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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 More
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 More
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 More
Suppose we have a value n, we have to find the number of pairs (a, b) [a < b], that exist such that the equation a*x + b*y = n, has at least one solution.So, if the input is like n = 4, then the output will be 2 because the valid pairs are (1, 2) and (1, 3).To solve this, we will follow these steps −Define a function divisors_gen() . This will take ndivs := a list of lists of size n+1. And each inner list is holding 1divs[0] := a list with only one element 0for i in ... Read More
Suppose we have a number n. We have to find smallest number m, such that factorial of m has at least n number of 0s.So, if the input is like n = 2, then the output will be 10 because 10! = 3628800 and 9! = 362880, minimum number with 2 zeros is 10.To solve this, we will follow these steps −Define a function count_fives() . This will take ncnt := 0while n > 0, don := floor of (n / 5)cnt := cnt + nreturn cntFrom the main method, do the following −left := 1right := 5^24while right - ... Read More
Suppose we have k number of candies. We have to distribute them among children. Now there are some rulesith child will get i^2 number of candiesany children at index i will not get any candy until all children from index 1 to i-i are servedIf ith children does not get i^2 number of candies, then that is not a valid serve.So, if the input is like k = 20, then the output will be 3 because, first one will get 1, second one will get 2^2 = 4, third one will get 3^2 = 9, but fourth one needs 4^2 ... Read More
Suppose we have two numbers n and m. We have to find the remainder after dividing n number of 1s by m.So, if the input is like n = 4 m = 27, then the output will be 4, because 1111 mod 27 = 4.To solve this, we will follow these steps −Define a function util() . This will take x, n, my := 1while n > 0, doif n is odd, theny := (y * x) mod mx := (x * x) mod mn := floor of n/2return yFrom the main method return floor of (util(10, n, 9 * ... Read More
Suppose we have a number n and another value k, consider we have an array A with first N natural numbers, we have to find the total number of pairs of elements A[i] and A[j] from A, such that, i < j and their sum is divisible by k.So, if the input is like n = 10 k = 4, then the output will be 10 because there are 10 pairs whose sum is divisible by 4. [(1, 3), (1, 7), (2, 6), (2, 10), (3, 5), (3, 9), (4, 8), (5, 7), (6, 10), (7, 9)]To solve this, we ... Read More
Suppose we have two long integer values maximum and minimum. We have to find a common fraction n/d such that min
Suppose we have a number n, we have to find out the probability that any proper divisor of n would be an even perfect square.So, if the input is like n = 36, then the output will be 1/8 because there are eight proper divisors of 36, these are {1, 2, 3, 4, 6, 9, 12, 18} and among them only one number (4) is perfect square and even.To solve this, we will follow these steps −if n mod 4 is not same as 0, thenreturn 0otherwise, nc := n, ptr := 2l := a new listwhile ptr 0, ... Read More