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Graph - Question List

Graph - Question List


Example

    Question 1: Clone Graph

Easy Level

Medium Level

    Question 1: Count Nodes With the Highest Score

    Question 2:  2192. All Ancestors of a Node in a Directed Acyclic Graph

Hard Level

    Question 1: 2246. Longest Path With Different Adjacent Characters

    Question 2: 2392. Build a Matrix With Conditions 

Interview Questions


More Questions


Upper Layer 

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Graph Question - Hard Level - Question 1

2246. Longest Path With Different Adjacent Characters You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1. The tree is represented by a 0-indexed array parent of size n, where parent[i] is the parent of node i. Since node 0 is the root, parent[0] == -1. You are also given a string s of length n, where s[i] is the character assigned to node i. Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same character assigned to them. Constraints: n == parent.length == s.length 1 <= n <= 10^5 0 <= parent[i] <= n - 1 for all i >= 1 parent[0] == -1 parent represents a valid tree. s consists of only lowercase English letters. Analysis For this question, we need to construct a graph, or more specifically, a n-nary tree. How to construct the graph/tree? What we only know is the parent array. So we need to go through the parent array and constr...

Bit Manipulation - Medium Level

 Leetcode 416 Partition Equal Subset Sum Given a non-empty array nums containing only positive integers, find if the array can be partitioned into two subsets such that the sum of elements in both subsets is equal. Constraints: 1 <= nums.length <= 200 1 <= nums[i] <= 100 Analysis: There are different ways to solve this problem, such as dp with a time complexity of O(N^2). Since N is small to this question, so it is Okay to pass OJ. Besides the small N, the value of each element is also very small. So this gives us some chance to use "space to trad off time". The data structure to be used is bitset. For bitset, each bit can be either 0 or 1. The index of that bit can be used as the corresponding sum. When the bit is 1, means there is a sum with the value of its index. When a new number comes, this number needs to be added to all the previous sums, to form new "previous" sums. Thus for each number, we need to go through all the previous sums, the time co...