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Last active Jan 19, 2021
ML
View LogisticRegression.py
 class LogisticRegression: def __init__(self, lr=0.001, n_iters= 1000): self.lr = lr self.n_iters = n_iters self.weights = None self.bias = None def fit(self, X, y): """ X: input vector with shape of [B, D]
Last active Jan 19, 2021
leetcode-directed graph
View 207. Course Schedule.py
 class Solution: """ Thought Process: The problem come down to find whether directed graphs is cyclic or not. DFS: step1: step2: do dfs on graph to .. The difference between find cycle in undirected graph is we require 3 states of node instead of 2 0: not visited yet
Last active Jan 19, 2021
leetcode-graph
View 305. Number of Islands II.py
 class UnionFindSet(object): def __init__(self): """ This is version, we don't know how many nodes we'll cope with in the begining. So, we use dict to store _parents and _rank instead of array. Also, we would know nb of connected components dynamically. """ self._parents = {} self._ranks = {} self.count = 0 # nb of connected components
Created Jan 18, 2021
leetcode-ml
View kmeans.py
 def eucliean_distance(x1, x2): return np.sqrt(np.sum((x1-x2)**2)) class Kmeans: def __init__(self, K=3, max_iters=100): """It's an iterative unsupervised algorithm. Basically, We'd like to cluster a dataset into k non-overlapping clusters such that total within-cluster variance summed over all k cluster is as small as possible. In the end, each sample in dataset will be assigned to the cluster with the nearest centroid. """
Created Jan 18, 2021
leetcode-graph
View 721. Accounts Merge.py
 class UnionFindSet: def __init__(self, n): #Initially, all elements are single element subsets. self._parents = [node for node in range(n)] # it's like height of tree but it's not always equal to height because path compression technique. self._ranks = [1 for i in range(n)] def find(self, u): """ The find method with path compression, return root of node u.
Last active Jan 18, 2021
leetcode-union-find
View 684. Redundant Connection.py
 class UnionFindSet: def __init__(self, n): #Initially, all elements are single element subsets. self._parents = [node for node in range(n)] # it's like height of tree but it's not always equal to height because path compression technique. self._ranks = [1 for i in range(n)] def find(self, u): """ The find method with path compression, return root of node u.
Created Jan 18, 2021
leetcode-graph
View 200. Number of Islands.py
 class UnionFindSet: def __init__(self, n): self._parents = [node for node in range(n)] self._ranks = [1 for _ in range(n)] def find(self, u): while u != self._parents[u]: self._parents[u] = self._parents[self._parents[u]] u = self._parents[u] return u
Created Jan 18, 2021
leetcode-array
View 277. Find the Celebrity.py
 # The knows API is already defined for you. # return a bool, whether a knows b # def knows(a: int, b: int) -> bool: class Solution: """ Problem Clarification: Definition of celibrity: everbody knows him but he does not know anybody Thought Process:
Last active Jan 17, 2021
leetcode-graph
View 323. Number of Connected Components in an Undirected Graph.py
 class Solution: """ PC: Can I assume that no duplicate edges will appear in edges """ def countComponents(self, n: int, edges: List[List[int]]) -> int: """ Given n nodes and edge list of undirected graph, to find number of connected components. Note:
Last active Jan 16, 2021
leetcode-union-find
View 323. Number of Connected Components in an Undirected Graph.py
 class UnionFindSet: def __init__(self, n): self._parents = [node for node in range(n)] self._ranks = [1 for _ in range(n)] def find(self, u): while u != self._parents[u]: self._parents[u] = self._parents[self._parents[u]] u = self._parents[u] return u
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