Hard
There is a group of n
members, and a list of various crimes they could commit. The ith
crime generates a profit[i]
and requires group[i]
members to participate in it. If a member participates in one crime, that member can’t participate in another crime.
Let’s call a profitable scheme any subset of these crimes that generates at least minProfit
profit, and the total number of members participating in that subset of crimes is at most n
.
Return the number of schemes that can be chosen. Since the answer may be very large, return it modulo 109 + 7
.
Example 1:
Input: n = 5, minProfit = 3, group = [2,2], profit = [2,3]
Output: 2
Explanation:
To make a profit of at least 3, the group could either commit crimes 0 and 1, or just crime 1.
In total, there are 2 schemes.
Example 2:
Input: n = 10, minProfit = 5, group = [2,3,5], profit = [6,7,8]
Output: 7
Explanation:
To make a profit of at least 5, the group could commit any crimes, as long as they commit one.
There are 7 possible schemes: (0), (1), (2), (0,1), (0,2), (1,2), and (0,1,2).
Constraints:
1 <= n <= 100
0 <= minProfit <= 100
1 <= group.length <= 100
1 <= group[i] <= 100
profit.length == group.length
0 <= profit[i] <= 100
public class Solution {
public int profitableSchemes(int n, int minProfit, int[] group, int[] profit) {
long[][] dp = new long[n + 1][minProfit + 1];
long modulus = 1000000007L;
for (int i = 0; i < dp.length; i++) {
dp[i][0] = 1;
}
for (int i = 0; i < group.length; i++) {
int currWorker = group[i];
int currProfit = profit[i];
for (int j = dp.length - 1; j >= currWorker; j--) {
for (int k = dp[j].length - 1; k >= 0; k--) {
dp[j][k] =
(dp[j][k] + dp[j - currWorker][Math.max((k - currProfit), 0)])
% modulus;
}
}
}
return (int) dp[n][minProfit];
}
}