2064. Minimized Maximum of Products Distributed to Any Store

Description

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 have been 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

Solutions

Binary search.

• 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;
}
}

• class Solution {
public:
int minimizedMaximum(int n, vector<int>& quantities) {
int left = 1, right = 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;
}
};

• class Solution:
def minimizedMaximum(self, n: int, quantities: List[int]) -> int:
def check(x):
return sum((v + x - 1) // x for v in quantities) <= n

return 1 + bisect_left(range(1, 10**6), True, key=check)


• 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;
}