Programming Articles - Page 1868 of 3366

In mathematics, the factors of a number are the integers that divide the number without leaving a remainder. For a given number, finding the minimum sum of its factors involves identifying a pair of factors whose sum is the smallest among all possible factor pairs. In this article, we will learn to find the minimum sum of factors of a number we will be using the Integer class and Math class in Java − Integer class The Java Integer class serves as a wrapper for the primitive int type, encapsulating a single integer value within an object. For Integer.MAX_VALUE, which provides ... Read More

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

In this article, we are given an integer value and the task is to count the total number of set bits of the given integer. For this task, we need to convert the given value into its corresponding binary representation. The binary number of an integer value is represented as the combination of 0's and 1's. Here, the digit 1 is known as the Set bit in the terms of the computer. Problem Statement Write a program in Java to count set bits in an integer − Input  int num = 10 Output binary representation = 1010 set bit ... Read More

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