Description#
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 <= 1000 <= minProfit <= 1001 <= group.length <= 1001 <= group[i] <= 100profit.length == group.length0 <= profit[i] <= 100
Solutions#
Solution 1: recursion with memoization#
We design a function $dfs(i, j, k)$, which means that we start from the $i$-th job, and have chosen $j$ employees, and the current profit is $k$, then the number of schemes in this case is $dfs(0, 0, 0)$.
The execution process of function $dfs(i, j, k)$ is as follows:
- If $i = n$, it means that all the jobs have been considered. If $k \geq minProfit$, the number of schemes is $1$, otherwise the number of schemes is $0$;
- If $i < n$, we can choose not to choose the $i$-th job, then the number of schemes is $dfs(i + 1, j, k)$; if $j + group[i] \leq n$, we can also choose the $i$-th job, then the number of schemes is $dfs(i + 1, j + group[i], \min(k + profit[i], minProfit))$. Here we limit the profit upper limit to $minProfit$, because the profit exceeding $minProfit$ has no effect on our answer.
Finally, return $dfs(0, 0, 0)$.
In order to avoid repeated calculation, we can use the method of memoization. We use a three-dimensional array $f$ to record all the results of $dfs(i, j, k)$. When we calculate the value of $dfs(i, j, k)$, we store it in $f[i][j][k]$. When we call $dfs(i, j, k)$, if $f[i][j][k]$ has been calculated, we return $f[i][j][k]$ directly.
The time complexity is $O(m \times n \times minProfit)$, and th e space complexity is $O(m \times n \times minProfit)$. Here $m$ and $n$ are the number of jobs and employees, and $minProfit$ is the minimum profit.
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| class Solution:
def profitableSchemes(
self, n: int, minProfit: int, group: List[int], profit: List[int]
) -> int:
@cache
def dfs(i: int, j: int, k: int) -> int:
if i >= len(group):
return 1 if k == minProfit else 0
ans = dfs(i + 1, j, k)
if j + group[i] <= n:
ans += dfs(i + 1, j + group[i], min(k + profit[i], minProfit))
return ans % (10**9 + 7)
return dfs(0, 0, 0)
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| class Solution {
private Integer[][][] f;
private int m;
private int n;
private int minProfit;
private int[] group;
private int[] profit;
private final int mod = (int) 1e9 + 7;
public int profitableSchemes(int n, int minProfit, int[] group, int[] profit) {
m = group.length;
this.n = n;
f = new Integer[m][n + 1][minProfit + 1];
this.minProfit = minProfit;
this.group = group;
this.profit = profit;
return dfs(0, 0, 0);
}
private int dfs(int i, int j, int k) {
if (i >= m) {
return k == minProfit ? 1 : 0;
}
if (f[i][j][k] != null) {
return f[i][j][k];
}
int ans = dfs(i + 1, j, k);
if (j + group[i] <= n) {
ans += dfs(i + 1, j + group[i], Math.min(k + profit[i], minProfit));
}
ans %= mod;
return f[i][j][k] = ans;
}
}
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| class Solution {
public:
int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
int m = group.size();
int f[m][n + 1][minProfit + 1];
memset(f, -1, sizeof(f));
const int mod = 1e9 + 7;
function<int(int, int, int)> dfs = [&](int i, int j, int k) -> int {
if (i >= m) {
return k == minProfit ? 1 : 0;
}
if (f[i][j][k] != -1) {
return f[i][j][k];
}
int ans = dfs(i + 1, j, k);
if (j + group[i] <= n) {
ans += dfs(i + 1, j + group[i], min(k + profit[i], minProfit));
}
ans %= mod;
return f[i][j][k] = ans;
};
return dfs(0, 0, 0);
}
};
|
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| func profitableSchemes(n int, minProfit int, group []int, profit []int) int {
m := len(group)
f := make([][][]int, m)
for i := range f {
f[i] = make([][]int, n+1)
for j := range f[i] {
f[i][j] = make([]int, minProfit+1)
for k := range f[i][j] {
f[i][j][k] = -1
}
}
}
const mod = 1e9 + 7
var dfs func(i, j, k int) int
dfs = func(i, j, k int) int {
if i >= m {
if k >= minProfit {
return 1
}
return 0
}
if f[i][j][k] != -1 {
return f[i][j][k]
}
ans := dfs(i+1, j, k)
if j+group[i] <= n {
ans += dfs(i+1, j+group[i], min(k+profit[i], minProfit))
}
ans %= mod
f[i][j][k] = ans
return ans
}
return dfs(0, 0, 0)
}
|
Solution 2: Dynamic Programming#
We define $f[i][j][k]$ to be the number of schemes to make a profit of at least $k$ with $i$ jobs and $j$ workers. Initially, we have $f[0][j][0] = 1$, which means that there is only one scheme to make a profit of $0$ without any jobs.
For the $i$-th job, we can choose to work or not to work. If we do not work, then $f[i][j][k] = f[i - 1][j][k]$; if we work, then $f[i][j][k] = f[i - 1][j - group[i - 1]][max(0, k - profit[i - 1])]$. We need to enumerate $j$ and $k$, and add up all the schemes.
The final answer is $f[m][n][minProfit]$.
The time complexity is $O(m \times n \times minProfit)$, and the space complexity is $O(m \times n \times minProfit)$. Here $m$ and $n$ are the numbers of jobs and workers, and $minProfit$ is the minimum profit to make.
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| class Solution:
def profitableSchemes(
self, n: int, minProfit: int, group: List[int], profit: List[int]
) -> int:
mod = 10**9 + 7
m = len(group)
f = [[[0] * (minProfit + 1) for _ in range(n + 1)] for _ in range(m + 1)]
for j in range(n + 1):
f[0][j][0] = 1
for i, (x, p) in enumerate(zip(group, profit), 1):
for j in range(n + 1):
for k in range(minProfit + 1):
f[i][j][k] = f[i - 1][j][k]
if j >= x:
f[i][j][k] = (f[i][j][k] + f[i - 1][j - x][max(0, k - p)]) % mod
return f[m][n][minProfit]
|
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| class Solution {
public int profitableSchemes(int n, int minProfit, int[] group, int[] profit) {
final int mod = (int) 1e9 + 7;
int m = group.length;
int[][][] f = new int[m + 1][n + 1][minProfit + 1];
for (int j = 0; j <= n; ++j) {
f[0][j][0] = 1;
}
for (int i = 1; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
for (int k = 0; k <= minProfit; ++k) {
f[i][j][k] = f[i - 1][j][k];
if (j >= group[i - 1]) {
f[i][j][k]
= (f[i][j][k]
+ f[i - 1][j - group[i - 1]][Math.max(0, k - profit[i - 1])])
% mod;
}
}
}
}
return f[m][n][minProfit];
}
}
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| class Solution {
public:
int profitableSchemes(int n, int minProfit, vector<int>& group, vector<int>& profit) {
int m = group.size();
int f[m + 1][n + 1][minProfit + 1];
memset(f, 0, sizeof(f));
for (int j = 0; j <= n; ++j) {
f[0][j][0] = 1;
}
const int mod = 1e9 + 7;
for (int i = 1; i <= m; ++i) {
for (int j = 0; j <= n; ++j) {
for (int k = 0; k <= minProfit; ++k) {
f[i][j][k] = f[i - 1][j][k];
if (j >= group[i - 1]) {
f[i][j][k] = (f[i][j][k] + f[i - 1][j - group[i - 1]][max(0, k - profit[i - 1])]) % mod;
}
}
}
}
return f[m][n][minProfit];
}
};
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| func profitableSchemes(n int, minProfit int, group []int, profit []int) int {
m := len(group)
f := make([][][]int, m+1)
for i := range f {
f[i] = make([][]int, n+1)
for j := range f[i] {
f[i][j] = make([]int, minProfit+1)
}
}
for j := 0; j <= n; j++ {
f[0][j][0] = 1
}
const mod = 1e9 + 7
for i := 1; i <= m; i++ {
for j := 0; j <= n; j++ {
for k := 0; k <= minProfit; k++ {
f[i][j][k] = f[i-1][j][k]
if j >= group[i-1] {
f[i][j][k] += f[i-1][j-group[i-1]][max(0, k-profit[i-1])]
f[i][j][k] %= mod
}
}
}
}
return f[m][n][minProfit]
}
|
