1697. Checking Existence of Edge Length Limited Paths

# Time:  O(nlogn + mlogm + n * α(n)) = O(nlogn + mlogm)
# Space: O(n)

class UnionFind(object):  # Time: O(n * α(n)), Space: O(n)
    def __init__(self, n):
        self.set = range(n)
        self.rank = [0]*n

    def find_set(self, x):
        stk = []
        while self.set[x] != x:  # path compression
            stk.append(x)
            x = self.set[x]
        while stk:
            self.set[stk.pop()] = x
        return x

    def union_set(self, x, y):
        x_root, y_root = map(self.find_set, (x, y))
        if x_root == y_root:
            return False
        if self.rank[x_root] < self.rank[y_root]:  # union by rank
            self.set[x_root] = y_root
        elif self.rank[x_root] > self.rank[y_root]:
            self.set[y_root] = x_root
        else:
            self.set[y_root] = x_root
            self.rank[x_root] += 1
        return True


class Solution(object):
    def distanceLimitedPathsExist(self, n, edgeList, queries):
        """
        :type n: int
        :type edgeList: List[List[int]]
        :type queries: List[List[int]]
        :rtype: List[bool]
        """
        for i, q in enumerate(queries):
            q.append(i)
        edgeList.sort(key=lambda x: x[2])
        queries.sort(key=lambda x: x[2])
        
        union_find = UnionFind(n)
        result = [False]*len(queries)
        curr = 0
        for u, v, w, i in queries: 
            while curr < len(edgeList) and edgeList[curr][2] < w: 
                union_find.union_set(edgeList[curr][0], edgeList[curr][1])
                curr += 1
            result[i] = union_find.find_set(u) == union_find.find_set(v)
        return result 

Solution from kamyu104/LeetCode-Solutions · MIT