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In this tutorial, we are going to write a program that finds the k-th prime number which is greater than the given number n.Initialise the number n.Find all the prime numbers till 1e6 and store it in a boolean array.Write a loop that iterates from n + 1 to 1e6.If the current number is prime, then decrement k.If k is equal to zero, then return i.Return -1.ExampleLet's see the code.#include using namespace std; const int MAX_SIZE = 1e6; bool prime[MAX_SIZE + 1]; void findAllPrimes() { memset(prime, true, sizeof(prime)); for (int p = 2; p * p Read More
In this tutorial, we are going to find the k-th smallest element after every insertion.We are going to use the min-heap to solve the problem. Let's see the steps to complete the program.Initialise the array with random data.Initialise the priority queue.Till k - 1 there won't be any k-th smallest element. So, print any symbol u like.Write a loop that iterates from k + 1 to n.Print the root of the min-heap.If the element is greater than the root of the min-heap, then pop the root and insert the element.ExampleLet's see the code.#include using namespace std; void findKthSmallestElement(int elements[], ... Read More
In this tutorial, we are going to write a program that finds the k-th smallest number in the unsorted array.Let's see the steps to solve the problem.Initialise the array and k.Sort the array using sort method.Return the value from the array with the index k - 1.Let's see the code.Example#include using namespace std; int findKthSmallestNumber(int arr[], int n, int k) { sort(arr, arr + n); return arr[k - 1]; } int main() { int arr[] = { 3, 5, 23, 4, 15, 16, 87, 99 }, k = 5; cout
In this tutorial, we are going to learn about largrange's four square theorem.The lagranges's four square theorem states that every natural number can be written as sum of squares of 4 numbers.The following code finds the 4 numbers which satisfies the above condition for the given number n.ExampleLet's see the code.#include using namespace std; void printSquareCombinations(int n) { for (int i = 0; i * i
In this tutorial, we are going to write a program that finds the result for lagranges's interpolation formula.You don't have write any logic for the program. Just convert the formula into code. Let's see the code.Example#include using namespace std; struct Data { int x, y; }; double interpolate(Data function[], int xi, int n) { double result = 0; for (int i = 0; i < n; i++) { double term = function[i].y; for (int j = 0; j < n; j++) { if (j != i) { ... Read More
In this tutorial, we are going to write a program that finds out the larger among the ab and baIt's a straightforward problem. Let's see the steps to solve it.Initialise the values of a and b.Take the log of both the values.Compute the values of $b\:\log\:a$ and $a\:\log\:b$Compare the both values.If $a\:\log\:b$ is greater than $b\:\log\:a$, then print bais greater.If $b\:\log\:a$ is greater than $a\:\log\:b$, then print abis greater.Else print both are equal.ExampleLet's see the code.#include using namespace std; int main() { int a = 4, b = 7; long double x = (long double) a * ... Read More
In this tutorial, we are going to write a program that finds the largest even and odd number of n digits number.Let's see the steps to solve the problem.Initialise the number n.The largest odd number is pow(10, n) - 1.The largest even number is odd - 1.ExampleLet's see the code.#include using namespace std; void findEvenAndOddNumbers(int n) { int odd = pow(10, n) - 1; int even = odd - 1; cout
In this tutorial, we are going to write a program that finds the largest number whose digits are all even and not greater than the given n.Let's see the steps to solve the problem.Initialise the number n.Write a loop from i = n .Check whether the digits of current number are all even or not.If the above condition satisfies, then print the number.Else decrement the i.ExampleLet's see the code.#include using namespace std; int allDigitsEven(int n) { while (n) { if ((n % 10) % 2){ return 0; } ... Read More
In this tutorial, we are going to write a program that finds the largest k-digit number that is divisible by x.Let's see the steps to solve the problem.Initialise the x and k.Find the value of pow(10, k) - 1 which is largest k-digit number.Now, remove the remainder value from the above value to get the largest k-digit number that is divisible by x.ExampleLet's see the code.#include using namespace std; int answer(int x, int k) { int max = pow(10, k) - 1; return max - (max % x); } int main() { int x = 45, k = 7; cout
In this tutorial, we are going to write a program that finds the largest n-digit number that is divisible by the given three numbers.Let's see the steps to solve the problem.Initialise three numbers along with n.Find the LCM of three numbers.Store the largest number with n-digits.If the largest number is divisible by n, then return it.Else check for the number obtained from subtracting remainder in the above step.ExampleLet's see the code.#include using namespace std; int LCM(int x, int y, int z) { int ans = ((x * y) / (__gcd(x, y))); return ((z * ans) / (__gcd(ans, ... Read More