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Articles by Hafeezul Kareem
Page 21 of 26
Next greater integer having one more number of set bits in C++
We are given a number n, we have to find the number that is greater than n with one more set bit than n in its binary representation.The digit 1 in the binary representation is called set bit.Let's see an example.Input124Output125AlgorithmInitialise the number n.Write a function get the count of number of set bits.Initialise the iterative variable with n + 1.Write an infinite loop.Check for the number of set bits for numbers greater than n.Return the number when you find it.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int getSetBitsCount(int n) { int count ...
Read MoreNext Greater Element in C++
The next greater element is the element that is first greater element after it. Let's see an example.arr = [4, 5, 3, 2, 1]The next greater element for 4 is 5 and the next greater element for elements 3, 2, 1 is -1 as there is no greater element after them.AlgorithmInitialise the array with random numbers.Initialise a stack.Add first element to the stack.Iterate through the element of the array.If the stack is empty, add the current element to the stack.While the current element is greater than the top element of the stack.Print the top element with the next greater element ...
Read MoreNewman-Shanks-Williams prime in C++
The newman-shanks-williams prime sequence is as follows1, 1, 3, 7, 17, 41...If we generalise the sequence items, we geta0=1 a1=1 an=2*a(n-1)+a(n-2)AlgorithmInitialise the number n.Initialise the first numbers of the sequence 1 and 1.Write a loop that iterates till n.Compute the next number using the previous numbers.Update the previous two numbers.Return the last number.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int getNthTerm(int n) { if(n == 0 || n == 1) { return 1; } int a = 1, b = 1; for(int i = 3; i
Read MoreNesbitt’s Inequality in C++
The nesbitt's inequality is (a/(b + c)) + (b/(c + a)) + (c/(a + b))>= 1.5, a > 0, b > 0, c > 0Given three number, we need to check whether the three numbers satisfy Nesbitt's inequality or not.We can test whether three number are satisfied nesbitt's inequality or not. It's a straightforward program.AlgorithmInitialise three numbers a, b, and c.Compute the values of each part from the equation.Add them all.If the total sum is greater than or equal to 1.5 then it satisfies the Nesbitt's inequality else it is not satisfied.ImplementationFollowing is the implementation of the above algorithm in ...
Read MoreNeon Number in C++
A neon number is a number where the sum of digits of the square of the number is equal to the number. Let's take an example.n = 9square = 81sum of digits of square = 8 + 1 = 9So, the number 9 is a neon number.We need to check whether the given number is a neon number or not. If the given number is a neon number, then print Yes else print No.AlgorithmInitialise the number n.Find the square of the number.Find the sum of the digits of the squareIf the sum of digits of the square is equal to ...
Read MoreNearest prime less than given number n C++
We are given a number n, we need to find the nearest prime number that is less than n. We can find the number easily if we start checking from the n - 1. Let's see some examples.Input10Output7AlgorithmInitialise the number n.Write a loop that iterates from n - 1 to 1Return the first prime number that you foundReturn -1 if you didn't find any prime that's less than given nImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; bool isPrime(int n) { if (n == 2) { return true; } ...
Read MoreNarayana number in C++
The Narayana numbers can be expressed in terms of binomial expression $1/n\binom{n}{k} \binom{n}{k-1}$ Learn more about the Narayana number here.You are given the numbers n and k. Find the Narayana number. It's a straightforward problem having the combinations formula. Let's see the code.AlgorithmInitialise the numbers n and k.Find the Narayana number using the given formula.Print the resultant number.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int factorial(int n) { int product = 1; for (int i = 2; i
Read MoreN’th Smart Number in C++
A smart number is a number that contains at least three distinct prime factors. You are given a number N. Find the n-th smart number.The smart number series are30, 42, 60, 66, 70, 78...AlgorithmInitialise the number N.Initialise the count to 0.Write a function that checks whether the given number is prime or not.Write a function that checks whether the number is smart or not.Write a loop that iterates from 30 as first smart number is 30.Check whether the current number is smart number or not using the prime number function.Increment the count by 1 when you find a smart number.Return ...
Read MoreN’th palindrome of K digits in C++
To find the n-th palindrome of k digits, we can iterate from the first k digits number till we find the n-th palindrome number. This approach is not efficient. You can try it yourself.Now, let's see the efficient approach to find the n-th palindrome of k digits.There are two halves in the numbers. The first half is equal to the reverse of the second half.The first half of the n-th number with k digits areIf k is odd then (n - 1) + 10k/2else(n-1)+10k/2-1The second half of the n-th number with k digits will be the reverse of the first ...
Read Moren-th term of series 1, 17, 98, 354…… in C++
The given series is 1, 17, 98, 354...If you clearly observe the series, you will find that the n-th number is equal to the 4 powers.Let's see the pattern. 1 = 1 ^ 4 17 = 1 ^ 4 + 2 ^ 4 98 = 1 ^ 4 + 2 ^ 4 + 3 ^ 4 354 = 1 ^ 4 + 2 ^ 4 + 3 ^ 4 + 4 ^ 4 ...AlgorithmInitialise the number N.Initialise the result to 0.Write a loop that iterates from 1 to n.Add 4th power current number to the result.Print the result.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; int getNthTerm(int n) { int nthTerm = 0; for (int i = 1; i
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