Today I have done leetcode's April LeetCoding Challenge with cpp.

April LeetCoding Challenge 21

Description

Profitable Schemes

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

Solution

class Solution {
  int len;
  const int MOD = 1e9 + 7;
  using pi = pair<int, int>;
  vector<pi> crimes;
  int dp[101][101][100];

  int solve(int restPeople, int restProfit, int position) {
    if(position == len || restPeople < crimes[position].first) {
      return restProfit <= 0;
    }
    restProfit = max(0, restProfit);
    if(dp[restPeople][restProfit][position] != -1) return dp[restPeople][restProfit][position];
    int result = solve(restPeople - crimes[position].first, restProfit - crimes[position].second, position + 1);
    result += solve(restPeople, restProfit, position + 1);
    result %= MOD;

    return dp[restPeople][restProfit][position] = result;
  }
public:
  int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
    len = group.size();
    memset(dp, -1, sizeof(dp));
    crimes.reserve(len);
    for(int i = 0; i < len; ++i) {
      crimes.push_back({group[i], profit[i]});
    }
    sort(crimes.begin(), crimes.end());

    return solve(n, minProfit, 0);
  }
};

// Accepted
// 45/45 cases passed (120 ms)
// Your runtime beats 91.04 % of cpp submissions
// Your memory usage beats 54.72 % of cpp submissions (13 MB)