript C#

Problem Description

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.

Example 1:

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.

Example 2:

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] <= 10⁷

Solution

Python Solution

class Solution:
    def PredictTheWinner(self, nums: List[int]) -> bool:
        n = len(nums)
        dp = {}
        
        def maxDiff(left: int, right: int) -> int:
            if left > right:
                return 0
            
            if (left, right) in dp:
                return dp[(left, right)]
            
            # Choose left end
            pickLeft = nums[left] - maxDiff(left + 1, right)
            # Choose right end
            pickRight = nums[right] - maxDiff(left, right - 1)
            
            dp[(left, right)] = max(pickLeft, pickRight)
            return dp[(left, right)]
        
        return maxDiff(0, n - 1) >= 0

Time Complexity: O(n²)

Where n is the length of the input array. We need to fill a n×n DP table.

Space Complexity: O(n²)

For storing the DP table.

Java Solution

class Solution {
    private int[][] dp;
    private int[] nums;
    
    public boolean PredictTheWinner(int[] nums) {
        int n = nums.length;
        this.nums = nums;
        this.dp = new int[n][n];
        
        for (int[] row : dp) {
            Arrays.fill(row, Integer.MIN_VALUE);
        }
        
        return maxDiff(0, n - 1) >= 0;
    }
    
    private int maxDiff(int left, int right) {
        if (left > right) {
            return 0;
        }
        
        if (dp[left][right] != Integer.MIN_VALUE) {
            return dp[left][right];
        }
        
        // Choose left end
        int pickLeft = nums[left] - maxDiff(left + 1, right);
        // Choose right end
        int pickRight = nums[right] - maxDiff(left, right - 1);
        
        dp[left][right] = Math.max(pickLeft, pickRight);
        return dp[left][right];
    }
}

Time Complexity: O(n²)

Where n is the length of the input array. We need to fill a n×n DP table.

Space Complexity: O(n²)

For storing the DP table.

C++ Solution

class Solution {
private:
    vector> dp;
    vector nums;
    
    int maxDiff(int left, int right) {
        if (left > right) {
            return 0;
        }
        
        if (dp[left][right] != INT_MIN) {
            return dp[left][right];
        }
        
        // Choose left end
        int pickLeft = nums[left] - maxDiff(left + 1, right);
        // Choose right end
        int pickRight = nums[right] - maxDiff(left, right - 1);
        
        dp[left][right] = max(pickLeft, pickRight);
        return dp[left][right];
    }
    
public:
    bool PredictTheWinner(vector& nums) {
        int n = nums.size();
        this->nums = nums;
        dp = vector>(n, vector(n, INT_MIN));
        
        return maxDiff(0, n - 1) >= 0;
    }
};

Time Complexity: O(n²)

Where n is the length of the input array. We need to fill a n×n DP table.

Space Complexity: O(n²)

For storing the DP table.

JavaScript Solution

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var PredictTheWinner = function(nums) {
    const n = nums.length;
    const dp = new Map();
    
    const maxDiff = (left, right) => {
        if (left > right) {
            return 0;
        }
        
        const key = `${left},${right}`;
        if (dp.has(key)) {
            return dp.get(key);
        }
        
        // Choose left end
        const pickLeft = nums[left] - maxDiff(left + 1, right);
        // Choose right end
        const pickRight = nums[right] - maxDiff(left, right - 1);
        
        const result = Math.max(pickLeft, pickRight);
        dp.set(key, result);
        return result;
    };
    
    return maxDiff(0, n - 1) >= 0;
};

Time Complexity: O(n²)

Where n is the length of the input array. We need to fill a n×n DP table.

Space Complexity: O(n²)

For storing the DP table.

C# Solution

public class Solution {
    private int[,] dp;
    private int[] nums;
    
    public bool PredictTheWinner(int[] nums) {
        int n = nums.Length;
        this.nums = nums;
        this.dp = new int[n,n];
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i,j] = int.MinValue;
            }
        }
        
        return MaxDiff(0, n - 1) >= 0;
    }
    
    private int MaxDiff(int left, int right) {
        if (left > right) {
            return 0;
        }
        
        if (dp[left,right] != int.MinValue) {
            return dp[left,right];
        }
        
        // Choose left end
        int pickLeft = nums[left] - MaxDiff(left + 1, right);
        // Choose right end
        int pickRight = nums[right] - MaxDiff(left, right - 1);
        
        dp[left,right] = Math.Max(pickLeft, pickRight);
        return dp[left,right];
    }
}

Time Complexity: O(n²)

Where n is the length of the input array. We need to fill a n×n DP table.

Space Complexity: O(n²)

For storing the DP table.

Approach Explanation

The solution uses dynamic programming with memoization:

  1. Key Insights:
    • Game theory
    • Dynamic programming
    • Optimal play
    • Score difference
  2. Algorithm Steps:
    • Initialize DP
    • Calculate differences
    • Track optimal choices
    • Compare final scores

Implementation Details:

  • Memoization table
  • Recursive solution
  • Score tracking
  • Optimal decision

Optimization Insights:

  • State reuse
  • Space optimization
  • Early termination
  • Efficient lookup

Edge Cases:

  • Single element
  • Two elements
  • Equal scores
  • Large numbers