3532. Path Existence Queries in a Graph I

Problem Link

Description

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You are given an integer n representing the number of nodes in a graph, labeled from 0 to n - 1.

You are also given an integer array nums of length n sorted in non-decreasing order, and an integer maxDiff.

An undirected edge exists between nodes i and j if the absolute difference between nums[i] and nums[j] is at most maxDiff (i.e., |nums[i] - nums[j]| <= maxDiff).

You are also given a 2D integer array queries. For each queries[i] = [ui, vi], determine whether there exists a path between nodes ui and vi.

Return a boolean array answer, where answer[i] is true if there exists a path between ui and vi in the ith query and false otherwise.

 

Example 1:

Input: n = 2, nums = [1,3], maxDiff = 1, queries = [[0,0],[0,1]]

Output: [true,false]

Explanation:

• Query [0,0]: Node 0 has a trivial path to itself.
• Query [0,1]: There is no edge between Node 0 and Node 1 because |nums[0] - nums[1]| = |1 - 3| = 2, which is greater than maxDiff.
• Thus, the final answer after processing all the queries is [true, false].

Example 2:

Input: n = 4, nums = [2,5,6,8], maxDiff = 2, queries = [[0,1],[0,2],[1,3],[2,3]]

Output: [false,false,true,true]

Explanation:

The resulting graph is:

• Query [0,1]: There is no edge between Node 0 and Node 1 because |nums[0] - nums[1]| = |2 - 5| = 3, which is greater than maxDiff.
• Query [0,2]: There is no edge between Node 0 and Node 2 because |nums[0] - nums[2]| = |2 - 6| = 4, which is greater than maxDiff.
• Query [1,3]: There is a path between Node 1 and Node 3 through Node 2 since |nums[1] - nums[2]| = |5 - 6| = 1 and |nums[2] - nums[3]| = |6 - 8| = 2, both of which are within maxDiff.
• Query [2,3]: There is an edge between Node 2 and Node 3 because |nums[2] - nums[3]| = |6 - 8| = 2, which is equal to maxDiff.
• Thus, the final answer after processing all the queries is [false, false, true, true].

 

Constraints:

• 1 <= n == nums.length <= 105
• 0 <= nums[i] <= 105
• nums is sorted in non-decreasing order.
• 0 <= maxDiff <= 105
• 1 <= queries.length <= 105
• queries[i] == [ui, vi]
• 0 <= ui, vi < n

Solution

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Python3

class DSU:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.rank = [0 for _ in range(n)]
 
    def find(self, x):
        if self.parent[x] == x:
            return self.parent[x]
        self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
 
    def union(self, u, v):
        pu = self.find(u)
        pv = self.find(v)
 
        if (pu == pv): return
 
        if self.rank[pu] < self.rank[pv]:
            pu, pv = pv, pu
 
        # ensure self.rank[pu] >= self.rank[pv]
        self.parent[pv] = pu
        if self.rank[pu] == self.rank[pv]:
            self.rank[pu] += 1
 
class Solution:
    def pathExistenceQueries(self, n: int, nums: List[int], maxDiff: int, queries: List[List[int]]) -> List[bool]:
        graph = defaultdict(set)
        i = 0
        uf = DSU(n)
 
        for j, x in enumerate(nums):
            while abs(nums[i] - nums[j]) > maxDiff:
                i += 1
 
            if abs(nums[i] - nums[j]) <= maxDiff:
                uf.union(i, j)
 
        res = []
        for x, y in queries:
            px, py = uf.find(x), uf.find(y)
            res.append(px == py)
 
        return res