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This approach involves using recursion to explore all possible decisions by each player. With each choice, the players reduce the size of the array by selecting an element from either the start or end. Memoization is used to store intermediate results to minimize computational overhead by avoiding repeated calculations.
Time Complexity: O(n²) - Each state (i, j) is computed once.
Space Complexity: O(n²) - Intermediate results are stored in the memoization table.
1public class Solution {
2 public bool PredictTheWinner(int[] nums) {
3 int n = nums.Length;
4 int[,] memo = new int[n, n];
5 for (int i = 0; i < n; i++)
6 for (int j = 0; j < n; j++)
7 memo[i, j] = int.MinValue;
8 return Calculate(nums, 0, n - 1, memo) >= 0;
9 }
10
11 private int Calculate(int[] nums, int start, int end, int[,] memo) {
12 if (start == end) return nums[start];
13 if (memo[start, end] != int.MinValue) return memo[start, end];
14
15 int pickStart = nums[start] - Calculate(nums, start + 1, end, memo);
16 int pickEnd = nums[end] - Calculate(nums, start, end - 1, memo);
17 memo[start, end] = Math.Max(pickStart, pickEnd);
18
19 return memo[start, end];
20 }
21}The C# solution utilizes recursion along with memoization. The recursive function Calculate determines the optimal score a player can achieve given the current array slice, storing intermediate results in the memo table to improve efficiency.
This approach utilizes dynamic programming to solve the problem iteratively. Instead of recursion, it fills up a DP table where each entry represents the best possible score difference a player can achieve for a subarray defined by its boundaries.
Time Complexity: O(n²) - The table is filled once for every distinct range.
Space Complexity: O(n²) - The DP table consumes space proportional to n².
1#include <algorithm>
class Solution {
public:
bool PredictTheWinner(std::vector<int>& nums) {
int n = nums.size();
std::vector<std::vector<int>> dp(n, std::vector<int>(n, 0));
for (int i = 0; i < n; ++i) {
dp[i][i] = nums[i];
}
for (int len = 2; len <= n; ++len) {
for (int i = 0; i <= n - len; ++i) {
int j = i + len - 1;
dp[i][j] = std::max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1]);
}
}
return dp[0][n - 1] >= 0;
}
};The C++ solution follows a similar paradigm to the previous solutions. The DP table dp is populated considering subarrays of increasing length, ensuring that each possible score choice is evaluated optimally for any subarray defined by maximum and minimum indices.