




Sponsored
Sponsored
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.
1#include <stdio.h>
2#include <limits.h>
3
4int calculate(int* nums, int i, int j, int** memo) {
5    if (i == j) return nums[i];
6    if (memo[i][j] != INT_MIN) return memo[i][j];
7    int pickI = nums[i] - calculate(nums, i + 1, j, memo);
8    int pickJ = nums[j] - calculate(nums, i, j - 1, memo);
9    memo[i][j] = (pickI > pickJ) ? pickI : pickJ;
10    return memo[i][j];
11}
12
13int PredictTheWinner(int* nums, int numsSize) {
14    int** memo = (int**) malloc(numsSize * sizeof(int*));
15    for (int i = 0; i < numsSize; ++i) {
16        memo[i] = (int*) malloc(numsSize * sizeof(int));
17        for (int j = 0; j < numsSize; ++j) memo[i][j] = INT_MIN;
18    }
19    int result = calculate(nums, 0, numsSize - 1, memo) >= 0;
20    for (int i = 0; i < numsSize; ++i) free(memo[i]);
21    free(memo);
22    return result;
23}This solution leverages recursion with memoization. It involves calculating the best score a player can secure optimally by recursively deciding whether to pick from the beginning or the end of the array. The results for each state (combination of i and j) of the nums array are stored in a memoization table to avoid redundant calculations.
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.
1This JavaScript solution involves creating a dynamic programming table. Each entry dp[i][j] is computed iteratively to determine the maximum attainable score difference for the subarray nums[i...j], considering both ends as potential picks.