LeetCode 3317: Find the Number of Possible Ways for an Event
LeetCode 3317 Solution 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:
- Initialize a 3D array
dpof size(n+1) x (x+1) x (y+1)with all values as 0. - Set
dp[0][0][0] = 1as there is 1 way to assign stages and scores to 0 performers with 0 stages and 0 scores. - 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 performerito a new stage and score)dp[i][j][k] = (dp[i][j][k] + dp[i-1][j][k]) * (i-j)(assigning performerito an existing stage and score)
- Update
- Iterate over scores from 1 to
- Iterate over stages from 1 to
- 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.
: :
LeetCode 3317 Solutions in Java, C++, Python
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];
}
}Interactive Code Editor for LeetCode 3317
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