990. Satisfiability of Equality Equations - notruilin/LeetCode GitHub Wiki

1. union-find algorithm in Python

        parent = [x for x in range(27)]
        rank = [0 for _ in range(27)]
        
        def find(x):
            if parent[x] == x:
                return x
            else:
                parent[x] = find(parent[x])
                return parent[x]
        
        def unite(x, y):
            x = find(x)
            y = find(y)
            if rank[x] < rank[y]:
                parent[x] = y
            else:
                parent[y] = x
                if rank[x] == rank[y]:
                    rank[x] += 1
        
        def same(x, y):
            if find(x) == find(y):
                return True
            return False

2. Initialize dictionary in Python

        dict = {char: char for char in string.lowercase}

{'a': 'a', 'c': 'c', 'b': 'b', 'e': 'e', 'd': 'd', 'g': 'g', 'f': 'f', 'i': 'i', 'h': 'h', 'k': 'k', 'j': 'j', 'm': 'm', 'l': 'l', 'o': 'o', 'n': 'n', 'q': 'q', 'p': 'p', 's': 's', 'r': 'r', 'u': 'u', 't': 't', 'w': 'w', 'v': 'v', 'y': 'y', 'x': 'x', 'z': 'z'}

        dict = {char: 0 for char in string.lowercase}

{'a': 0, 'c': 0, 'b': 0, 'e': 0, 'd': 0, 'g': 0, 'f': 0, 'i': 0, 'h': 0, 'k': 0, 'j': 0, 'm': 0, 'l': 0, 'o': 0, 'n': 0, 'q': 0, 'p': 0, 's': 0, 'r': 0, 'u': 0, 't': 0, 'w': 0, 'v': 0, 'y': 0, 'x': 0, 'z': 0}