Programming Articles - Page 1749 of 3363

We are given with an array of numbers. The goal is to find the count of prime numbers in that array.A prime number is the one which is divisible by 1 and the number itself. It has only two factors. We will check if the number is prime starting from the first element till the last and increase the count of prime numbers found so far.To check if the number N is prime, check if numbers between the range [2 to N/2], fully divides N. If yes then it is non-prime. Else it is prime.Let’s understand with examples.Input − arr[]= ... Read More

We are given with five integers N, A, B, X and Y. The goal is to maximize the profit by checking that between numbers in range [ 1 to N ] , ifA number is divisible by A, then profit increases by X.A number is divisible by B then profit increases by Y.A profit can be added once only, for a particular number in range.Let’s understand with examples.Input − N=4, A=2, B=3, X=2, Y=3Output − Maximized profit is − 7Explanation −2, 4 are divisible by A ( 2 ). Profit increases from 0 to 2, then 2 to 4 ( ... Read More

We are given with an array of N numbers with elements lying in range 0 and N-1. The elements are unsorted. The goal is to find the maximum number of partitions of the array which can be sorted individually and then can be concatenated to make a whole sorted array of length N.Each partition is chosen such that elements in it are unsorted. For N numbers ranging between 0 and N-1, sorted elements are at index equal to the value. Arr[i] = i.We will solve this by comparing each element by maximum value found so far on its left. When ... Read More

A matrix can have one or more than one minimum and maximum values. Also, the size of the matrix can be just one column and multiple rows or thousands of columns and thousands of rows. The row number and column number for the minimum and maximum values in a matrix can be found by using the following syntax −For Maximumwhich(“Matrix_Name”==min(“Matrix_Name”),arr.ind=TRUE)For Minimum>which(“Matrix_Name”==max(“Matrix_Name”),arr.ind=TRUE)Example M1

The pairwise maximum refer to the values that are largest between the vectors. For example, if we have a vector that contains 1, 2, 3 and a second vector contains 2, 1, 4 then the pairwise maximum will be 2, 2, 4 because the maximum between 1 and 2 is 2, the maximum between 2 and 1 is 2, and the maximum between 3 and 4 is 4. In R, we can find these maximum values for many vectors using pmax function.Example> x1 y1 pmax(x1, y1) [1] 27 28 65 25 17 21 29 > x2 x2 [1] 7 ... Read More

Suppose we have an array of numbers; we have to find a number B which is the divisor of all except for exactly one element in the given array. We have to keep in mind that the GCD of all the elements is not 1.So, if the input is like {8, 16, 4, 24}, then the output will be 8 as this is the divisor of all except 4.To solve this, we will follow these steps −n := size of arrayif n is same as 1, thenreturn(array[0] + 1)prefix := an array of size n, and fill with 0suffix := ... Read More

Suppose we have two strings S and T, we have to check whether a string of the same length which is lexicographically bigger than S and smaller than T. We have to return -1 if no such string is available. We have to keep in mind that S = S1S2… Sn is termed to be lexicographically less than T = T1T2… Tn, provided there exists an i, so that S1= T1, S2= T2, … Si – 1= Ti – 1, Si < Ti.So, if the input is like S = "bbb" and T = "ddd", then the output will be ... Read More

Suppose we have a set of elements; we have to find which permutation of these elements would result in worst case of Merge Sort? As we know asymptotically, merge sort always consumes O (n log n) time, but some cases need more comparisons and consumes more time. Here we have to find a permutation of input elements that will require higher number of comparisons when sorted implementing a typical Merge Sort algorithm.So, if the input is like [11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26] , then the output will be [11, ... Read More

Suppose we have a balanced binary search tree and a target sum, we have to define a method that checks whether it is a pair with sum equals to target sum, or not. In this case. We have to keep in mind that the Binary Search Tree is immutable.So, if the input is likethen the output will be (9 + 26 = 35)To solve this, we will follow these steps −Define stacks s1, s2done1 := false, done2 := falseval1 := 0, val2 := 0curr1 := root, curr2 := rootinfinite loop, do −while done1 is false, do −if curr1 is not ... Read More

Suppose we have an array A, we have to find a number X such that (A[0] XOR X) + (A[1] XOR X) + … + A[n – 1] XOR X is as minimum as possible.So, if the input is like [3, 4, 5, 6, 7], then the output will be X = 7 , Sum = 10To solve this, we will follow these steps −Define a function search_res() . This will take arr, nelement := arr[0]for i in range 0 to size of arr, doif arr[i] > element, thenelement := arr[i]p := integer of (log of element base 2) + ... Read More