Programming Articles - Page 2359 of 3363

The DeserializationProblemHandler class can be registered to get called when a potentially recoverable problem is encountered during the deserialization process. We can handle the errors generated while deserializing the JSON by implementing the handleUnknownProperty() method of DeserializationProblemHandler class.Syntaxpublic boolean handleUnknownProperty(DeserializationContext ctxt, JsonParser p, JsonDeserializer deserializer, Object beanOrClass, String propertyName) throws IOExceptionExampleimport java.io.*; import com.fasterxml.jackson.core.*; import com.fasterxml.jackson.databind.*; import com.fasterxml.jackson.databind.deser.*; public class DeserializationErrorTest {    public static void main(String[] args) throws JsonMappingException, JsonGenerationException, IOException {       String jsonString = "{\"id\":\"101\", \"name\":\"Ravi Chandra\", \"address\":\"Pune\", \"salary\":\"40000\" }";       ObjectMapper objectMapper = new ObjectMapper();       DeserializationProblemHandler deserializationProblemHandler = new UnMarshallingErrorHandler();     ... Read More

The ObjectMapper class provides functionality for converting between Java objects and matching JSON constructs. We can achieve mapping of JSON data represented by an Object Model to a particular Java object using a tree-like data structure that reads and stores the entire JSON content in memory. In the first step, read the JSON data into the JsonNode object then mapped it to another instance by calling the treeToValue() method of ObjectMapper class.Syntaxpublic T treeToValue(TreeNode n, Class valueType) throws JsonProcessingExceptionExampleimport java.io.*; import com.fasterxml.jackson.core.*; import com.fasterxml.jackson.databind.*; public class JsonTreeModelDemo {    public static void main(String[] args) throws JsonProcessingException, IOException {       String jsonString ... Read More

The JsonParserSequence is a helper class that can be used to create a parser containing two sub-parsers placed in a particular sequence. We can create a sequence using the static method createFlattened() of the JsonParserSequence class.Syntaxpublic static JsonParserSequence createFlattened(JsonParser first, JsonParser second)Exampleimport java.io.*; import com.fasterxml.jackson.core.*; import com.fasterxml.jackson.core.util.*; public class JsonParserSequenceTest {    public static void main(String[] args) throws JsonParseException, IOException {       String jsonString1 = "{\"id\":\"101\", \"name\":\"Ravi Chandra\", \"address\":\"Pune\"}";       String jsonString2 = "{\"id\":\"102\", \"name\":\"Raja Ramesh\", \"address\":\"Hyderabad\", \"contacts\":[{\"mobile\":\"9959984805\", \"home\":\"7702144400\"}]}";       JsonFactory jsonFactory = new JsonFactory();       JsonParser jsonParser1 = jsonFactory.createParser(jsonString1);       JsonParser jsonParser2 = jsonFactory.createParser(jsonString2);       ... Read More

Suppose we have a container with size X. It has a mixture of water and other liquid, the mixture has W% of water in it. We have to find how many water must be added to increase the ratio of water to Y%? If X = 125, W = 20 and Y = 25, then output will be 8.33 liters.Suppose we have to add A amount of water with the previous mixture, so new amount will be X + A. So the amount of water in the mixture will follow this formula.Old Amount+A=((W% of X) + A)Also the amount of ... Read More

Suppose we have an array A, from that array, we have to choose two pairs (a, b) and (c, d), such that ab = cd. Let the array A = [3, 4, 7, 1, 2, 9, 8]. The output pairs are (4, 2) and (1, 8). To solve this, we will follow these steps −For i := 0 to n-1, dofor j := i + 1 to n-1, doget product = arr[i] * arr[j]if product is not present in the hash table, then Hash[product] := (i, j)if product is present in the hash table, then print previous and current elements.Example Live ... Read More

Suppose we have an array A, from that array, we have to get all pairs (a, b) such that the a%b = k. Suppose the array is A = [2, 3, 4, 5, 7], and k = 3, then pairs are (7, 4), (3, 4), (3, 5), (3, 7).To solve this, we will traverse the list and check whether the given condition is satisfying or not.Example Live Demo#include using namespace std; bool displayPairs(int arr[], int n, int k) {    bool pairAvilable = true;    for (int i = 0; i < n; i++) {       for (int j = 0; j < n; j++) {          if (arr[i] % arr[j] == k) {             cout

Here we will see how to print all factorial numbers less than or equal to n, a number N is said to be factorial number if it is a factorial of a positive number. So some factorial numbers are 1, 2, 6, 24, 120.To print factorial numbers, we do not need to find the factorial directly. Starting from i = 1, print factorial*i. Initially factorial is 1. Let us see the code for better understanding.Example Live Demo#include using namespace std; void getFactorialNumbers(int n) {    int fact = 1;    int i = 2;    while(fact

Consider we have a binary tree. We have to find if there are some duplicate subtrees in the tree or not. Suppose we have a binary tree like below −There are two identical subtrees of size 2. In each subtree D, BD and BE both are also duplicate subtrees We can solve this problem by using tree serialization and hashing process. We will store the inorder traversal of subtrees in the hash table. We will insert opening and closing parenthesis for empty nodes.Example Live Demo#include #include #include #include using namespace std; const char MARKER = '$'; struct ... Read More

Here we will see how to generate one symmetric matrix of order N, and the elements of each row will contain numbers from 0 to N – 1. The diagonal elements will be 0 always.This task is easy, we will form a matrix of N x N, then for each row i and for each column j, if i and j are same, then mark it as 0, otherwise increase one counter from 1 to N – 1, place the values for each individual row.Example#include using namespace std; void makeSymmetricMatrix(int n) {    int matrix[n][n];    for(int i = 0; i

Here we have an array A with some elements. Our task is to find the subset where the geometric mean is maximum. Suppose A = [1, 5, 7, 2, 0], then the subset with greatest geometric mean will be [5, 7].To solve this, we will follow one trick, we will not find the mean, as we know that the largest two elements will form the greatest geometric mean, so the largest two elements will be returned as subset.Example#include using namespace std; void largestGeoMeanSubset(int arr[], int n) {    if (n < 2) {       cout max) ... Read More