Ravindra Ranwala's Blog

 We return today to graph search. Last time we saw breadth-first search, today we're going to do depth-first search. It's a simple algorithm, but you can do lots of cool things with it. DFS is often used to explore the whole graph, and so we are going to see how to do that today. So high-level description is we're going to just recursively explore the graph, backtracking as necessary, kind of like how you solve a maze. Let’s now consider a real world application of the DFS algorithm. Number of Closed Islands - Problem Definition Given a 2D grid consists of $0s$ (land) and $1s$ (water).  An island is a maximal 4-directionally connected group of $0s$ and a closed island is an island totally (all left, top, right, bottom) surrounded by $1s$. Return the number of closed islands. Output: 5 Output: 4 Note that closed islands are colored in the tables above. Output: 2 Output: 1 Number of Closed Islands - Approach We use the table $d$ to keep track of the discovery state of each bl...

A word, phrase, or sequence that reads the same backwards as forwards is a palindrome. For example the sequence "Step on no pets" is a palindrome. Today, let's focus on converting a sequence into a palindrome by inserting characters at arbitrary positions in the original input sequence. Minimum Insertion Steps to Make a String Palindrome - Problem Statement Given an input sequence $s$, in one step you can insert any character at any position of the input sequence. Return the minimum number of steps to make $s$ palindrome. For example, given the input sequence "leetcode", inserting five characters the string becomes "leetcodocteel". This is the minimum number of insertions required to make the input sequence a palindrome. Minimum Insertion Steps to Make a String Palindrome - Definition Let $s$ be an input sequence, $i$ and $j$ denote the start and end positions of a subsequence. Let $p[i][j]$ be the minimum number of insertions required to make subseque...

Graph problems pervade computer science, and algorithms to working with them are fundamental to the field. Hundreds of interesting computational problems are couched in terms of graphs. In this part, we touch the Breadth First Search. Breadth-first search is one of the simplest algorithms for searching a graph and the archetype for many important graph algorithms. Given a graph $G = (V, E)$ and a distinguished source vertex $s$, breadth-first search systematically explores the edges of $G$ to discover every vertex that is reachable from $s$. It computes the distance (smallest number of edges) from $s$ to each reachable vertex. [1] Now let’s consider a real world application of the BFS algorithm.  Treasure Island - Problem Definition You have a map that marks the location of a treasure island. Some of the map area has jagged rocks and dangerous reefs. Other areas are safe to sail in. There are other explorers trying to find the treasure. So you must figure out a shortest route to th...

 Today, let’s focus on a very good application of the famous algorithm binary search. Most of you may be familiar with binary search since it is widely used to implement different algorithms. In fact, I have used it in a few articles before.  Minimum capacity to ship within given days - Problem statement A conveyor belt has packages that must be shipped from one port to another within $d$ days. The $i^{th}$ package on the conveyor belt has a weight of $w[i]$. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship. Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within $d$ days. Also note that we are not allowed to split the items apart and then ship. Minimum capacity to ship within given days - Definition We first need to define the window of possible capacities. Let $w$ be the weight array of item...

Today, let’s focus on implementing a regular expression engine. Matching a given input string against a regular expression pattern is a very common application in the real world. Most regular expression engines are far too complex, and are often based on Finite automata state machines. However, our naive implementation doesn’t use Finite automata state machines.  Regular Expression Matching - Problem Statement Given an input string s and a pattern p, implement regular expression matching with support for '.' and '*' where: '.' Matches any single character. '*' Matches zero or more of the preceding element. The matching should cover the entire input string (not partial). Regular Expression Matching - Definition For the input string $s$ and the pattern $p$, each with length $m$ and $n$ characters respectively, let $t[i][j]$ be the evaluation of the string $s_1 \ldots s_i$ against the pattern $p_1 \ldots p_j$ such that $0 \le i \le m$ and $0 \le j \le n$. H...

 Today, let’s take a look at another interesting exercise given in the CLRS book [1]. Problem Statement Exercise 15.4-6 Give an $O(n \cdot log_2 n)-$ time algorithm to find the longest monotonically increasing subsequence of a sequence of $n$ numbers. (Hint: Observe that the last element of a candidate subsequence of length $i$ is at least as large as the last element of a candidate subsequence of length $i - 1$. Maintain candidate subsequences by linking them through the input sequence.) Longest Monotonically Increasing Subsequence - Approach There may be many different approaches to solve this problem. However, we are going to describe one such approach that we came up with. Each approach has its own strengths and weaknesses. The book [1] suggests a different approach to what we are going to present here. Nonetheless, both the approaches have the same asymptotic time and space complexity. Our approach uses an auxiliary array $m$ to store intermediate results. Finally, this array ...

Introduction In the fractional knapsack problem, the setup is the same in the 0-1 knapsack problem, but the thief can take fractions of items, rather than having to make a binary (0, 1) choice for each item. You can think of an item in the 0-1 knapsack problem as being like a gold ingot and an item in the fractional knapsack problem as more like gold dust. [1] We can solve the fractional knapsack problem by a greedy strategy. To solve the fractional problem, we first compute the value per pound, $v_i/w_i$ for each item. Obeying a greedy strategy, the thief begins by taking as much as possible of the item with the greatest value per pound. If the supply of that item is exhausted, and he can still carry more, he takes as much as possible of the item with the next greatest value per pound, and so forth. [1] Now let’s draw our attention to an exercise in the CLRS book [1]. Exercise 16.2-6 Show how to solve the fractional knapsack problem in $O(n)$ time. [1] By sorting the items by value pe...