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We can identify elements by partially comparing to its attributes in Selenium with the help of regular expression. In xpath, there is contains () method. It supports partial matching with the value of the attributes. This method comes as useful while dealing with elements having dynamic values in their attributes.Syntax driver.findElement(By.xpath("//tagname[contains(@attributes, ’value’)]"))In CSS, we can identify elements by partially comparing to its attributes by using regular expressions. There can be three scenarios −Using ^ to target attributes starting with a particular text.Syntax −driver.findElement(By.cssSelector("tagname[attribute^=’value’]"))Using $ to target attributes ending with a particular text.Syntax −driver.findElement(By.cssSelector("tagname[attribute$=’value’]"))Using * to target attributes containing a particular text.Syntax −driver.findElement(By.cssSelector("tagname[attribute*=’value’]"))Exampleimport org.openqa.selenium.By; ... Read More
Suppose we have a specific rectangular web page area, our job is to design a rectangular web page, whose length L and width W that satisfies the following requirements −The area of the web page must equal to the given target area.The width W should not be larger than the length L, and L >= W.The difference between L and W should be as small as possible.So, if the input is like 4, then the output will be [2, 2], as the target area is 4, and all the possible ways to construct it are [1, 4], [2, 2], [4, ... Read More
Suppose we have to design a standard heater with a fixed warm radius to warm all the houses. Now, we have given positions of houses and heaters on a horizontal line, we have to find the minimum radius of heaters so that all houses could be covered by those heaters. So, we will provide houses and heaters separately, and our expected output will be the minimum radius standard of heaters.So, if the input is like [1, 2, 3, 4], [1, 4], then the output will be 1 as the two heaters was placed in position 1 and 4. We have ... Read More
Suppose we are trying to distribute some cookies to children. But, we should give each child at most one cookie. Now each child i has a greed factor gi, which is the minimum size of a cookie that the child will be content with; and each cookie j has a size sj. When sj >= gi, we can assign the cookie j to the child i, and the child i will be content. Our goal is to maximize the number of content children and output the maximum number.So, if the input is like [1, 2], [1, 2, 3], then the ... Read More
Suppose we have an array of size n, we have to find the minimum number of moves required to make all array elements the same, where a move means incrementing n - 1 elements by 1.So, if the input is like [3, 2, 3, 4], then the output will be 4.To solve this, we will follow these steps −n := size of numsif n is same as 0, then −return 0sort the array numsans := 0for initialize i := 0, when i < n, update (increase i by 1), do −ans := ans + nums[i] - nums[0]return ansExample Let us see ... Read More
Suppose we have n points in the plane that are all pairwise distinct. Now a "boomerang" is a tuple of points like (i, j, k) such that the distance between i and j is the same as the distance between i and k. We have to find the number of boomerangs.So, if the input is like [[0, 0], [1, 0], [2, 0]], then the output will be 2, as two boomerangs are [[1, 0], [0, 0], [2, 0]] and [[1, 0], [2, 0], [0, 0]].To solve this, we will follow these steps −counter_of_boomerangs := 0for each point_1 in points array, ... Read More
Suppose we have a string s. We have to count the number of segments in a string, where a segment is defined to be a contiguous sequence of characters (no whitespace).So, if the input is like "Hello, I love programming", then the output will be 4, as there are 4 segments.To solve this, we will follow these steps −n := 0for initialize i := 0, when i < size of s, update (increase i by 1), do −if s[i] is not equal to white space, then −(increase n by 1)while (i < size of s and s[i] is not equal ... Read More
The various types of locators that Selenium support are listed below −ID − This attribute is unique for every element.Syntax − driver.findElement(By.id("")).Name − This attribute is not unique for every element.Syntax − driver.findElement(By.name("")).CSS Selector − This can be derived from element tags and attributes.Syntax −driver.findElement(By.cssSelector("")).xpath − This can be derived from element tags and attributes.Syntax −driver.findElement(By.xpath("")).TagName − This can be derived from the HTML tags to identify the elements.Syntax − driver.findElement(By.tagName("")). LinkText − This can be derived from the anchor text to identify the elementszSyntax − driver.findElement(By.linkText("")).PartialLinkText − This can be derived from the partial anchor text to identify the elementsSyntax − driver.findElement(By.partialLinkText("")).Classname ... Read More
Suppose we have a non-empty array of integers; we have to find the third maximum number in this array. If there is no 3rd max number, then return the maximum one. The challenge is, we have to solve this using linear time complexity.So, if the input is like [5, 3, 8, 9, 1, 4, 6, 2], then the output will be 6.To solve this, we will follow these steps −initialize a, b, c with NULLfor initialize i := 0, when i < size of nums, update (increase i by 1), do −if a is null or nums[i] >= value of ... Read More
Suppose we have a string which consists of lowercase or uppercase letters, we have to find the length of the longest palindromes that can be built with those letters. Now the string is case sensitive, so "Aa" is not considered a palindrome here.So, if the input is like "abccccdd", then the output will be 7, as one longest palindrome that can be built is "dccaccd", whose length is 7.To solve this, we will follow these steps −Define one map mpfor each character i in s(increase mp[i] by 1)ma := 0, c := 0, ans := 0for each key-value pair i ... Read More