Check Duplicate Parenthesis in String

Given a balanced expression, find if it contains duplicate parenthesis or not. A set of parenthesis are duplicate if the same subexpression is surrounded by multiple parenthesis.


 import java.util.*;

class GFG {
    static String check(String s)
    {
        Stack<Character>st=new Stack<>();
        for(int i=0;i<s.length();i++)
        {
            if(s.charAt(i)==')')
            {
                int ele=0;
                while(!st.isEmpty()&&st.peek()!='(')
                {
                    st.pop();
                    ele++;
                }
                if(ele<1)
                  return "Duplicate Parenthesis";
                else
                {
                    if(!st.isEmpty())
                     st.pop();
                }
            }
            else
            st.push(s.charAt(i));
        }
        return "Not Duplicate";
    }
    public static void main (String[] args) {
        Scanner sc=new Scanner(System.in);
        String s=sc.next();
        System.out.println(check(s));
    }
}

Maximum depth of left node in Binary Tree

Given a Binary tree, print the maximum depth of a left node.( the node needs to be a left child ) (if the node is right child of the left child of the root node then it wont count as a left node)

import java.io.*;
class Node
{
    int data;
    Node left,right;
    Node(int data)
    {
        this.data=data;
        left=right=null;
    }
}
class TreeQuestion {
    Node root;
    static int max=0;
    void maxDepth(Node root,int direction,int depth)
    {
        if(root==null)
          return;
        if(direction==1)
        {
            max=Math.max(max,depth);
        }
        maxDepth(root.left,1,depth+1);
        maxDepth(root.right,2,depth+1);
    }
    public static void main (String[] args) {
        TreeQuestion tree=new TreeQuestion();
        tree.root=new Node(1);
        tree.root.left=new Node(2);
        tree.root.left.left=new Node(3);
        tree.root.left.left.left=new Node(4);
        tree.maxDepth(tree.root,1,0);
        System.out.println(max+1);
       
    }
}

Balanced Binary Tree

Given a binary tree, determine whether or not it is height-balanced. A height-balanced binary tree can be defined as one in which the heights of the two subtrees of any node never differ by more than one.

Algorithm:-

 import java.util.*;
class Node
{
    int data;
    Node left,right;
    Node(int data)
    {
        this.data=data;
        left=right=null;
    }
}
class CheckBT {
    Node root;
    boolean isBalanced(Node root)
    {
        if(root==null)
          return true;
        int l,r;
        l=height(root.left);
        r=height(root.right);
        if(Math.abs(l-r)<=1&&isBalanced(root.left)&&isBalanced(root.right))
             return true;      
      return false;       
    }
    int height(Node node)
    {
        if(node==null)
         return 0;
        return 1+Math.max(height(node.left),height(node.right));
    }
    public static void main (String[] args) {
        CheckBT tree=new CheckBT();
        tree.root=new Node(1);
        tree.root.left=new Node(2);
        tree.root.left.left=new Node(3);
        if(tree.isBalanced(tree.root))
           System.out.println("Balanced");
         else
           System.out.println("Not Balanced");
    }
}

Absolute Path Coding Problem with Solution

Given an absolute pathname that may have . or .. as part of it, return the shortest standardized path.

For example, given "/usr/bin/../bin/./scripts/../", return "/usr/bin/".

import java.util.*;
class GFG {
    static String getAbsolutePath(String s)
    {
        String path[]=s.split("/");
        ArrayList<String>al=new ArrayList<>();
        for(int i=0;i<path.length;i++)
        {
            if(path[i].equals("."))
              continue;
            else if(path[i].equals(".."))
              {
                  if(al.size()>0)
                   al.remove(al.size()-1);
              }
              else
              al.add("/"+path[i]);
        }
        if(al.size()>0)
          al.add("/");
        s="";
       for(int i=1;i<al.size();i++)
         s+=al.get(i);
       return s;        
    }
    public static void main (String[] args) {
            Scanner sc=new Scanner(System.in);
            String s=sc.next();
            System.out.println(getAbsolutePath(s));
    }
}

Finding sum of digits of a number until sum becomes single digit

Given a number n we need to find the sum of it's digit until it becomes single digit and expected time complexity O(1).

Input:-  12345

Output-   6


Solution 1:- Brute Force

import java.util.*;

class GFG {
    static int getSum(int n)
    {
        int sum=0;
        while(n>0||sum>9)
        {
            if(n==0)
            {
                n=sum;
                sum=0;
            }
           
                sum+=n%10;
                n=n/10;
           
        }
        return sum;
    }
    public static void main (String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        System.out.println(getSum(n));
    }
}


Solution 2:- Best Way


import java.util.*;
class GFG {
    static int getSum(int n)
    {
        if(n==0)
          return n;
        return (n%9==0?9:n%9); 
    }
    public static void main (String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        System.out.println(getSum(n));
    }
}

Time Complexity- O(1)

Samsung Noida R&D Coding Round Question minimum distance between source and destination

There is dedicated Samsung software for coding test the question is given below:
There is one spaceship. X and Y co-odinate of source of spaceship and destination spaceship is given. There are N number of warmholes each warmhole has 5 values.
First 2 values are starting co-ordinate of warmhole and after that value no. 3 and 4 represents ending co-ordinate of warmhole and last 5th value is represents cost to pass through this warmhole. Now these warmholes are bi-direction.
Now the to go from (x1,y1) to (x2,y2) is abs(x1-x2)+abs(y1-y2).
The main problem here is to find minimum distance to reach spaceship from source to destination co-ordinate using any number of warm-hole. It is ok if you wont use any warmhole.


Solution: 

import java.util.Scanner;

class Samsung{
    static int ans=Integer.MAX_VALUE;
    static int distance(int sx,int sy,int dx,int dy)
    {
        return Math.abs(sx-dx)+Math.abs(sy-dy);
    }
    static void calCulateUtil(int mat[][],int n,int sx,int sy,int dx,int dy,int dis,boolean visited[])
    {
        ans=Math.min(ans,distance(sx,sy,dx,dy)+dis);
        for(int i=0;i<n;i++)
        {
            if(visited[i]==false)
            {
                visited[i]=true;
                int temp=distance(sx,sy,mat[i][0],mat[i][1])+dis+mat[i][4];
                calCulateUtil(mat,n,mat[i][2],mat[i][3],dx,dy,temp,visited);
                temp=distance(sx,sy,mat[i][2],mat[i][3])+dis+mat[i][4];
                 calCulateUtil(mat,n,mat[i][0],mat[i][1],dx,dy,temp,visited);
                 visited[i]=false;
            }
        }
    }
    static int calCulate(int mat[][],int n,int sx,int sy,int dx,int dy,boolean visited[])
    {
        calCulateUtil(mat,n,sx,sy,dx,dy,0,visited);
        return ans;
    }
    public static void main (String[] args) {
          Scanner sc=new Scanner(System.in);
          int t=sc.nextInt();
          while(t-->0)
          {
              int n=sc.nextInt();
              int sx=sc.nextInt();
              int sy=sc.nextInt();
              int dx=sc.nextInt();
              int dy=sc.nextInt();
              int mat[][]=new int[n][5];
              boolean visited[]=new boolean[n];
              for(int i=0;i<n;i++)
              {
                  for(int j=0;j<5;j++)
                  {
                      mat[i][j]=sc.nextInt();
                  }
              }
              System.out.println(calCulate(mat,n,sx,sy,dx,dy,visited));
          }
    }
}

Shortest Path between source to destination in matrix

Given a Boolean 2D matrix (0-based index), find whether there is a path from (0,0) to (x,y) and if there is one path, print the minimum no of steps needed to reach it, else print -1 if the destination is not reachable. Moves are possible in only four directions i.e. up, down, left and right. The path can only be created out of a cell if its value is 1.



Solution:-

BFS traversal of matrix will give us minimum distance from source to destination in matrix.

Idea behind Solution:-  Step1:-  make a boolean matrix of same dimention and assign true every 1 and false at every 0 on the basis of given matrix.

step2:- start from source and do BFS traversal of matrix and every step increase the distance by 1.

step3:- when we reach destination then return the distance.



import java.util.Scanner;
class Point
{
    int x,y,d;
    Point(int x,int y,int d)
    {
        this.x=x;
        this.y=y;
        this.d=d;
    }
}
class Queue
{
    Point p[]=new Point[10000];
    int front=-1,end=-1;
    void add(int x,int y,int d)
    {
        if(front==-1)
            front=0;
        if(end<9999)
        {
            p[++end]=new Point(x,y,d);
        }
    }
    Point poll()
    {
        Point x=p[front++];
        return x;
    }
    boolean isEmpty()
    {
        if(front>end)
          return true;
        if(front==-1||end==-1)
          return true;
      return false;     
    }

}
class GFG {
    static int getMinValue(int mat[][],int n,int m,int sx,int sy,int dx,int dy)
    {
        boolean visited[][]=new boolean[n][m];
          for(int i=0;i<n;i++)
          {
              for(int j=0;j<m;j++)
              {
                  if(mat[i][j]==0)
                  visited[i][j]=true;
                  else
                  visited[i][j]=false;
              }
          }
          
          visited[sx][sy]=true;
          visited[dx][dy]=false;
          //Point p=new Point(sx,sy,0);
          Queue q=new Queue();
          q.add(sx,sy,0);
          while(!q.isEmpty())
          {
              Point p=q.poll();
              if(p.x==dx&&p.y==dy)
                return p.d;
            //left
              if(p.y-1>=0&&visited[p.x][p.y-1]==false)
              {
                  visited[p.x][p.y-1]=true;
                  q.add(p.x,p.y-1,p.d+1);
              }
              //right
              if(p.y+1<m&&visited[p.x][p.y+1]==false)
              {
                  visited[p.x][p.y+1]=true;
                  q.add(p.x,p.y+1,p.d+1);
              }
              //up
              if(p.x-1>=0&&visited[p.x-1][p.y]==false)
              {
                  visited[p.x-1][p.y]=true;
                  q.add(p.x-1,p.y,p.d+1);
              }
              //down
              if(p.x+1<n&&visited[p.x+1][p.y]==false)
              {
                  visited[p.x+1][p.y]=true;
                  q.add(p.x+1,p.y,p.d+1);
              }
             
          }
          return -1;
         
    }
    public static void main (String[] args) {
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int m=sc.nextInt();
        int mat[][]=new int[n][m];
          for(int i=0;i<n;i++)
          {
              for(int j=0;j<m;j++)
              {
                  mat[i][j]=sc.nextInt();
              }
          }
        int sx=sc.nextInt();
        int sy=sc.nextInt();
        int dx=sc.nextInt();
        int dy=sc.nextInt();
        System.out.println(getMinValue(mat,n,m,sx,sy,dx,dy));
    }
}



 https://ide.geeksforgeeks.org/pqvGANXWLT

Find whether there is path between two cells in matrix

Solution:-

import java.util.*;
class Point
{
    int x,y;
    Point(int x,int y)
    {
        this.x=x;
        this.y=y;
    }
}
class Queue
{
    Point p[]=new Point[10000];
    int front=-1,end=-1;
    void add(int x,int y)
    {
        if(front==-1)
            front=0;
        if(end<9999)
        {
            p[++end]=new Point(x,y);
        }
    }
    Point poll()
    {
        Point x=p[front++];
        return x;
    }
    boolean isEmpty()
    {
        if(front>end)
          return true;
        if(front==-1||end==-1)
          return true;
      return false;     
    }
}

class GFG {
    static String check(int mat[][],int n,int m)
    {
        boolean visited[][]=new boolean[n][m];
        Point p=new Point(0,0);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(mat[i][j]==1)
                {
                    p.x=i;
                    p.y=j;
                }
                else if(mat[i][j]==0)
                 visited[i][j]=true;
                 else
                 visited[i][j]=false;
            }
        }
        Queue q=new Queue();
        q.add(p.x,p.y);
        visited[p.x][p.y]=true;
        while(!q.isEmpty())
        {
            p=q.poll();
            if(mat[p.x][p.y]==2)
             return "YES";
             //left
            if(p.y-1>=0&&visited[p.x][p.y-1]==false)
            {
                q.add(p.x,p.y-1);
                visited[p.x][p.y-1]=true;
            }
            //right
             if(p.y+1<m&&visited[p.x][p.y+1]==false)
            {
                q.add(p.x,p.y+1);
                visited[p.x][p.y+1]=true;
            }
            //bottom
             if(p.x+1<n&&visited[p.x+1][p.y]==false)
            {
                q.add(p.x+1,p.y);
                visited[p.x+1][p.y]=true;
            }
            //top
             if(p.x-1>=0&&visited[p.x-1][p.y]==false)
            {
                q.add(p.x-1,p.y);
                visited[p.x-1][p.y]=true;
            }
        }
        return "NO";
    }
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int m=sc.nextInt();
int mat[][]=new int[n][m];
for(int i=0;i<n;i++)
{
    for(int j=0;j<m;j++)
    {
        mat[i][j]=sc.nextInt();
    }
}
System.out.println(check(mat,n,m));
}
}

Mobikwik Interview Experience |( OffCampus Drive Quality Assurance SDET-I)

MobiKwik is an Indian company founded in 2009 that provides a mobile phone based payment system and digital wallet. Customers add money to an online wallet that can be used for payments.

A couple of days back(22nd July,2019) MobiKwik hire as Quality Assurance Engineer SDET-I for passout 2019 batch.
Selection process go thru various round let's discuss about that.

Round 1:- Offline Wriiten Coding and MCQ and Test Cases   

This round was 40 minuts and contains 15 MCQ which was technical,aptitude ,Data Structure and 2 Coding Questions,and 1 Test Case writting.

Coding 1:-  Given a string find all permutaions of given string in lexicographical order.

Coding 2:-  Given an array remove duplicate elements from the array.

ex:-  a[]={1,2 3,1,2,3,1,2}

Output:-   {1,2,3,0,0,0,0,0}

Round 2:-  Technical Interview-I

After written test almost 34 students got selected for next round and I was one of them.
In this round Interviewer came and ask me tell me something about yourself. I was very nervous but interviewer was very cool and after introduction he asked me 3 coding problem.
 1. Given an array sort array using recursion.
 2. Given string print all the string that accure more than one times in the given string.
     ex:  my name is sushil mall my name is.

     Output:-   my name is
 3. Find Second highest element in the array.

after asking coding he asked me some puzzles, databases queries and testing related questions.

Round 3:- Technical Interview-II

I was very happy to clear the first round and in the second round interviewer came and asked me tell me something about yourself. she was very cool and very friendy.
she asked me a coding question and write code about the given question. queastion was given a array find second minimum element in the gievn array.after that she asked me puzzles and queries.

Round 3:- Testing interviews-III

Again i was very happy because I have clear the second technical round and on left testing and was last round.Interviewer came and asked me why should we hire you convince me.I explain but he was not convinced. I was very nervous because he was not friendy and his attitude was very dangrous.then after discussion at leas 10 minut on why should we hire you. he said me to write test case of wallet transaction. again I was fear and nervous because he was not friendly and everyone knows that 'fear is not good for fair' I write some test cases he said think something and write some another test cases. and then he asked me to write Test cases of shoes I write but he gave me time limit and said to write 50 test cases. Finaly result announce and I was not selected only 4 student got selected.But when I was going to my house HR call me and said that come tomarrow with some preperation about testing.

 Round 4:- Technical Interview-IV

Next day I went again Mobikwik office for last interview. Then inetrviewer came and asked my why you did not select previuos day. I explain all about previous day then he gave me a coding problem to solve and write the code. problem was given a 2-D n*n matrix rotate it 90 degree. I wrote code firstly and explain him. He said write test case of every steps of given problem I wrote he was very happy. he asked me to write test case of lift I wrote and tell him but he want to know more test cases. then he asked me how much companies you have already give Interviews then last hes asked me a puzlle and said if you will solve this puzzle then you will got selected today otherwise some difficulties will got. and you have infinite time and you can try many time. I solve that with in 10 min with 5th tery he becomy happy and handshake with me congratulate me. then HR came and asked about my family and said congratulation and we are going to hire you said your family.
I was very happy to get a offer from Mobikwik. Thanks a lot Mobikwik and their Team.

*******************************Enjoy ********************************************

Best Data Science and Big Data Analytic LinkedIn Group You should Join them

We are familiar with linkedin which is social networking site where many people are connected to each other and share their experience and opportunity to other. These all people are professional, experienced,expert in their era,Employee of the best companies.

 

1. Big Data and Analytics

Members: 354,502

Founded in 2012 by Rob Howes, Linkedin Group: Big Data and Analytics has grown to be the largest and most active group of it's kind. With over 300,000 members, it has become the go-to community for all things data. You will find a wide range of topics including AI, Machine learning, Blockchain, etc. This group is moderated and managed by Sarah Howes. Contact :Sarah@howesmedia.com

2. Big Data, Analytics, Business Intelligence & Visualization Experts Community

Members: 252,466

A premier community for both existing expert professionals and companies researching the convergence of big data analytics and discovery, Hadoop, data warehousing, cloud, unified data architectures, digital marketing, visualization and business intelligence.

3. Business Intelligence Professionals (BI, Big Data, Analytics)

Members: 231,388

This group covers the broad area of data analytics & intelligence for businesses. Tools, frameworks, white papers, industry news & best practices! PLEASE READ THE GROUP RULES before participating! Visit Aigents.co to discover career opportunities for Data-, AI- & Machine Learning Engineers.

4. Data Mining, Statistics, Big Data, Data Visualization, and Data Science

Members: 181,538

This is a group for data mining and statistical professionals who wish to expand our network of people and share ideas. Methodological issues are fair game, as well as discussion of software (SAS, R, WEKA, etc), technology (Hadoop, relational databases, etc) conferences, and job postings. Please note that job postings should only be submitted in the Jobs space, while advertising of services should only be submitted.

5. Big Data, Analytics, IoT (Internet of Things) & Blockchain

Members: 75,012

The group promotes discussion around Big Data, NoSql and Cloud Computing. Like Big Data Analytics, BI, SAS, Big Data Hadoop, Big Data MapReduce, YARN, PIG, Hive, HQL, Oozie, Sqoop, Flume, Kafka, Spark, Big Data NoSQL, HBase, Casandra, MongoDB,

6. IBM Big Data and Analytics

Members: 47,702

Everyday, we create 2.5 Quintilion bytes of data–so much that 90% of the data in the world today has been created in the last two years alone. This data comes from everywhere: from sensors used to gather climate information, posts to social media sites, digital pictures and videos posted online, transaction records of online purchases, and from cell phone GPS signals to name a few. This data is big data.

 7. Advanced Analytics and Data Science

Members: 12,054

Our mission is to provide a resource for people who want to learn about and use advanced analytics and data science capabilities. Meet other people involved with predictive analytics, statistics, machine learning, decision optimization, real-time analytics and big data to have discussions and make connections. Whether you are a data professional who wants to extract meaning from and interpret data, or an expert analyst responsible for making decisions in areas such as marketing, operations, finance, HR or sales, this is the place for you.

Which Programming Language is best for Competitive Coding

As we know competitive programming is best for learn a lot of thing related to code and every time  we learn a new problem in the Competitive Programming. A lot of languages available with us like C,C++,java,Python,C#,Perl,PHP,Scala etc but we need to choose best of these which would be suitable for us to write the code easily and that take less time to execute.So let's learn which one is best for the Competitive Programminng.

1:-The best programming language for competitive Coding is C++ because it is object oriented and take less time to execute it's code and extra STL(Standard Template Library) available in C++ which help to reduce the line of code and make language easy to code.The Standard Template Library (STL) is a set of C++ template classes to provide common programming data structures and functions such as lists, stacks, arrays, etc. It is a library of container classes, algorithms, and iterators. It is a generalized library and so, its components are parameterized. A working knowledge of template classes is a prerequisite for working with STL. 

          "So C++ is best for Competitive Coding every Coding lover should learn it".

2:- Java is also best for competitive programming because it is Object Oriented and provide Collections to write the code.But it is lesser than C++. But it is better to learn and easily to write the code.

         "So Java is best for Competitive Coding every Coding lover should learn it".  

Algorithm is also put's matter on writing your code. Both java and C++ are better for competitive programming. Everyone whole write the code in C or C# please switch either C++ or Java because it is best for competitive coding.
Python is also better but Python is more efficient for development,Machine Learning and Artificial Intelligence.
So it's your choice which language you are suitable to learn but again C++ and Java is best for Competitive Coding.    

My Pathshala Interview Experiences

My Pathshala is a startup that provide a platform for education and preparation of various exam like SSC, Bank, UPSC etc.
Today My Pathshala visit in our Campus for recruit as php web developer They conduct following processes.

1)Written Test:- In this Round there are four section Aptitude,Reasoning,English and Technical
They give more preference to technical portion I cleared easly this round.

2)HR Round:- Second Round was HR Round in this round they asked simply question that are based on resume and some general question based on Behavior I have cleared this round.

3)Technical interview :- This round was last round and based on technical questions Firstly Interviewer asked how you solve technical portion of the written test I explain.
 Then he asked me which programming language which have more strength I said Java.Then he asked me questions on Java.Questions was...

a)What is anonymous class in Java?

b)What is Map?

c)A coding question Question was Given List of name and id of Employee how you will sort it using Id.

I did not give the answer of first question rest question I gave the answer but I was rejected.
Finally my three friends are selected.(30 March,2019).


 

Google Kick-Start 2019 Practice Round Problem 2 with solution

Problem 2 Mural(15pts,23pts)

Thanh wants to paint a wonderful mural on a wall that is N sections long. Each section of the wall has a beauty score, which indicates how beautiful it will look if it is painted. Unfortunately, the wall is starting to crumble due to a recent flood, so he will need to work fast!

At the beginning of each day, Thanh will paint one of the sections of the wall. On the first day, he is free to paint any section he likes. On each subsequent day, he must paint a new section that is next to a section he has already painted, since he does not want to split up the mural.

At the end of each day, one section of the wall will be destroyed. It is always a section of wall that is adjacent to only one other section and is unpainted (Thanh is using a waterproof paint, so painted sections can't be destroyed).

The total beauty of Thanh's mural will be equal to the sum of the beauty scores of the sections he has painted. Thanh would like to guarantee that, no matter how the wall is destroyed, he can still achieve a total beauty of at least B. What's the maximum value of B for which he can make this guarantee?

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with a line containing an integer N. Then, another line follows containing a string of N digits from 0 to 9. The i-th digit represents the beauty score of the i-th section of the wall.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the maximum beauty score that Thanh can guarantee that he can achieve, as described above.

Limits

1 ≤ T ≤ 100.
Time limit: 20 seconds per test set.
Memory limit: 1 GB.

Small dataset (Test set 1 - Visible)

2 ≤ N ≤ 100.

Large dataset (Test set 2 - Hidden)

For exactly 1 case, N = 5 × 10⁶; for the other T - 1 cases, 2 ≤ N ≤ 100.

Sample

Input	Output
4 4 1332 4 9583 3 616 10 1029384756	Case #1: 6 Case #2: 14 Case #3: 7 Case #4: 31

In the first sample case, Thanh can get a total beauty of 6, no matter how the wall is destroyed. On the first day, he can paint either section of wall with beauty score 3. At the end of the day, either the 1st section or the 4th section will be destroyed, but it does not matter which one. On the second day, he can paint the other section with beauty score 3.

In the second sample case, Thanh can get a total beauty of 14, by painting the leftmost section of wall (with beauty score 9). The only section of wall that can be destroyed is the rightmost one, since the leftmost one is painted. On the second day, he can paint the second leftmost section with beauty score 5. Then the last unpainted section of wall on the right is destroyed. Note that on the second day, Thanh cannot choose to paint the third section of wall (with beauty score 8), since it is not adjacent to any other painted sections.

In the third sample case, Thanh can get a total beauty of 7. He begins by painting the section in the middle (with beauty score 1). Whichever section is destroyed at the end of the day, he can paint the remaining wall at the start of the second day.

[Solution in Java]

import java.util.*;
import java.io.*;
class Solution {
    static long getValue(String s,int n)
    {
        int k=0;
        long sum=0;
        if(n%2==0)
          k=n/2;
        else
         k=n/2+1;
         for(int i=0;i<k;i++)
         {
             sum+=Integer.parseInt(Character.toString(s.charAt(i)));
         }
      long curr_sum = sum;
        for (int i=k; i<n; i++)
        {
           curr_sum += Integer.parseInt(Character.toString(s.charAt(i))) -Integer.parseInt(Character.toString(s.charAt(i-k)));
           sum = Math.max(sum, curr_sum);
        }
      
        return sum;
       
    }
  public static void main(String[] args) {
    Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    int t = in.nextInt(); 
    for (int i = 1; i <= t; ++i) {
      int n = in.nextInt();
      String s=in.next();
      long x=getValue(s,n);
      System.out.println("Case #" + i + ": " + x);
    }
  }
}

Check a triplet that sum is given value

Given an array A[] of N numbers and another number x, determine whether or not there exist three elements in A[] whose sum is exactly x.

Input:
First line of input contains number of testcases T. For each testcase, first line of input contains n and x. Next line contains array elements.

Output:
Print 1 if there exist three elements in A whose sum is exactly x, else 0.

Constraints:
1 ≤ T ≤ 100
1 ≤ N ≤ 10³
1 ≤ A[i] ≤ 10⁵

Example:
Input:
2
6 13
1 4 45 6 10 8
5 10
1 2 4 3 6

Output:
1
1

Explanation:
Testcase 1: There is one triplet with sum 13 in the array. Triplet elements are 1, 4, 8, whose sum is 13.

Solution 1:(Brute-Force or Simple Method) in O(n^3)

import java.lang.*;
import java.io.*;
class Sushil
 {
     static int getCount(int a[],int x)
     {
         Arrays.sort(a);
         int n=a.length;
         for(int i=0;i<n;i++)
         {
             for(int j=i+1;j<n;j++)
             {
                 for(int k=j+1;k<n;k++)
                 {
                     if((a[i]+a[j]+a[k])==x)
                         return 1;
                 }
             }
         }
               return 0;
     }
    public static void main (String[] args)
     {
         Scanner sc=new Scanner(System.in);
         int t=sc.nextInt();
         while(t-->0)
         {
             int n=sc.nextInt();
             int x=sc.nextInt();
             int a[]=new int[n];
             for(int i=0;i<n;i++)
               a[i]=sc.nextInt();
           System.out.println(getCount(a,x));   
         }
        
   
     }
}

Solution 2:(Sorting Technique) in O(n^2)

 import java.lang.*;
import java.io.*;
class Sushil
 {
     static int getCount(int a[],int x)
     {
       
         Arrays.sort(a);
         int n=a.length;
         for(int i=0;i<n-2;i++)
         {
             int l=i+1;
             int r=n-1;
             while(l<r)
             {
                 if(a[i]+a[l]+a[r]==x)
                   return 1;
                 if(a[i]+a[l]+a[r]<x)
                    l++;
                  else
                   r--;
             }
         }
         return 0;
     }
    public static void main (String[] args)
     {
         Scanner sc=new Scanner(System.in);
         int t=sc.nextInt();
         while(t-->0)
         {
             int n=sc.nextInt();
             int x=sc.nextInt();
             int a[]=new int[n];
             for(int i=0;i<n;i++)
               a[i]=sc.nextInt();
           System.out.println(getCount(a,x));  
         }
       
    
     }
}

Pune Visit of Me and My Friends

5G Internet and Mean of 5G

Superfast "fifth generation 5G" mobile internet could be launched as early as next year in some countries, promising download speeds 10 to 20 times faster than we have now.

So let's learn and understand about the 5G Internet.

What is 5G?

It's the next - fifth-generation of mobile internet connectivity promising much faster data download and upload speeds, wider coverage and more stable connections.

5G performance targets high data rate, reduced latency, energy saving, cost reduction, higher system capacity, and massive device connectivity. The first phase of 5G specifications in Release-15 will be completed by April 2019 to accommodate the early commercial deployment. The second phase in Release-16 is due to be completed by April 2020 for submission to the International Telecommunication Union (ITU) as a candidate of IMT-2020 technology.

It's all about making better use of the radio spectrum and enabling far more devices to access the mobile internet at the same time.

What it will do for Us?

 Currently whatever we are doing to the current 4G network after launching the 5G network all these things can be faster and better.

Think of smart glasses featuring augmented reality, mobile virtual reality, much higher quality video, the internet of things making cities smarter. and all these things can be operates and mange by 5G network very efficiently and fast.

 

How Does it works? 

There are a number of new technologies likely to be applied - but standards haven't been hammered out yet for all 5G protocols. Higher-frequency bands - 3.5GHz (gigahertz) to 26GHz and beyond - have a lot of capacity but their shorter wavelengths mean their range is lower - they're more easily blocked by physical objects.

So we may see clusters of smaller phone masts closer to the ground transmitting so-called "millimetre waves" between much higher numbers of transmitters and receivers. This will enable higher density of usage. But it's expensive and telecoms companies are not wholly committed yet.

How it is different from 4G

5G will offer much higher bandwidth and capacity than 4G services. 5G not only unlocks more wireless frequencies to use between 5G NR (new radio) and mmWave (millimeter wave), it’s also designed to makes better use of what’s available between potentially many thousands of users in urban environments.

5G can offer a much lower latency than 4G. Ultra-low latency is desirable to consumers who expect a real-time response (video calling, streaming, etc), however 5G can even support some scenarios where an immediate response is not just preferred but mission critical: such as medical applications.

5G is a brand new technology, but you might not notice vastly higher speeds at first because 5G is likely to be used by network operators initially as a way to boost capacity on existing 4G (LTE - Long-Term Evolution) networks, to ensure a more consistent service for customers. The speed you get will depend on which spectrum band the operator runs the 5G technology on and how much your carrier has invested in new masts and transmitters.

Why we need 5G?

The world is going mobile and we're consuming more data every year, particularly as the popularity of video and music streaming increases. Existing spectrum bands are becoming congested, leading to breakdowns in service, particularly when lots of people in the same area are trying to access online mobile services at the same time. 5G is much better at handling thousands of devices simultaneously, from mobiles to equipment sensors, video cameras to smart street lights. 

Will it work in rural Areas?

Lack of signal and low data speeds in rural areas is a common complaint in the India and many other countries. But 5G won't necessarily address this issue as it will operate on high-frequency bands - to start with at least - that have a lot of capacity but cover shorter distances. 5G will primarily be an urban service for densely populated areas.