Sunday, May 1, 2022

How to convert glide app to an android apk

 How to convert glide app to an android apk


Step1: 

Login to https://www.kodular.io/ 


Step 2: 

Click on CREATE APPS. Sign up your account. Now you will see the below screen. click create project and proceed.







Step 3: 

On the left side pane type web in the search box , now drag and drop the Web viewer component in to the mobile screen.


Step 4: 

Click on the Web_Viewer1 component highlighted as 1 in the below image.

Properties pane opens at the right side, Give your glide url in the Home Url as highlighted in the below image as 2






Step 5:

As highlighted in the below image click on the "Blocks" tab in the top right corner 




Step 6: 

You will see the below page. on the left pane scroll down and select screen1 and the select the first item that is higlighted as 2.


Step 7: 

Click the Control option and drag and drop the first option highlighted as 2. fit if then component inside the previous screen1 component.


Step 8: 

From the left pane scroll down and click the Web_Viewer1 component and choose the call Web_Viewer1.Can Go Back Or Forward component highlighted below.grag and drop inside the if then component created in the previous step.




Step 9:

Now choose the Math option from the left pane and drag and drop the first option highlighted as 2. After drag and drop inside the box type "-2"





Step 10:

Choose the Go back component as shown in below image and drag and drop. 


Step 11: 

Now click the setting button left to if and select else option from the box below. add that below the "then ".





Step 12:

Finally choose the close appication option from the Control component and drag and drop.






Step 13: 

Arrange al those components as in below image.


Step 13:

Click on Export button at the top and click on Android App(apk) to download the apk file.






Thank you...


Monday, April 25, 2022

How to add an element in an array in given position

     How to add an element in an array in given position


public class Demo {

public static void main(String args[]) {

int arr[] = new int[5];

arr[0]=1;

arr[1]=2;

arr[2]=3;

arr[3]=4;

int x = 5,pos=3; // x = number to be added in the array , pos = position

int n=4; //length of array

n++;

for(int i =n-1;i>=pos;i--) {

arr[i] = arr[i-1];

}

arr[pos-1] = x;

//print the array

for (int i=0;i<arr.length;i++) {

System.out.println(arr[i]);

}

}

}


How to Find the Sub Array in Java

 

How to Find the Sub Array in Java



import java.util.ArrayList;


import java.util.List;

import java.util.TreeSet;


public class Demo {

public static void main(String args[]) {

Demo psm = new Demo();

int arr[] = { 1, 2, 3 };

psm.printSubArray(arr);

}


void printSubArray(int arr[]) {

List<List<Integer>> intList = new ArrayList<>();

int n = arr.length;

for (int i = 0; i < n; i++) // This loop will select start element

{

for (int j = i; j < n; j++) // This loop will select end element

{


List<Integer> list1 = new ArrayList<>();

for (int k = i; k <= j; k++) // This loop will print element from start to end


{

list1.add(arr[k]);


}

intList.add(list1);


}

}

System.out.println(intList);

TreeSet<Integer> tSet = new TreeSet<>();

for (List<Integer> list : intList) {

tSet.add((list.stream().mapToInt(i->i).sum())%2);

}

System.out.println(tSet);

System.out.println(tSet.last());

}


}



Output

[[1], [1, 2], [1, 2, 3], [2], [2, 3], [3]]

[0, 1]

1


Tuesday, April 12, 2022

Java String Compare

 Hackerrak Java String Compare By Java Tech Solutionz


Problem:

We define the following terms:

  • Lexicographical Order, also known as alphabetic or dictionary order, orders characters as follows:

    For example, ball < catdog < dormHappy < happyZoo < ball.

  • substring of a string is a contiguous block of characters in the string. For example, the substrings of abc are abcabbc, and abc.

Given a string, , and an integer, , complete the function so that it finds the lexicographically smallest and largest substrings of length .

Function Description

Complete the getSmallestAndLargest function in the editor below.

getSmallestAndLargest has the following parameters:

  • string s: a string
  • int k: the length of the substrings to find

Returns

  • string: the string ' + "\n" + ' where and are the two substrings

Input Format

The first line contains a string denoting .
The second line contains an integer denoting .

Constraints

  •  consists of English alphabetic letters only (i.e., [a-zA-Z]).

Sample Input 0

welcometojava
3

Sample Output 0

ava
wel

Explanation 0

String  has the following lexicographically-ordered substrings of length :

We then return the first (lexicographically smallest) substring and the last (lexicographically largest) substring as two newline-separated values (i.e., ava\nwel).

The stub code in the editor then prints ava as our first line of output and wel as our second line of output.



Solution:


public class Solution {

    public static String getSmallestAndLargest(String s, int k) {
        String smallest = "";
        String largest = "";
        
        // Complete the function
        // 'smallest' must be the lexicographically smallest substring of length 'k'
        // 'largest' must be the lexicographically largest substring of length 'k'
        
      
        
        int n=0;
        for(int i=0 ; i<=s.length()-k;i++){
            n++;
        }
        String arr[] = new String[n];
        
        for(int i=0 ; i<=s.length()-k;i++){
          
            arr[i]=s.substring(i,i+k);
        }
        arr = mySort(arr);
        System.out.println(arr[0]);
        System.out.println(arr[n-1]);
        
        return smallest + "\n" + largest;
    }
    
    public static String[] mySort(String[] ar) {
        
        String temp;
        
        for(int i=ar.length-1; i>0 ; i--) {
            for(int j=0;j<i;j++) {
                if(ar[j].compareTo(ar[j+1] )>0) {
                    temp = ar[j];
                    ar[j] = ar[j+1];
                    ar[j+1] = temp;
                }
            }
        }
    
        return ar;
    }


Concurrent Hashmap in java

  A ConcurrentHashMap is a thread-safe implementation of the Map interface in Java. Unlike a HashMap , which is not thread-safe and can ca...