grid47 Exploring patterns and algorithms
Sep 19, 2024
6 min read
Solution to LeetCode 486: Predict the Winner Problem
You are given an integer array nums. Two players, Player 1 and Player 2, take turns to pick numbers from either end of the array. Each player adds the selected number to their score. The goal is to determine if Player 1 can win the game. Player 1 wins if they have a higher score or if the scores are tied.
Problem
Approach
Steps
Complexity
Input: The input consists of an array `nums`, representing the available numbers that the players can select from.
Example: nums = [3, 9, 1, 2]
Constraints:
• 1 <= nums.length <= 20
• 0 <= nums[i] <= 10^7
Output: Return `true` if Player 1 can win the game, or `false` otherwise.
Example: Output: true
Constraints:
• The result will be a boolean value indicating whether Player 1 can win the game.
Goal: Implement a strategy to simulate the game and check if Player 1 can win using dynamic programming or recursive techniques.
Steps:
• 1. Use dynamic programming to calculate the best score each player can achieve based on optimal choices.
• 2. Start with Player 1 making the first move.
• 3. Evaluate both possible moves from either end of the array.
• 4. Use memoization to avoid recalculating the same game states.
Goal: The constraints on the input are as follows.
Steps:
• 1 <= nums.length <= 20
• 0 <= nums[i] <= 10^7
Assumptions:
• The array will always have at least one element.
• Both players are playing optimally, making the best choice at each step.
• Input: nums = [3, 9, 1, 2]
• Explanation: Player 1 picks 3 first, leaving Player 2 with 9 and 1. Player 1 picks 9 next, leaving Player 2 to pick 1. Player 1 ends up with a score of 12, while Player 2 has a score of 3. Therefore, Player 1 wins.
• Input: nums = [5, 3, 4, 5]
• Explanation: Player 1 picks 5 first. Player 2 can pick either 5 or 4, but Player 1 will always be able to pick the higher number. In the end, Player 1 wins.
Approach: The approach involves simulating the game and using dynamic programming or recursion to calculate the best score each player can achieve based on the current state of the array.
Observations:
• Both players can make optimal decisions by looking ahead at the possible outcomes.
• Memoization is essential to avoid recalculating the same game states.
• The problem can be solved using a top-down dynamic programming approach with memoization to optimize the score calculation.
Steps:
• 1. Use a helper function `dp(l, r)` to calculate the best score difference between the two players starting from index `l` to `r`.
• 2. Use recursion to calculate the result for the next possible game state by selecting either the leftmost or rightmost number from the array.
• 3. Store the intermediate results in a memo table to avoid recomputation.
Empty Inputs:
• The input will always contain at least one element, so there is no need to handle empty arrays.
Large Inputs:
• For larger input sizes, ensure that the solution is optimized with memoization to handle the recursive calls efficiently.
Special Values:
• When the array has only one element, Player 1 will always win by selecting that element.
Constraints:
• Ensure that the algorithm can handle input arrays up to the maximum length of 20 elements.
vector<int> nums;
vector<vector<map<int, int>>> memo;
intdp(int l, int r, int net) {
if(l == r) return nums[l];
if(memo[l][r].count(net)) return memo[l][r][net];
int ans = net - dp(l +1, r, net - nums[l]);
ans = max(ans, net - dp(l, r -1, net - nums[r]));
return memo[l][r][net] = ans;
}
boolpredictTheWinner(vector<int>& nums) {
this->nums = nums;
int n = nums.size();
memo.resize(n +1, vector<map<int,int>>(n +1));
int net = accumulate(nums.begin(), nums.end(), 0);
int ret = dp(0, n -1, net);
return ret >= net - ret;
}
1 : Variable Declaration
vector<int> nums;
Declares a vector `nums` to store the game values that will be used to compute the scores.
2 : Memoization Structure
vector<vector<map<int, int>>> memo;
Declares a 3D vector of maps `memo` for memoization, storing the results of subproblems to avoid redundant calculations in the dynamic programming solution.
3 : Function Definition
intdp(int l, int r, int net) {
Defines the `dp` function that calculates the maximum score difference for a given range of the game array.
4 : Base Case
if(l == r) return nums[l];
Base case: when only one element is left, return its value as it represents the score the current player will get.
5 : Memoization Check
if(memo[l][r].count(net)) return memo[l][r][net];
Checks if the result for the current subproblem has already been computed. If yes, it returns the stored result to avoid redundant calculations.
6 : Recursion
int ans = net - dp(l +1, r, net - nums[l]);
Recursively calculates the score difference by choosing the leftmost element and subtracting it from the net score.
7 : Recursion
ans = max(ans, net - dp(l, r -1, net - nums[r]));
Compares the result of choosing the leftmost element with the result of choosing the rightmost element. The maximum of these two choices is stored as the final answer.
8 : Memoization
return memo[l][r][net] = ans;
Stores the computed result in the `memo` structure to avoid redundant calculations in future calls.
9 : Function Definition
boolpredictTheWinner(vector<int>& nums) {
Defines the `predictTheWinner` function which determines if the first player can win given the array `nums`.
10 : Input Assignment
this->nums = nums;
Assigns the input vector `nums` to the class-level `nums` variable for use in the dynamic programming solution.
11 : Size Calculation
int n = nums.size();
Calculates the size of the input vector `nums` and stores it in the variable `n`.
12 : Memoization Initialization
memo.resize(n +1, vector<map<int,int>>(n +1));
Initializes the memoization structure to handle all subproblems of size up to `n`.
13 : Net Score Calculation
int net = accumulate(nums.begin(), nums.end(), 0);
Calculates the total score (`net`) by summing all elements in `nums`.
14 : Recursive Call
int ret = dp(0, n -1, net);
Calls the `dp` function to calculate the score difference between the two players starting from the first and last elements of the array.
15 : Return Statement
return ret >= net - ret;
Returns `true` if the first player can win or tie, otherwise returns `false`.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The time complexity is O(n^2) because the algorithm involves recursively calculating the best score for each subarray, and memoization ensures that each state is calculated only once.
Best Case: O(n^2)
Worst Case: O(n^2)
Description: The space complexity is O(n^2) due to the memoization table storing results for every possible subarray.
