Server Side Programming Articles - Page 1719 of 2650

To expand a String if range is given, the Java code is as follows −Example Live Demopublic class Demo {    public static void expand_range(String word) {       StringBuilder my_sb = new StringBuilder();       String[] str_arr = word.split(", ");       for (int i = 0; i < str_arr.length; i++){          String[] split_str = str_arr[i].split("-");          if (split_str.length == 2){             int low = Integer.parseInt(split_str[0]);             int high = Integer.parseInt(split_str[split_str.length - 1]);             while (low

To count trailing zeroes in factorial of a number, the Java code is as follows −Example Live Demoimport java.io.*; public class Demo{    static int trailing_zero(int num){       int count = 0;       for (int i = 5; num / i >= 1; i *= 5){          count += num / i;       }       return count;    }    public static void main (String[] args){       int num = 1000000;       System.out.println("The number of trailing zeroes in " + num +" factorial is " + ... Read More

To count the occurrence of each character in a string using Hashmap, the Java code is as follows −Example Live Demoimport java.io.*; import java.util.*; public class Demo{    static void count_characters(String input_str){       HashMap my_map = new HashMap();       char[] str_array = input_str.toCharArray();       for (char c : str_array){          if (my_map.containsKey(c)){             my_map.put(c, my_map.get(c) + 1);          }else{             my_map.put(c, 1);          }       }       for (Map.Entry entry : my_map.entrySet()){ ... Read More

To count the characters in each word in a given sentence, the Java code is as follows −Example Live Demoimport java.util.*; public class Demo{    static final int max_chars = 256;    static void char_occurence(String my_str){       int count[] = new int[max_chars];       int str_len = my_str.length();       for (int i = 0; i < str_len; i++)       count[my_str.charAt(i)]++;       char ch[] = new char[my_str.length()];       for (int i = 0; i < str_len; i++){          ch[i] = my_str.charAt(i);          int find = 0;          for (int j = 0; j

To convert Iterator to Spliterator, the Java code is as follows −Example Live Demoimport java.util.*; public class Demo{    public static Spliterator getspiliter(Iterator iterator){       return Spliterators.spliteratorUnknownSize(iterator, 0);    }    public static void main(String[] args){       Iterator my_iter = Arrays.asList(56, 78, 99, 32, 100, 234).iterator();       Spliterator my_spliter = getspiliter(my_iter);       System.out.println("The values in the spliterator are : ");       my_spliter.forEachRemaining(System.out::println);    } }OutputThe values in the spliterator are : 56 78 99 32 100 234A class named Demo contains a function named ‘getspiliter’ that returns a spliterator. In ... Read More

To check whether it is possible to make a divisible by 3 number using all digits in an array, the Java code is as follows −Example Live Demoimport java.io.*; import java.util.*; public class Demo{    public static boolean division_possible(int my_arr[], int n_val){       int rem = 0;       for (int i = 0; i < n_val; i++)       rem = (rem + my_arr[i]) % 3;       return (rem == 0);    }    public static void main(String[] args){       int my_arr[] = { 66, 90, 87, 33, 123}; ... Read More

To check if the count of divisors is even or odd, the Java code is as follows −Example Live Demoimport java.io.*; import java.math.*; public class Demo{    static void divisor_count(int n_val){       int root_val = (int)(Math.sqrt(n_val));       if (root_val * root_val == n_val){          System.out.println("The number of divisors is an odd number");       }else{          System.out.println("The number of divisors is an even number");       }    }    public static void main(String args[]) throws IOException{       divisor_count(25);    } }OutputThe number of divisors is an ... Read More

The given article involves determining if all digits of a positive integer can divide the number without leaving a remainder. If any digit is zero or does not divide the number the result is false; otherwise, it is true using Java. This can be solved using two approaches: the Naive-Based Approach, which relies on arithmetic operations, and the String-Based Approach, which uses string manipulation for digit processing. Both methods ensure each digit meets the divisibility condition efficiently. Approaches to Check If All Digits of a Number Divide It Following are the different approaches to check if all digits of a ... Read More

In this article, you will learn to write a program in Java to calculate the sum of squares of first n natural numbers. This program efficiently computes the sum using a mathematical formula − (val * (val + 1) / 2) * (2 * val + 1) / 3 Problem Statement Write a program in Java to calculate the sum of squares of first n natural numbers, where n is a user-defined value. Input val = 8 Output The sum of squares of first 8 natural numbers is204 Steps to calculate sum of squares of first n natural numbers Following are ... Read More

To find the smallest K digit number divisible by X, the Java code is as follows −Example Live Demoimport java.io.*; import java.lang.*; public class Demo{    public static double smallest_k(double x_val, double k_val){       double val = 10;       double MIN = Math.pow(val, k_val - 1);       if (MIN % x_val == 0)       return (MIN);       else       return ((MIN + x_val) - ((MIN + x_val) % x_val));    }    public static void main(String[] args){       double x_val = 76;       double k_val ... Read More