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Java-based LeetCode algorithm problem solutions, regularly updated
package g0801_0900.s0879_profitable_schemes;
// #Hard #Array #Dynamic_Programming #2022_03_28_Time_80_ms_(43.42%)_Space_42.9_MB_(71.71%)
/**
* 879 - Profitable Schemes\.
*
* 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];
}
}