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Formatted question description: https://leetcode.ca/all/2064.html

# 2064. Minimized Maximum of Products Distributed to Any Store (Medium)

You are given an integer n indicating there are n specialty retail stores. There are m product types of varying amounts, which are given as a 0-indexed integer array quantities, where quantities[i] represents the number of products of the ith product type.

You need to distribute all products to the retail stores following these rules:

• A store can only be given at most one product type but can be given any amount of it.
• After distribution, each store will be given some number of products (possibly 0). Let x represent the maximum number of products given to any store. You want x to be as small as possible, i.e., you want to minimize the maximum number of products that are given to any store.

Return the minimum possible x.

Example 1:

Input: n = 6, quantities = [11,6]
Output: 3
Explanation: One optimal way is:
- The 11 products of type 0 are distributed to the first four stores in these amounts: 2, 3, 3, 3
- The 6 products of type 1 are distributed to the other two stores in these amounts: 3, 3
The maximum number of products given to any store is max(2, 3, 3, 3, 3, 3) = 3.


Example 2:

Input: n = 7, quantities = [15,10,10]
Output: 5
Explanation: One optimal way is:
- The 15 products of type 0 are distributed to the first three stores in these amounts: 5, 5, 5
- The 10 products of type 1 are distributed to the next two stores in these amounts: 5, 5
- The 10 products of type 2 are distributed to the last two stores in these amounts: 5, 5
The maximum number of products given to any store is max(5, 5, 5, 5, 5, 5, 5) = 5.


Example 3:

Input: n = 1, quantities = [100000]
Output: 100000
Explanation: The only optimal way is:
- The 100000 products of type 0 are distributed to the only store.
The maximum number of products given to any store is max(100000) = 100000.


Constraints:

• m == quantities.length
• 1 <= m <= n <= 105
• 1 <= quantities[i] <= 105

Similar Questions:

Intuition:

1. Given k (the maximum number of products given to any store), we can compute the number of stores we are able to assign in O(Q) time where Q is the length of quantities.
2. k is in range [1, MAX(quantities)] and this answer range has perfect monotonicity: There must exist a K that for every k >= K, we can do the assignment with k; for every k < K, we can’t do the assignment because the number of stores are not enough.

So, we can use binary search to find this K in O(Qlog(MAX(quantities))) time.

• // OJ: https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/
// Time: O(QlogM) where Q is the length of quantities and M is the max element in quantities.
// Space: O(1)
class Solution {
public:
int minimizedMaximum(int n, vector<int>& Q) {
long long L = 1, R = accumulate(begin(Q), end(Q), 0LL);
auto valid = [&](int M) {
int ans = 0;
for (int n : Q) ans += (n + M - 1) / M; // ceil(n / M)
return ans <= n;
};
while (L <= R) {
long long M = (L + R) / 2;
if (valid(M)) R = M - 1;
else L = M + 1;
}
return L;
}
};

• class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
left, right = 1, int(1e5)
while left < right:
mid = (left + right) >> 1
s = sum([(q + mid - 1) // mid for q in quantities])
if s <= n:
right = mid
else:
left = mid + 1
return left

############

# 2064. Minimized Maximum of Products Distributed to Any Store
# https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/

class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
total = sum(quantities)

def good(x):
return sum(ceil(quantity / x) for quantity in quantities) <= n

left, right = 1, max(quantities)

while left < right:
mid = left + (right - left) // 2
x = good(mid)

if x:
right = mid
else:
left = mid + 1

return left


• class Solution {
public int minimizedMaximum(int n, int[] quantities) {
int left = 1, right = (int) 1e5;
while (left < right) {
int mid = (left + right) >> 1;
int cnt = 0;
for (int v : quantities) {
cnt += (v + mid - 1) / mid;
}
if (cnt <= n) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}

• func minimizedMaximum(n int, quantities []int) int {
return 1 + sort.Search(1e5, func(x int) bool {
x++
cnt := 0
for _, v := range quantities {
cnt += (v + x - 1) / x
}
return cnt <= n
})
}

• function minimizedMaximum(n: number, quantities: number[]): number {
let left = 1;
let right = 1e5;
while (left < right) {
const mid = (left + right) >> 1;
let cnt = 0;
for (const v of quantities) {
cnt += Math.ceil(v / mid);
}
if (cnt <= n) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}



Or use L < R template

• // OJ: https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/
// Time: O(QlogM) where Q is the length of quantities and M is the max element in quantities.
// Space: O(1
class Solution {
public:
int minimizedMaximum(int n, vector<int>& Q) {
long long L = 1, R = *max_element(begin(Q), end(Q));
auto valid = [&](int M) {
int ans = 0;
for (int n : Q) {
ans += (n + M - 1) / M;
}
return ans <= n;
};
while (L < R) {
long long M = (L + R) / 2;
if (valid(M)) R = M;
else L = M + 1;
}
return L;
}
};

• class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
left, right = 1, int(1e5)
while left < right:
mid = (left + right) >> 1
s = sum([(q + mid - 1) // mid for q in quantities])
if s <= n:
right = mid
else:
left = mid + 1
return left

############

# 2064. Minimized Maximum of Products Distributed to Any Store
# https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/

class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
total = sum(quantities)

def good(x):
return sum(ceil(quantity / x) for quantity in quantities) <= n

left, right = 1, max(quantities)

while left < right:
mid = left + (right - left) // 2
x = good(mid)

if x:
right = mid
else:
left = mid + 1

return left


• class Solution {
public int minimizedMaximum(int n, int[] quantities) {
int left = 1, right = (int) 1e5;
while (left < right) {
int mid = (left + right) >> 1;
int cnt = 0;
for (int v : quantities) {
cnt += (v + mid - 1) / mid;
}
if (cnt <= n) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
}

• func minimizedMaximum(n int, quantities []int) int {
return 1 + sort.Search(1e5, func(x int) bool {
x++
cnt := 0
for _, v := range quantities {
cnt += (v + x - 1) / x
}
return cnt <= n
})
}

• function minimizedMaximum(n: number, quantities: number[]): number {
let left = 1;
let right = 1e5;
while (left < right) {
const mid = (left + right) >> 1;
let cnt = 0;
for (const v of quantities) {
cnt += Math.ceil(v / mid);
}
if (cnt <= n) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}



## Discuss

https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/discuss/1563749/C%2B%2B-Binary-Search