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We can perform browser plugin testing with Selenium webdriver. We can have single or multiple extensions of the Chrome browser while we manually open the browser and work on it.However, while the Chrome browser is opened through Selenium webdriver, those extensions which are available to the local browser will not be available. To configure an extension, we have to obtain the .crx extension file of the extension.Then we have to add the extension to the Chrome browser which is launched by Selenium. To get all the extensions available to the browser type chrome://extensions on the browser address bar.To get add ... Read More
We can maximize the browser window with Selenium webdriver in C#. This can be done with the help of the Maximize method. We shall launch the browser and then apply this method on the driver object.Syntaxdriver.Manage().Window.Maximize();For implementation we shall be using the NUnit framework.Exampleusing NUnit.Framework; using OpenQA.Selenium; using OpenQA.Selenium.Firefox; using System; namespace NUnitTestProject1{ public class Tests{ String u = "https://www.tutorialspoint.com/index.htm"; IWebDriver d; [SetUp] public void Setup(){ //creating object of FirefoxDriver d = new FirefoxDriver(); } ... Read More
Given an array arr[] containing integer elements. The goal is to count the number of Arithmetic Progression subsequences inside arr[]. The range of elements inside arr[] are [1, 1000000].Empty sequence or single element will also be counted.Let us understand with examples.For ExampleInput - arr[] = {1, 2, 3}Output - Count of AP (Arithmetic Progression) Subsequences in an array are: 8Explanation - The following subsequences will form AP:-{}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}Input - arr[] = {2, 4, 5, 8}Output - Count of AP (Arithmetic Progression) Subsequences in an array are: 12Explanation - The following subsequences ... Read More
The following problem is an example of daily newspaper crossword, here we are given a 2-Dimensional character array and the problem statement is to find out the given word from the 2-Dimensional character array maze.The search algorithm includes finding the individual characters from Top-to-Bottom, Right-to-Left and vice-versa but not diagonally.Let us understand with examplesInput- String word-LAYS2D String[ ] - { "LOAPYS", "KAYSOT", "LAYSST", "MLVAYS", "LAYSAA", "LAOYLS" };Output- Count of number of given strings in 2D character array: 7Explanation - As we are given with the string array of words and from them, we have to find the word “LAYS” which can ... Read More
We are given a tree which can have endless levels, a variable child which will store the number of children a node can have, a variable weight which will store the weight associated with the path and a variable path that will store the path and task is to calculate the count of number of paths which has weights equals to the X and there must be an at least one edge with the given weight.For ExampleInput - int child = 4, weight = 4, path = 4; Output - Count of number of paths whose weight is exactly X and ... Read More
Given three variables size, max_val, last_element as input. The goal is to find the count of different arrays that can be formed in a manner such that they have size elements, have elements between 1 and max_val and the first element is always 1 and the last element is always max_val. Also make sure that no two consecutive elements are the same.Let us understand with examples.For ExampleInput - size = 5, max_val = 3, last_element = 3Output - Count of arrays having consecutive element with different values are: 5Explanation - The arrays will be:-[ 1, 2, 3, 1, 3 ], ... Read More
We are given with an integer number let's say, N which is considered as a divisor and it will be divided with the numbers starting from the 1 - N and the task is to calculate the count of those divisors which have more number of set bits than the quotient when divided with the given number N.For ExampleInput - int N = 6Output - Count of divisors having more set bits than quotient on dividing N are: 5Explanation - Firstly, we will divide the number N with the numbers starting from 1 - N and calculate the set bits of ... Read More
We are given with two binary strings let's say str_1 and str_2 containing the combination of 1's and 0's and the task is to firstly form the set let's say “SET” of different permutations possible from the string str_1 and then we will perform XOR operations of the elements in set with the binary string str_2 and then check whether the XOR is returning 0 or not. If yes, then consider the case else ignore it.Let us understand with examples.For ExampleInput - string str_1 = "1111", string str_2 = "1111"Output - Count of cyclic permutations having XOR with other binary string ... Read More
We are given with an integer range starting from a variable let's say start till the variable end and a variable k and d. The task is to calculate the count of digits d in a range such that d occurs exactly the k times.For ExampleInput - int start = 10, int end = 100, d = 4 and K = 2Output - Count of Numbers in a Range where digit d occurs exactly K times are: 1Explanation - The range is starting from 10 to 100. So, the possible numbers with digit d i.e. 4 occurs exactly k i.e. 2 ... Read More
We are given with an integer range, a variable m which is used as divisor and a variable d which is used to check whether the digit 'd' is at even position or not and the task is to calculate the count of those numbers in a range which are divisible by the variable m and have digit d in even positions.For ExampleInput - int start = 20, end = 50, d = 8 and m = 4Output - Count of Numbers in a Range divisible by m and having digit d in even positions are: 2Explanation - The range ... Read More