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We can fetch href links in a page in Selenium by using the method find_elements(). All the links in the webpage are designed in a html document such that they are enclosed within the anchor tag.To fetch all the elements having tagname, we shall use the method find_elements_by_tag_name(). It will fetch a list of elements of anchor tag name as given in the method argument. If there is no matching tagname in the page, an empty list shall be returned.ExampleCode Implementation.from selenium import webdriver driver = webdriver.Chrome (executable_path="C:\chromedriver.exe") driver.maximize_window() driver.get("https://www.google.com/") # identify elements with tagname lnks=driver.find_elements_by_tag_name("a") # traverse ... Read More
Suppose we have a bitonic sequence, we have to find the Bitonic Point in it. As we know a Bitonic Sequence is a sequence of numbers which is first strictly increasing then after a certain point it is strictly decreasing. This point is bitonic point. For only increasing or only decreasing sequences, bitonic points are not available.So, if the input is like [7, 8, 9, 12, 10, 6, 3, 2], then the output will be 12To solve this, we will follow these steps −define a function binary_search(array, l, r)if l array[m + 1], then −return mif array[m] < array[m ... Read More
Suppose we have two numbers a and b, we have to find an array containing values in range [1, a] and requires exactly b number of calls of recursive merge sort function.So, if the input is like a = 10, b = 15, then the output will be [3, 1, 4, 6, 2, 8, 5, 9, 10, 7]To solve this, we will follow these steps −Define a function solve() . This will take left, right, array, bif b < 1 or left + 1 is same as right, thenreturnb := b - 2mid := (left + right) / 2temp := ... 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 an array of size N; we have to find an element which divides the array into two different sub-arrays with equal product. Return -1 if no such partition is possible.So, if the input is like [2, 5, 3, 2, 5], then the output will be 3 then subarrays are: {2, 5} and {2, 5}To solve this, we will follow these steps −n := size of arraymultiply_pref := a new listinsert array[0] at the end of multiply_preffor i in range 1 to n, doinsert multiply_pref[i-1]*array[i] at the end of multiply_prefmultiply_suff := a list of size n, and fill ... Read More
Suppose we have an array of positive numbers; we have to check a point/item up to which items create a strictly decreasing sequence first followed by a sequence of strictly increasing integers. These are the following properties: We have to keep in mind that the sequences must be of minimum length 2Also, we have taken care that the last value of the decreasing sequence is the first value of the increasing sequence.So, if the input is like {5, 4, 3, 4}, then the output will be 3, as {5, 4, 3} is strictly decreasing then {3, 4} is strictly increasing.To ... Read More
Suppose we have an array of words, we have to find any alphabetical order in the English alphabet so that the given words can be considered sorted in ascending order, if there is any such order exists, otherwise return "impossible".So, if the input is like words = ["efgh", "wxyz"], then the output will be zyxvutsrqponmlkjihgfewdcbaTo solve this, we will follow these steps −ALPHABET := 26n := size of vif n is same as 1, then −display "abcdefghijklmnopqrstuvwxyz"returnDefine an array adj of size ALPHABETDefine an array in of size ALPHABET and fill with 0pre := v[0]for initialize i := 1, when ... Read More
Suppose we have a character mapping as follows, here each digit, from 1 to 9, maps to few characters.1 -> ['A', 'B', 'C'] 2 -> ['D', 'E', 'F'] 3 -> ['G', 'H', 'I'] 4 -> ['J', 'K', 'L'] 5 -> ['M', 'N', 'O'] 6 -> ['P', 'Q', 'R'] 7 -> ['S', 'T', 'U'] 8 -> ['V', 'W', 'X'] 9 -> ['Y', 'Z']If we have a number, we have to change its digits with corresponding characters in given mapping list and show all generated strings. We should consider same character for every occurrence of a digit in the number. The given ... Read More
Suppose we have a binary 2D matrix, now we have to find the beginning point and terminating point of all rectangles filled with 0s. We have to keep in mind that rectangles are separated and do not touch each other however they can touch the array boundary. A rectangle with only single element is also possible.So, if the input is like −10111011101111101100110110011011011101000011100011011101then the output will be [[0, 1, 0, 1], [0, 5, 0, 5], [1, 2, 1, 2], [2, 3, 2, 4], [3, 1, 5, 1], [3, 4, 6, 5], [5, 3, 6, 5], [7, 1, 7, 1], [7, 5, ... Read More
Suppose we have one undirected graph and a set of vertices; we have to find all reachable nodes from every vertex present in the given set.So, if the input is likethen the output will be [1, 2, 3] and [4, 5] as these are two connected components.To solve this, we will follow these steps −nodes := number of nodes in the graphDefine an array visited of size: nodes+1. And fill with 0Define one map mcomp_sum := 0for initialize i := 0, when i < n, update (increase i by 1), do −u := arr[i]if visited[u] is false, then −(increase comp_sum ... Read More