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
.
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.
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).
1 <= n <= 100
0 <= minProfit <= 100
1 <= group.length <= 100
1 <= group[i] <= 100
profit.length == group.length
0 <= profit[i] <= 100
impl Solution {
pub fn profitable_schemes(n: i32, min_profit: i32, group: Vec<i32>, profit: Vec<i32>) -> i32 {
let n = n as usize;
let min_profit = min_profit as usize;
let mut dp = vec![vec![0; n + 1]; min_profit + 1];
dp[0][n] = 1;
for i in 0..group.len() {
for p in (0..=min_profit).rev() {
for m in group[i] as usize..=n {
dp[min_profit.min(p + profit[i] as usize)][m - group[i] as usize] += dp[p][m];
dp[min_profit.min(p + profit[i] as usize)][m - group[i] as usize] %=
1_000_000_007;
}
}
}
dp[min_profit]
.iter()
.fold(0, |acc, x| (acc + x) % 1_000_000_007)
}
}