LeetCode 3317: Find the Number of Possible Ways for an Event

Explanation:

To solve this problem, we can use dynamic programming. We will create a 3D array dp where dp[i][j][k] represents the number of ways to assign stages and scores to the first i performers, with j stages and k scores.

We can iterate over the performers and for each performer, we can iterate over the stages and scores, updating the dp array based on the previous state. The final answer will be the sum of all possible ways to assign stages and scores to all performers.

Algorithm:

1. Initialize a 3D array dp of size (n+1) x (x+1) x (y+1) with all values as 0.
2. Set dp[0][0][0] = 1 as there is 1 way to assign stages and scores to 0 performers with 0 stages and 0 scores.
3. Iterate over performers from 1 to n:
   • Iterate over stages from 1 to x:
     • Iterate over scores from 1 to y:
       • Update dp[i][j][k] based on the previous state:
         • dp[i][j][k] = (dp[i][j][k] + dp[i-1][j-1][k-1]) * j (assigning performer i to a new stage and score)
         • dp[i][j][k] = (dp[i][j][k] + dp[i-1][j][k]) * (i-j) (assigning performer i to an existing stage and score)
4. Return the total number of ways as dp[n][x][y] % (10^9 + 7).

Time Complexity:

The time complexity of this solution is O(n * x * y) where n, x, and y are the given integers.

Space Complexity:

The space complexity is also O(n * x * y) due to the 3D array dp.

: :

class Solution {
    public int countWays(int n, int x, int y) {
        int MOD = 1000000007;
        int[][][] dp = new int[n + 1][x + 1][y + 1];
        dp[0][0][0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= x; j++) {
                for (int k = 1; k <= y; k++) {
                    dp[i][j][k] = (int) (((long) dp[i - 1][j - 1][k - 1] * j + (long) dp[i - 1][j][k] * (i - j)) % MOD);
                }
            }
        }
        return dp[n][x][y];
    }
}