There is a donuts shop that bakes donuts in batches of batchSize
. They have a rule where they must serve all of the donuts of a batch before serving any donuts of the next batch. You are given an integer batchSize
and an integer array groups
, where groups[i]
denotes that there is a group of groups[i]
customers that will visit the shop. Each customer will get exactly one donut.
When a group visits the shop, all customers of the group must be served before serving any of the following groups. A group will be happy if they all get fresh donuts. That is, the first customer of the group does not receive a donut that was left over from the previous group.
You can freely rearrange the ordering of the groups. Return the maximum possible number of happy groups after rearranging the groups.
Example 1:
Input: batchSize = 3, groups = [1,2,3,4,5,6] Output: 4 Explanation: You can arrange the groups as [6,2,4,5,1,3]. Then the 1st, 2nd, 4th, and 6th groups will be happy.
Example 2:
Input: batchSize = 4, groups = [1,3,2,5,2,2,1,6] Output: 4
Constraints:
1 <= batchSize <= 9
1 <= groups.length <= 30
1 <= groups[i] <= 109
class Solution:
def maxHappyGroups(self, batchSize: int, groups: List[int]) -> int:
@cache
def dfs(state, mod):
res = 0
x = int(mod == 0)
for i in range(1, batchSize):
if state >> (i * 5) & 31:
t = dfs(state - (1 << (i * 5)), (mod + i) % batchSize)
res = max(res, t + x)
return res
state = ans = 0
for v in groups:
i = v % batchSize
ans += i == 0
if i:
state += 1 << (i * 5)
ans += dfs(state, 0)
return ans
class Solution:
def maxHappyGroups(self, batchSize: int, groups: List[int]) -> int:
@cache
def dfs(state, x):
if state == mask:
return 0
vis = [False] * batchSize
res = 0
for i, v in enumerate(g):
if state >> i & 1 == 0 and not vis[v]:
vis[v] = True
y = (x + v) % batchSize
res = max(res, dfs(state | 1 << i, y))
return res + (x == 0)
g = [v % batchSize for v in groups if v % batchSize]
mask = (1 << len(g)) - 1
return len(groups) - len(g) + dfs(0, 0)
class Solution {
private Map<Long, Integer> f = new HashMap<>();
private int size;
public int maxHappyGroups(int batchSize, int[] groups) {
size = batchSize;
int ans = 0;
long state = 0;
for (int g : groups) {
int i = g % size;
if (i == 0) {
++ans;
} else {
state += 1l << (i * 5);
}
}
ans += dfs(state, 0);
return ans;
}
private int dfs(long state, int mod) {
if (f.containsKey(state)) {
return f.get(state);
}
int res = 0;
for (int i = 1; i < size; ++i) {
if ((state >> (i * 5) & 31) != 0) {
int t = dfs(state - (1l << (i * 5)), (mod + i) % size);
res = Math.max(res, t + (mod == 0 ? 1 : 0));
}
}
f.put(state, res);
return res;
}
}
class Solution {
public:
int maxHappyGroups(int batchSize, vector<int>& groups) {
using ll = long long;
unordered_map<ll, int> f;
ll state = 0;
int ans = 0;
for (auto& v : groups) {
int i = v % batchSize;
ans += i == 0;
if (i) {
state += 1ll << (i * 5);
}
}
function<int(ll, int)> dfs = [&](ll state, int mod) {
if (f.count(state)) {
return f[state];
}
int res = 0;
int x = mod == 0;
for (int i = 1; i < batchSize; ++i) {
if (state >> (i * 5) & 31) {
int t = dfs(state - (1ll << (i * 5)), (mod + i) % batchSize);
res = max(res, t + x);
}
}
return f[state] = res;
};
ans += dfs(state, 0);
return ans;
}
};
func maxHappyGroups(batchSize int, groups []int) (ans int) {
state := 0
for _, v := range groups {
i := v % batchSize
if i == 0 {
ans++
} else {
state += 1 << (i * 5)
}
}
f := map[int]int{}
var dfs func(int, int) int
dfs = func(state, mod int) int {
if v, ok := f[state]; ok {
return v
}
res := 0
x := 0
if mod == 0 {
x = 1
}
for i := 1; i < batchSize; i++ {
if state>>(i*5)&31 != 0 {
t := dfs(state-1<<(i*5), (mod+i)%batchSize)
res = max(res, t+x)
}
}
f[state] = res
return res
}
ans += dfs(state, 0)
return
}
func max(a, b int) int {
if a > b {
return a
}
return b
}