Leetcode Easy: Longest Common Prefix

Tea-time snacking!

PROBLEM: https://leetcode.com/problems/longest-common-prefix/

TIME-COMPLEXITY: O(n*m) where n is the number of strings in input and m is the length of longest common prefix.

Sample Java Solution for tea-time snacking:

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs.length == 1)
            return strs[0];
        
        int minLen = Integer.MAX_VALUE;
        for (int i=0; i<strs.length; i++) {
            minLen = Math.min(minLen, strs[i].length());
        }
        int lastPastIdx = minLen;
        outer:
        for (int i=0; i<minLen; i++) {
            char ch = strs[0].charAt(i);
            for (int str=0; str<strs.length; str++) {
                if (strs[str].charAt(i) != ch) {
                    lastPastIdx = i;
                    break outer;
                }
            }
        }
        return strs[0].substring(0, lastPastIdx);
    }
}

Enjoy and Happy Coding!!

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LeetCode Medium Trick Problem: Longest Consecutive Sequence

Came across this problem on Leetcode, which requires one to focus on optimization.

PROBLEM LINK: https://leetcode.com/problems/longest-consecutive-sequence

TIME COMPLEXITY: O(n) at worst case where n is the number of elements in the input array (precisely the constant factor for n is 2)

APPROACH: The most naive approach to solve this problem would have been to sort the input array using comparison sort in O(n lg n) worst case time complexity and then in single pass find the length of the consecutive sequences in the same, resulting in overall worst case time complexity on O(n lg n).

But, if we notice carefully, we can make use of a trick to solve the problem in O(n) worst case time complexity. The trick makes a space trade off of O(n) and makes use of a HashSet to store the values to quickly check if any number exists in the input array. Then, it checks for all values in input, if it is the start of a consecutive sequence by checking presence of 1 lesser value than that on the number line, in the number Set. If any start-of-consecutive-range is found in the array elements, the length of the range is found by consecutively checking all the numbers in the increasing order on the number line consecutively.

In each such iteration of the numbers in input array, the length of the max-range is updated and the largest value is returned at the end of all iterations.

Here's a sample code for the same:

class Solution {
    
    public int longestConsecutive(int[] nums) {
        Set numberSet = new HashSet ();
        
        // Add numbers to Set while updating length of the longest consecutive elements sequence
        int maxLLCES = 0, currLLCES = 0;
        for (int i=0; i<nums.length; i++) {
            numberSet.add (nums[i]);
        }
        
        for (int i=0; i<nums.length; i++) {
            if (!numberSet.contains(nums[i]-1)) {
                // nums[i] is not a start of a consecutive sequence
                currLLCES = 1;
                int checkNum = nums[i]+1;
                while(numberSet.contains(checkNum)) {
                    currLLCES++;
                    checkNum++;
                }
                maxLLCES = Math.max(maxLLCES, currLLCES);
                // System.out.println("Updated maxLLCES:" + maxLLCES);
            }
        }
        return maxLLCES;
    }
}

Do share your thoughts and feel free to talk about the alternatives/optimisations you feel can be done in the comment section!

Love! ❤️ 
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Recursion Series: Deleting the middle element from a stack

This post marks the start point of the much awaited recursion series on Let'sCode_ =) 

We start it with a GeeksForGeeks problem : https://www.geeksforgeeks.org/delete-middle-element-stack/

Position of Middle element of all elements from top of stack (1 based):

stack.size()/2 + 1

Now, to build an effective solution we can use:

Methodology: Recursion

Time Complexity: O(n) where n is the number of elements in the stack

Space Complexity: O(n) considering the accumulation of constant space in each level of the recursive stack. O(1) if we dis-consider the stack space accumulation.

Here is one such solution:

<-- --todeletepos="" base="" case="" deletemidrecur="" int="" nteger="" printstack="" private="" return="" stack.pop="" stack.push="" stack="" static="" tack="" top="" void="">/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

class GFG {
	public static void main (String[] args) {
		System.out.println("GfG!");
		Stack<Integer> stack = new Stack<>();
		
		stack.push(6);
		stack.push(5);
		stack.push(4);
		stack.push(3);
		stack.push(2);
		stack.push(1);
		deleteMid(stack);
		
	}
	
	private static void deleteMid(Stack<Integer> stack) {
	    if (stack == null)
	        return;
	    if (stack.isEmpty())
	        return;
	        
	    int mid = stack.size()/2+1;
	    deleteMidRecur(stack, mid);
	    printStack(stack);
	}
	
	private static void deleteMidRecur (Stack<Integer> stack, int toDeletePos) {
	    if (toDeletePos == 1) {
	        // delete this element <-- base case
	        stack.pop();
	        return;
	    }
	    
	    int top = stack.pop();
	    deleteMidRecur(stack, --toDeletePos);
	    stack.push(top);
	}
	
	private static void printStack (Stack<Integer> stack) {
	    while (!stack.isEmpty()) {
	        System.out.println(stack.pop());
	    }
	}
}

IMO, such are the most efficient solutions to this problem. So share your thoughts and let me know your point of views.

Happy Coding!

Love! ❤️ 
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Beware of Multiple internet versions

Some time ago, I submitted an email to malwarebytes with data depicting the presence of multiple websites for an original tricking people into targetted scams.

In light of yesterday's (12 October, 2021's) news where malwareBytes admits and warns users of the same, I'd like to point out that not only for sites like the MalwareBytes.com site where one can be tricked into a duplicate site with malformed hyperlinks, even the Search Giant, Google's site has many duplicate websites, including 1 which I found my own internet pointing to on nslookup.

Here are the relevant screenshots worth the surprise:

Below you can see that my www.google.com entry points to the IP address whose reverse DNS lookup is the site-name which I have copied and hit with my web browser as seen in the above screenshot.

For your reference my /etc/resolv.conf, which is used for local entries for DNS lookup remained auto-generated, and looked like this (notice trail):

All-in-all:

Also, with today's announcement of the Honourable Prime Minister of India, where he warns everyone of cyber misinterpretation, such facts come to the rescue of the citizens of the world and provoke them to be more cautious by making them wary and aware!

Any such websites should be immediately reported and taken down from the internet's DNSes and a regular blacklisting of these should be automated and checked every very frequently on all the World Wide Web's DNSes in my honest opinion. Honestly said, the internet is not safe any more and it's even more evident with such data!

In light of the same, and for the lack thereof, to create more awareness and an actionable log of referable observations, Let's Code_ will be launching a dedicated platform where people will be able to report cybersecurity incidents.

Keep aware and stay safe!

Best of Love and Luck,

Yours Truly,
Chandni Verma
Chief Editor,
Let's Code_

#Lets #Code

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LeetCode Medium: Course Schedule

 The next problem in the current coding spree was:

Problem Name: Course Schedule

Problem Description: https://leetcode.com/problems/course-schedule/

Problem Approach used: Detecting cycles on the directed graph pf dependencies using DFS to solve in O(V+E) where V is the number of courses in the input and E is the number of edges between them.

Time Complexity: O(lg n) worst-case time and O(1) auxiliary space complexity

Java Solution:

//Cycle detection

// package com.projects.cv.course_schedule;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {

    // private static Logger logger = Logger.getLogger("MyLogger");
    int nodes;


    public boolean canFinish(int numCourses, int[][] prerequisites) {

        if (prerequisites == null)
            return false;
        int pL = prerequisites.length;
        nodes = numCourses;

        // Build directed graph
        Map<Integer, List<Integer>> g = new HashMap<>();
        for (int i=0; i<pL; i++) {
            if (!g.containsKey(prerequisites[i][0]))
                g.put(prerequisites[i][0], new ArrayList<>());
            g.get(prerequisites[i][0]).add(prerequisites[i][1]);
        }

        System.out.println(g);

        //Create visited to prevent re-visiting
        int visited[] = new int[numCourses];

        for (int i=0; i<numCourses; i++) {
            if (visited[i] == 0) {
                if (hasCycleDfs(g, visited, i, -1))
                    return false;
            }
        }

        return true;

    }

    private boolean hasCycleDfs( Map<Integer, List<Integer>> g, int[] visited, int n, int parent) {
        if (visited[n] == -1) {
            //current exploration path
            System.out.println("cycle found at u(" + parent + ")->v(" + n + ")");
            return true;
        }
        if (visited[n] == 1) {
            return false;
        }

        visited[n] = -1;

        if (g.get(n) == null) { // tackle bad callers
            visited[n] = 1;
            return false;
        }

        for (int neighBr : g.get(n)) {
            if (hasCycleDfs(g, visited, neighBr, n))
                return true;
        }

        visited[n] = 1;

        return false;
    }

    // public static void main(String[] args) {
    //     Solution s = new Solution();
    //     int[][] deps = {{1, 0}};
    //     s.canFinish(2, deps);
    // }

}

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GeeksforGeeks Medium: Find the Number of Islands

Yesterday I solved a few hands-on coding problems using Java programming language.

I came across this easy problem(called Medium there) over GfG, to turn on the inertia:

Problem: Find the Number of Islands

Problem Description: https://practice.geeksforgeeks.org/problems/find-the-number-of-islands 

Problem Approach: DFS

Time Complexity: O(V) where v = number of cells in the input grid

One Java Solution:

// { Driver Code Starts
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(br.readLine().trim());
        while(T-->0)
        {
            String[] s = br.readLine().trim().split(" ");
            int n = Integer.parseInt(s[0]);
            int m = Integer.parseInt(s[1]);
            char[][] grid = new char[n][m];
            for(int i = 0; i < n; i++){
                String[] S = br.readLine().trim().split(" ");
                for(int j = 0; j < m; j++){
                    grid[i][j] = S[j].charAt(0);
                }
            }
            Solution obj = new Solution();
            int ans = obj.numIslands(grid);
            System.out.println(ans);
        }
    }
}// } Driver Code Ends



class Solution
{
    int r, c;
    byte[] xOffset = {1, 1, 1, 0, -1, -1, -1, 0};
    byte[] yOffset = {1, 0, -1, -1, -1, 0, 1, 1};
    
    //Function to find the number of islands.
    public int numIslands(char[][] grid)
    {
        // Code here
        r = grid.length;
        c = grid[0].length;
        
        boolean[][] visited = new boolean[r][c];
        int cnt = 0;
        for (int i=0; i<r; i++) {
            for(int j=0; j<c; j++) {
                if (grid[i][j]=='1' && !visited[i][j]) {
                    dfs(grid, visited, i, j);
                    cnt++;
                }
            }
        }
        
        return cnt;
    }
    
    private void dfs (char[][] grid, boolean[][] visited, int x, int y) {
        //input validation
        if (!valid (grid, x, y) || visited[x][y]==true) {
            return;
        }
        
        visited[x][y] = true;
        
        for (int i=0; i<8; i++) {
            int newX = x + xOffset[i];
            int newY = y + yOffset[i];
            
            if(valid(grid, newX, newY) && !visited[newX][newY]) {
                dfs(grid, visited, newX, newY);
            }
        }
    }
    
    boolean valid (char[][]grid, int x, int y) {
        if (x<0 || x>=r || y<0 || y>=c || grid[x][y] == '0'){
            return false;
        }
        return true;
    }
}

Do share your thoughts and feel free to talk about the alternatives/optimisations you feel can be done in the comment section!

Love! ❤️ 
#Lets #Code

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Restarting Shorter and More Intense Problem-Solving Sprints

Hello Fellas,

After a long break, I have re-started #Problem #Solving #Tutorials using DS/Algorithms over (this) Let's Code Blog, but this time, it's gonna be consisting more intense but shorter #Coding #Sprints over weekends and somewhat lesser live-videos!! :D

Join me and get the weekend puzzle-solver activated in you!

✌️

#LetsCode