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Given with an array of natural numbers and the task is to calculate the mean absolute deviation and for that we must require the knowledge of mean, variance and standard deviation.There are steps that need to be followed for calculating the mean absolute deviationCalculate the meanCalculate absolute deviationAdd all the calculated deviationsApply the formulaInput arr[] = { 34, 21, 56, 76, 45, 11}Output mean absolute deviation is : 18.5Input arr[] = {10, 15, 15, 17, 18, 21}Output mean absolute mean absolute deviation is : 2.66used in the given program is as followsInput the elements of an arrayCalculate the mean of an arrayCalculate deviation using ... Read More
Given with an array of an unsorted array and the task is to calculate the mean and median of an unsorted array.For calculating the meanMean is calculated for finding out the average. We can use the given formula to find out the meanMean = (sum of all the elements of an array) / (total number of elementsFor calculating the medianIf an array is sorted, median is the middle element of an array in case of odd number of elements in an array and when number of elements in an array is even than it will be an average of two ... Read More
Given with a value ‘n’ and the task is to generate the centered Icosahedral number for n and centered Icosahedral series till n and display the results.What is centered Icosahedral number?Centered Icosahedral number is a centered number used for representing an icosahedrons(it is a polyhedron figure with 20 faces).The first few centered icosahedral number series till n = 1000 are −1, 13, 55, 147, 309, 561, 923Centered Icosahedral number can be calculated using the formula −$$(2n+1)\times\frac{5n^{2}+5n+3}{3}$$Input number: 20Output Centered Icosahedral Number is : 28741Input number: 12Output Centered Icosahedral Number is : 6525AlgorithmStart Step 1→ declare function to calculate centered iscosahedral number int calculate(int ... Read More
Given with a value ‘n’ and the task is to generate the centered nonagonal number for n and centered nonagonal series till n and display the results.What is centered nonagonal number?Centered nonagonal number contains the nonagonal layers formed by the dots and one corresponding dot in the center.Given above is the figure of centered nonagonal number 𝑁2.It can be calculated using the formula −$$Nc(n)=\frac{(3n-2)(3n-1)}{2}$$Input number: 20Output centered nonagonal number : 1711Input number: 10Output centered nonagonal series : 1 10 28 55 91 136 190 253 325 406AlgorithmStart Step 1→ declare function to calculate centered nonagonal number int calculate_number(int num) return ... Read More
Given a string str, we have to print the given string str in ‘+’ pattern in the matrix. To form plus pattern in a matrix the matrix must be a square matrix. A square matrix is that matrix which has same number of rows and column.Like we have a string “Tutor” our task is to print the string horizontally and vertically intersecting each other from the center, and make the rest of the elements of the matrix as “x” like in the given figure −Input str[] = {“Point”}Output Input str[] = {“this”}Output Pattern not possibleApproach used below is as follows to solve the problemGet ... Read More
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 ... Read More
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 ... 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 −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}; ... Read More
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
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 ... Read More