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author | Jack Sangdahl <[email protected]> | 2024-11-13 22:07:00 -0700 |
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committer | Jack Sangdahl <[email protected]> | 2024-11-13 22:07:08 -0700 |
commit | ecc2fcf61ab52d10b402d8d87146c06c53c4c423 (patch) | |
tree | 7db88fd93e87bd70be7e4a1718581f0c818796f3 /mini_maxi_products_distrib.py | |
parent | 6ce80ea59406da5284927cffc5cf770fb1f09bcd (diff) | |
download | leetcode-ecc2fcf61ab52d10b402d8d87146c06c53c4c423.tar.gz leetcode-ecc2fcf61ab52d10b402d8d87146c06c53c4c423.zip |
2064. minimised maximum of products distributed to any store
Diffstat (limited to 'mini_maxi_products_distrib.py')
-rw-r--r-- | mini_maxi_products_distrib.py | 38 |
1 files changed, 38 insertions, 0 deletions
diff --git a/mini_maxi_products_distrib.py b/mini_maxi_products_distrib.py new file mode 100644 index 0000000..3baae0a --- /dev/null +++ b/mini_maxi_products_distrib.py @@ -0,0 +1,38 @@ +# 2064. Minimized Maximum of Products Distributed to Any Store +from math import ceil + +""" +you are given an integer 'n' indicating that there are 'n' specialty retail +stores. there are 'm' product types of varying amounts, which are given as a +0-indexed integer array 'quantities' owhere 'quantities[i]' represents the +number of products of the i'th product type. you need to distribute all the +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, and after +distribution, each store will have been given some number products. let 'x' +represent the maximum number of products given to any store. you want 'x' to +be as small as possible. return the minimum possible 'x'. +""" + + +class Solution(object): + def minimizedMaximum(self, n, quantities): + """ + :type n: int + :type quantities: List[int] + :rtype: int + """ + right, left = 0, max(quantities) + while right + 1 < left: + mid = (right + left) // 2 + if sum(ceil(i / mid) for i in quantities) <= n: + left = mid + else: + right = mid + return left + + +if __name__ == "__main__": + obj = Solution() + print(obj.minimizedMaximum(n=6, quantities=[11, 6])) + print(obj.minimizedMaximum(n=7, quantities=[15, 10, 10])) + print(obj.minimizedMaximum(n=1, quantities=[100000])) |