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The Motzkin number series starts with 1, 1, 4, 9, etc.., We can get the generalised nth term with the sequence. The Motzkin number sequence is as follows.a0 = 1a1 = 1a2 = 4a3 = 9an = ((2 * n + 1)/ n + 2) * M(n-1) +((3 * n - 3)/ n + 2) * M(n - 2)AlgorithmInitialise the number n.Iterate till n.Update the previous two numbersReturn the last number.ExampleImplementationFollowing 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 = 2; i
We are given a array and we need to find the most frequent element from it. Let's see an example.Inputarr = [1, 2, 3, 3, 2, 2, 1, 1, 2, 3, 4]Output2In the above array, 2 occurs 4 times which is most frequent than any other in the array.Algorithm - 1Initialise the array.Initialise a map to store the frequency of each element.Count the frequency of each element and store it in the map.Iterate over the map and find the element with most frequency.Return the element.Algorithm - 2Initialise the array.Sort the given array.Maintain the variables for max count, result and current ... Read More
We are given a number N. We need to find the prime palindrome that is greater than N. Let's see an example.InputN = 10Output11AlgorithmInitialise the number N.Write a function to check whether the given number is prime or not.Write a function to check whether the given number is palindrome.Write a loop that iterates from N + 1 till you find the next prime palindrome.Check if the number is prime and palindrome.If the number is prime and palindrome.Return the number.ImplementationFollowing is the implementation of the above algorithm in C++#include using namespace std; bool isPrime(int n) { if (n < 2) ... Read More
The next smaller element is the element that is the first smaller element after it. Let's see an example.arr = [1, 2, 3, 5, 4]The next smaller element for 5 is 4 and the next smaller element for elements 1, 2, 3 is -1 as there is no smaller element after them.AlgorithmInitialise the array with random numbersInitialise 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 smaller than the top element of the stack.Print the top element with the next smaller ... Read More
The n-ary tree is the tree with n children for each node. We are given a number n and we have to find the next larger element from the n-ary tree.We can find the solution by traversing through the n-ary tree and maintaining the result.AlgorithmCreate n-ary tree.Initialise a result.Write a function to get next larger element.Return if the current node is null.Check the current node data is greater than the expected element or not.If yes, then check whether the result is empty or the result is greater than the current node data.If the above condition satisfies, then update the result.Get ... Read More
We are given a number n, we have to find the number that is greater than n with same number of set bits as n in its binary representation.The digit 1 in the binary representation is called set bit.Let's see an example.Input124Output143AlgorithmInitialise 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 equal to the number of set bits of n.Return the number when you find it.Increment the number.ImplementationFollowing is the implementation of the above algorithm in C++#include ... Read More
Given a number n, swap any two digits of the number so that the resulting number is greater than the number n. If it's not possible then print -1. Let's see an example.Input12345Output12354We have swapped the digits 4 and 5. And we got the higher number with one swap.AlgorithmIt's not possible to form the number if the digits of the number are in decreasing order.Find the index of the digit from the right of the number which is less than the last digit.Find the index of the digit which is greater than the previous digit and less than all digits.Swap ... Read More
Given N, A, and B. Find the number which is greater than N with the same number of A and B digits. Let's see an example.N = 1234 A = 2 B = 3We need to check for every possibility of the given number of digits. There are two digits to form the number. And each digit count in the number should be the same.AlgorithmInitialise A, B, and N.Write a recursive function.Check whether the current number is greater than N and has equal number of A and B digits.Return the number if the above condition satisfies.Add the digit A to the result.Add the digit B ... Read More
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 More
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 More