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Java API call Example using GSON, org.json.json and Jackson [ Simple Get Call] and parsing result as JSON

import com.fasterxml.jackson.databind.JsonNode ; import com.fasterxml.jackson.databind.ObjectMapper ; import com.google.gson.* ; import org.json.JSONArray ; import org.json.JSONObject ; import java.io.* ; import java.net.HttpURLConnection ; import java.net.URL ; public class APICALL { public static void main (String[] args) throws IOException { // String url="https://mocki.io/v1/19a50724-c2e5-46a1-b457-543462cdfde2"; String url= "https://jsonplaceholder.typicode.com/users" ; String line ; StringBuilder resp= new StringBuilder() ; System. out .println(url) ; HttpURLConnection con= (HttpURLConnection) new URL(url).openConnection() ; con.setRequestMethod( "GET" ) ; con.setRequestProperty( "Accept" , "application/json" ) ; System. out .println(con.getResponseMessage()) ; System. out .println(con.getContentType()) ; InputStream inputStream=con.getInput

python program to Print Starting Series OF Indian Mobile Number for a State or operator or both

import requests import urllib.request import time from bs4 import BeautifulSoup as bs import re url = ' https://en.wikipedia.org/wiki/Mobile_telephone_numbering_in_India' state_to_extract = "UE" #if set to None all state is considered telecom_to_extracted = None #if set to none all operator from particular city is extracted response = requests . get(url) print (response) soup = bs(response . text, "html.parser" ) one_a_tag = soup . findAll( 'tr' )[ 35 :] lst = [] for k in one_a_tag: s = k . findAll( 'td' ) limit = len (s) i = 0 while True : if i == limit: break no = s[i] . text i += 1 if i == limit: break operator = s[i] . text i += 1 if i == limit: break state = s[i] . text i += 1 if i == limit: break res = f "{no} {operator} {state}" if state_to_extract is None : if telecom_to_extracted is None : lst . append(no) elif telecom_to_e

Binary Search Tree in Java implementation (reference based, dynamic memory)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 import java.util.Scanner ; class BST { static BST . Node root = null ; public void insert ( int num ) { if ( root == null ) { root = new BST . Node ( num ); } else { // root node is not empty BST . Node temp = root ; while ( temp != null ) { if ( num <= temp . getVal ()) { if ( temp . getLeft () != null ) temp = temp . getLeft ();

Binary Search Tree in C++( dynamic memory based )

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 #include<bits/stdc++.h> using namespace std; struct bst { int val; bst * left, * right; }; bst * root = nullptr; void srch ( int num,bst * head) { if (head == nullptr){ cout << " \n Number is not present \a " << endl; return ; } if (head -> val == num) { cout << " \n Number is present \n\a " ; return ; } else { if (num < head -> val) srch(num,head -> left); else srch(num,head -> right);