Skip to main content

Greedy Algorithm

Greedy Algorithm is one of the intuitive algorithms that appear not complicated to understand. As indicated by its name, greedy algorithm picks up the extreme values for each step since the beginning. Thus, greedy algorithm is only valid when the optimal of every step, or the optimal on the fly, is guarantee to be true, which is usually not.

The interesting part is that, when the greedy algorithm is valid to a question, it is usually not difficulty to "find it". However, it is not rare that the proving of the valid is not straightforward. One of the ways to prove the correctness is proof by contradiction.

To make a brief summary:

1. greedy algorithm requires the correctness to pick up the extreme value for each step;

2. the proof of the validity of greedy algorithm usually is not easy.


Question list


Upper Layer

Comments

Post a Comment

Popular posts from this blog

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...
Post a Comment
Read more

Priority_queue - Question 4

973 K Closest Points to Origin Given an array of points where points[i] = [xi, yi] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0). The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x1 - x2)2 + (y1 - y2)2). You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in). Constraints: 1 <= k <= points.length <= 10^4 -10^4 < xi, yi < 10^4 Analysis: It is quite straightforward to find the K points that have the closest distances to the original point. Since we need to find the smallest values, the root of the priority_queue (pq) is the largest one for all the elements inside the pq. Once the size of pq is larger than K, we need to pop out one. See the code below: class Solution { public: typedef pair<int, int> pp; vector<vector<int>> kClosest(vector<vector<int>>& points, int k) { vector...
Post a Comment
Read more