서랍장

분산 시스템, 학습, 추론, ... data parallel, model parallel Herring: Rethinking the Parameter Server at Scale for the Cloud parameter server -> gpu step by step -> ... (in aws, parameter server) Instead of sharding variables across parameter servers, Herring shards fusion buffers that are usually a concatenation of a number of gradients across all parameter servers. At different time periods during the ba..

Summary HDFS is a scalable distributed file system for large, distributed data-intensive applications. HDFS uses commodity hardware to reduce infrastructure costs. HDFS provides APIs for usual file operations like create, delete, open, close, read, and write. Random writes are not possible; writes are always made at the end of the file in an append-only fashion. HDFS does not support multiple co..

Summary Kafka provides low-latency, high-throughput, fault-tolerant publish and subscribe pipelines and can process huge continuous streams of events. Kafka can function both as a message queue and a publisher-subscriber system. At a high level, Kafka works as a distributed commit log. Kafka server is also called a broker. A Kafka cluster can have one or more brokers. A Kafka topic is a logical ..

분산시스템은 디자인하기 어렵고, Network asynchrony, Partial Failures, Concurrency 이슈가 있다. Partial Failures는 분산 시스템의 일부 컴포넌트들이 실패하는 경우를 의미하는데, 컴포넌트들 간의 atomicity를 보장하기가 어려운 문제가 있다. Concurrency 이슈는 same piece of data에 동시에 다수의 연산의 실행이 일어날 때 발생할 수 있다.

class Solution: def maximizeSweetness(self, sweetness: List[int], k: int) -> int: # Initialize the left and right boundaries. # left = 1 and right = (total sweetness) / (number of people). number_of_people = k + 1 left = min(sweetness) right = sum(sweetness) // number_of_people while left < right: mid = (left + right + 1) // 2 cur_sweetness = 0 people_with_chocolate = 0 # Start assigning chunks ..

1060. Missing Element in Sorted Array n = [1, 2, 4], k = 3일 때를 생각해보면 좋다. class Solution: def missingElement(self, nums: List[int], k: int) -> int: missing = lambda idx: nums[idx] - nums[0] - idx n = len(nums) # for the case [1, 2, 4] # 4 + 3 - 1 = 6 if k > missing(n-1): return nums[-1] + k - missing(n-1) idx = 1 while missing(idx) < k: idx += 1 return nums[idx-1] + k - missing(idx-1) binary sera..

1662. Check If Two String Arrays are Equivalent 매우 간단한 문제다. class Solution: def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool: mergedWord1 = ''.join(word1) mergedWord2 = ''.join(word2) return mergedWord1 == mergedWord2

Rotate Array, 문제는 정말 간단하다. 그렇지만 시간복잡도를 O(1)으로 풀기 위해서는 약간의 아이디어가 필요한데, 배열의 item 값을 바꿀 때, item을 가르키는 prev을 지정해야하고, cnt로 전체 item을 탐색하고 있는지 추적해야한다. start는 n//k값이다. class Solution: def rotate(self, nums: List[int], k: int) -> None: """ Do not return anything, modify nums in-place instead. """ n = len(nums) k %= n # start defines start = cnt = 0 while cnt < n: # prev는 바뀌기 이전의 value를 지정 curr, prev = st..

1239. Maximum Length of a Concatenated String with Unique Characters backtracking이랑, 이중 for문 2가지로 해결할 수 있다. 1. 백트래킹 class Solution: def maxLength(self, arr: List[str]) -> int: self.maxLength = 0 # s is 현재까지의 문자열 # 탐색하는 문자열의 idx def backtrack(s, j): if len(s) != len(set(s)): return self.maxLength = max(len(s), self.maxLength) for idx in range(j, len(arr)): backtrack(s + arr[idx], idx) backtrack('..

요새 코테 공부를 다시 매일 한 문제 씩 풀어나가고 있다. 프로그래머스 이분 탐색 문제도 풀면서, while l bool: pass # could be [0, n], [1, n] etc. depends on problem left, right = min(searchSpace), max(searchSpace) while left < right: mid = left + (right - left) // 2 if condition(mid): right = mid else: left = mid..