Profitable schemes

Array Dynamic Programming

48.2% acceptance

Feb 22, 2026

1944

126

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.

Solution

Rust

Time O(n * m)

Space O(n * m)

LeetCode

solution.rs

/*
 * 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
 */

impl Solution {
  pub fn profitable_schemes(n: i32, min_profit: i32, group: Vec<i32>, profit: Vec<i32>) -> i32 {
    const MOD: i64 = 1_000_000_007;
    let n = n as usize;
    let p = min_profit as usize;
    // dp[j][k] = ways using j people generating >= k profit
    let mut dp = vec![vec![0i64; p + 1]; n + 1];
    dp[0][0] = 1;
    for (g, pr) in group.iter().zip(profit.iter()) {
      let (g, pr) = (*g as usize, *pr as usize);
      // iterate in reverse to avoid using same crime twice
      for j in (0..=n).rev() {
        if j < g { continue; }
        for k in (0..=p).rev() {
          let k2 = (k + pr).min(p);
          dp[j][k2] = (dp[j][k2] + dp[j - g][k]) % MOD;
        }
      }
    }
    let ans: i64 = (0..=n).map(|j| dp[j][p]).sum();
    (ans % MOD) as i32
  }
}