Programming Articles

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Java Program to check if count of divisors is even or odd

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 268 Views

To check if the count of divisors is even or odd, the Java code is as follows −Exampleimport 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 odd ...

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Product of first N factorials in C++

Sunidhi Bansal
Sunidhi Bansal
Updated on 11-Mar-2026 443 Views

Given a number N, the task is to find the product of first N factorials modulo by 1000000007. . Factorial implies when we find the product of all the numbers below that number including that number and is denoted by ! (Exclamation sign), For example − 4! = 4x3x2x1 = 24.So, we have to find a product of n factorials and modulo by 1000000007..Constraint 1 ≤ N ≤ 1e6.Input n = 9Output 27Explanation 1! * 2! * 3! * 4! * 5! * 6! * 7! * 8! * 9! Mod (1e9 + 7) = 27Input n = 3Output 12Explanation 1! * 2! * 3! mod (1e9 ...

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Java Program to check whether it is possible to make a divisible by 3 number using all digits in an array

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 291 Views

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 −Exampleimport 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};   ...

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Product of factors of number in C++

Sunidhi Bansal
Sunidhi Bansal
Updated on 11-Mar-2026 511 Views

Given a number n we have to find its all factors and find the product of those factors and return the result, i.e, the product of factors of a number. Factors of a number are those numbers which can divide the number completely including 1. Like factors of 6 are − 1, 2, 3, 6.Now according to the task we have to find the product all the factors of the number.Input − n = 18Output − 5832Explanation − 1 * 2 * 3 * 6 * 9 * 18 = 5832Input − n = 9Output − 27Explanation − 1 * 3 * 9 = 27Approach used below is as follows to solve the problem −Take the input num .Loop from i = 1 till i*i

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Java Program to Convert Iterator to Spliterator

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 186 Views

To convert Iterator to Spliterator, the Java code is as follows −Exampleimport 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 the ...

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Java program to count the characters in each word in a given sentence

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 603 Views

To count the characters in each word in a given sentence, the Java code is as follows −Exampleimport 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

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Java Program to Count trailing zeroes in factorial of a number

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 898 Views

To count trailing zeroes in factorial of a number, the Java code is as follows −Exampleimport 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 " +   ...

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Java program to expand a String if range is given?

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 428 Views

To expand a String if range is given, the Java code is as follows −Examplepublic 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

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Java Program to find reminder of array multiplication divided by n

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 283 Views

To find reminder of array multiplication divided by n, the Java code is as follows −Exampleimport java.util.*; import java.lang.*; public class Demo{    public static int remainder(int my_arr[], int arr_len, int val){       int mul_val = 1;       for (int i = 0; i < arr_len; i++)       mul_val = (mul_val * (my_arr[i] % val)) % val;       return mul_val % val;    }    public static void main(String argc[]){       int[] my_arr = new int []{ 35, 100, 69, 99, 27, 88, 12, 25 };   ...

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Java Program to Find the closest pair from two sorted arrays

AmitDiwan
AmitDiwan
Updated on 11-Mar-2026 396 Views

To find the closest pair from two sorted array, the Java code is as follows −Examplepublic class Demo {    void closest_pair(int my_arr_1[], int my_arr_2[], int arr_1_len, int arr_2_len, int sum){       int diff = Integer.MAX_VALUE;       int result_l = 0, result_r = 0;       int l = 0, r = arr_2_len-1;       while (l=0){          if (Math.abs(my_arr_1[l] + my_arr_2[r] - sum) < diff){             result_l = l;             result_r = r;             diff ...

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