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

Master LeetCode problem 3317 (Find the Number of Possible Ways for an Event), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3317. Find the Number of Possible Ways for an Event

Problem Explanation

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.

: :

Solution Code

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];
    }
}

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