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Articles by Hafeezul Kareem
Page 19 of 26
Number of digits in 2 raised to power n in C++
The power of a number can be computed using the iterative multiplication or function that the language provides. It's a straightforward thing.Here, we have to find the 2 raised to power n. And the number of digits in the result. Let's see some examples.Input5Output2Input10Output4 AlgorithmInitialise the number n.Find the value of 2n.The ceil of log10(n) will give you number of digits in the number n.Find it and return it.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int getDigitsCount(int n) { return ceil(log10(pow(2, n))); } int main() { int n = 8; ...
Read MoreNumber of arrays of size N whose elements are positive integers and sum is K in C++
We are given two numbers n and k. We need to find the count of arrays that can be formed using the n numbers whose sum is k.The number of arrays of size N with sum K is $\dbinom{k - 1}{n - 1}$.This a straightforward formula to find the number arrays that can be formed using n elements whose sum k. Let's see an example.Inputn = 1 k = 2Output1 The only array that can be formed is [2]Inputn = 2 k = 4Output3 The arrays that can be formed are [1, 3], [2, 2], [3, 1].AlgorithmInitialise the numbers n ...
Read MoreNumber of anomalies in an array in C++
In this tutorial, we are going to write a program that finds the number of anomalies in the given array.A number is an anomaly in the given array if the absolute difference between the number and all other numbers is greater than the given number k. Let's see an example.Inputarr = [3, 1, 5, 7] k = 1Output4The absolute difference between all numbers with others is greater than k.AlgorithmInitialise the array.Iterate over the array.Take the element and iterate over the array.Find the absolute difference between the two numbers.If no absolute difference is less than or equal to k than increment ...
Read MoreNumber is divisible by 29 or not in C++
It's a straightforward problem. We can use the modulo (%) operator to check whether the given number is divisible by 29 or not. Let's see some examples.Input29 254Output1 0AlgorithmImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; bool isDivisibleBy29(long long n) { return n % 29 == 0; } int main() { cout
Read MoreMultiply two polynomials in C++
The coefficients of each term of the polynomial are given in an array. We need to multiply the two polynomials. Let's see an example.InputA = [1, 2, 3, 4] B = [4, 3, 2, 1]Output4x6 + 11x5 + 20x4 + 30x3 + 20x2 + 11x1 + 4AlgorithmInitialise two polynomials.Create a new array with a length of two polynomials.Iterate over the two polynomials.Take one term from the first polynomial and multiply it with all the terms in the second polynomial.Store the result in the resultant polynomial.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int *multiplyTwoPolynomials(int ...
Read MoreMultiply two numbers represented as linked lists into a third list in C++
Given two linked lists with digits in it. We need to multiply two numbers formed by the linked list. It can be done easily by forming the numbers from the two linked lists. Let's see an example.Input1 -> 2 -> NULL 2 -> 3 -> NULLOutput2 -> 7 -> 6 -> NULLAlgorithmInitialise the two linked lists.Initialise two variables with 0 to store the two numbers.Iterate over the two linked lists.Add each digit to the respective number variable at the end.Multiply the resultant numbers and store the result in a variable.Create a new list with the result.Print the new list.ImplementationFollowing is ...
Read MoreMultiply any Number with using Bitwise Operator in C++
In this tutorial, we are going write a program that multiplies the given two numbers using bitwise operators.The left shift () is used for the division.The multiplication of two numbers x, y can be written as x * y = (x * 2) * (y / 2) if y is even else it's equal to x * y = (x * y) * (y / 2) + x.So whenever the second number becomes odd, add the first number to the result. Let's see the steps to solve the problem.AlgorithmInitialise two numbers.Write a loop that iterates till the second number becomes ...
Read MoreMultiply a given Integer with 3.5 in C++
To get the result of n * 3.5 we need to calculate (n * 2) + n + (n / 2). Moving the bits to left by 1 will give you n * 2 and moving the bits to right by will you n / 2. Add those to get the result.n * 3.5 = (n * 2) + n + (n / 2)You can submit different values of n to verify the above equation. Let's see some examples.Input2 7 10Output7 24 35AlgorithmInitialise the number n.Find the n * 2 using left shift bitwise operatorFind the n / 2 using ...
Read MoreMultiples of 3 or 7 in C++
Given a number n, we need to find the count of multiples of 3 or 7 till n. Let's see an example.Input100Output43There are total of 43 multiples of 3 or 7 till 100.AlgorithmInitialise the number n.Initialise the count to 0.Write a loop that iterates from 3 to n.Increment the count if the current number is divisible by 3 or 7.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int getMultiplesCount(int n) { int count = 0; for (int i = 3; i
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