answer:Sorting by multiple columns does not drastically affect time complexity compared to sorting by one column. The time complexity of sorting an array can be generally considered O(n log n) via efficient algorithms like quicksort or mergesort. When you sort by multiple columns, you are essentially just adding a constant time operation for each addition in column to compare, but the time complexity remains O(n log n).Regarding the user experience, not all websites or applications feature multi-column sorting because the interface for such a feature may not be intuitive, or the feature may not be frequently requested or used by users. While time complexity is a factor to consider when deciding to implement a feature, there are usually other usability and demand considerations as well.

answer:To display by combining two SVGs together, you can simply append the new SVG code after the first one. However, remember to understand and maintain the structure and properties of the original SVG.For example, let's combine the SVG provided above with a SVG for a circle:```HTML<svg height="100" width="100"> <circle cx="50" cy="50" r="40" stroke="black" stroke-width="3" fill="red" /></svg>```Combined SVG would be:```HTML<svg width="32" height="32" viewBox="0 0 32 32" fill="none" xmlns="http://www.w3.org/2000/svg"> <path d="M15 11C15 10.4477 15.4477 10 16 10H19.4995C21.4982 10 23.0469 11.4456 23.0469 13.4972C23.0469 14.4867 22.6808 15.373 22.0193 16.0101C21.3612 16.6439 20.4671 16.9777 19.4995 16.9777H17.125C17.056 16.9777 17 17.0337 17 17.1027V21C17 21.5523 16.5523 22 16 22C15.4477 22 15 21.5523 15 21V11ZM17.125 14.9777H19.4995C20.005 14.9777 20.3846 14.8078 20.6319 14.5696C20.8757 14.3347 21.0469 13.9808 21.0469 13.4972C21.0469 12.6027 20.4469 12 19.4995 12H17.125C17.056 12 17 12.056 17 12.125V14.8527C17 14.9217 17.056 14.9777 17.125 14.9777Z" fill="#212121"/> <path d="M6 10.0781C6.55228 10.0781 7 10.5258 7 11.0781V17.9531C7 19.0958 7.91105 20 9.00781 20C10.1046 20 11.0156 19.0958 11.0156 17.9531V11.0781C11.0156 10.5258 11.4633 10.0781 12.0156 10.0781C12.5679 10.0781 13.0156 10.5258 13.0156 11.0781V17.9531C13.0156 20.1759 11.2334 22 9.00781 22C6.78223 22 5 20.1759 5 17.9531V11.0781C5 10.5258 5.44772 10.0781 6 10.0781Z" fill="#212121"/> <path d="M26.0117 10C26.564 10 27.0117 10.4477 27.0117 11V17.9062C27.0117 18.4585 26.564 18.9062 26.0117 18.9062C25.4594 18.9062 25.0117 18.4585 25.0117 17.9062V11C25.0117 10.4477 25.4594 10 26.0117 10Z" fill="#212121"/> <path d="M27.1367 20.6016C27.1367 21.2229 26.633 21.7266 26.0117 21.7266C25.3904 21.7266 24.8867 21.2229 24.8867 20.6016C24.8867 19.9802 25.3904 19.4766 26.0117 19.4766C26.633 19.4766 27.1367 19.9802 27.1367 20.6016Z" fill="#212121"/> <path d="M6 1C3.23858 1 1 3.23858 1 6V26C1 28.7614 3.23858 31 6 31H26C28.7614 31 31 28.7614 31 26V6C31 3.23858 28.7614 1 26 1H6ZM3 6C3 4.34315 4.34315 3 6 3H26C27.6569 3 29 4.34315 29 6V26C29 27.6569 27.6569 29 26 29H6C4.34315 29 3 27.6569 3 26V6Z" fill="#212121"/></svg><svg height="100" width="100"> <circle cx="50" cy="50" r="40" stroke="black" stroke-width="3" fill="red" /></svg>```Keep in mind:1. SVGs will appear in the order you write them.2. Each SVG will remain an independent entity. They will not interact with each other unless scripting (like JavaScript) or interactivity (like CSS) is added to it.3. If you want to superimpose one SVG over another, use the same viewBox for both and ensure that one of them has a transparent background.

answer:The solution for this problem would involve either creating a new SQL table or adding additional columns to an existing one. In either case, you would need to store data related to individual game sessions for each user.Here's a simple example of how such SQL Schema might look like:Table - `game_sessions`:| Column Name | Type || --------------- | ------ || session_id | INT || user_id | INT || game_id | INT || session_start | DATETIME || session_end | DATETIME |This table stores each game session with an ID, the user who played it, the game played, and the start and end times for the session.Once you have stored the session start and end times, you can calculate session duration easily like this:``` SELECT user_id, AVG(TIME_TO_SEC(TIMEDIFF(session_end, session_start))) as average_session_durationFROM game_sessionsGROUP BY user_id;``` This SQL query calculates the average session duration in seconds for each user. It calculates the duration of a session using `TIMEDIFF` function, converts it into seconds and find out the average using `AVG` for each user.Note: The `TIME_TO_SEC` and `TIMEDIFF` functions are available in MySQL.Pls, note - SQL capabilities can vary significantly between different platforms, so make sure these functions or equivalent are available in your specific SQL variant. Similarly, the schema design might need to be adjusted to fit your specific needs, but this example should provide a good starting point.

answer:The given code is almost correct, except for 2 points:1. The for loop needs to run from 10 to 101 (not 100), because if you want to include 100, Python's `range()` function operates from the first number (included) till the second number (excluded).2. It's not a good practice to use Python reserved keywords as variable names. `sum` is a built-in Python function, so we should use another name for our variable.Here is the correct version:```pythontotal = 0for i in range(10, 101): #changed this from 100 to 101 if i % 4 == 0: total += i #changed this from sum to total else: passprint(total)```

answer:This problem can be solved using a hash map to keep a record of the mappings from the characters in string `s` to the characters in string `t`, and another hash map for the reverse mapping. We iterate through the characters of the strings and whenever a new character is found, we set a new mapping in both the hash maps. For each character, we check if the mapping in either hash map contradicts the current character pair. If it does, we return false immediately, otherwise we continue to the next character.Here is one possible Python solution:```pythondef isIsomorphic(s: str, t: str) -> bool: s_to_t = {} t_to_s = {} for c_s, c_t in zip(s, t): if (c_s in s_to_t and s_to_t[c_s] != c_t) or (c_t in t_to_s and t_to_s[c_t] != c_s): return False s_to_t[c_s] = c_t t_to_s[c_t] = c_s return True```This solution runs in O(n) time, where n is the length of string `s` (which is also the length of string `t`), as we just go through each character in the string once. The space complexity is also O(n) as we store two hash maps of the same size as the number of unique characters in the input strings. Test this function with the provided examples:```pythonprint(isIsomorphic("egg", "add")) # Output: Trueprint(isIsomorphic("foo", "bar")) # Output: Falseprint(isIsomorphic("paper", "title")) # Output: Trueprint(isIsomorphic("hello", "world")) # Output: False```

answer:This problem can be solved using 3 nested loops to check for every possible combination of 3 points to form a triangle. The area of triangle with vertices at (x1, y1), (x2, y2) and (x3, y3) can be computed as: |Area| = 1/2 * |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|The algorithm will go as follows:Algorithm:1) Define a function to calculate the area of a triangle given three points.2) Initialize maximumArea as 0 and save the points of the maximum triangle as maxTriangle.3) Use three for loops to check all combinations of three points. * For the current set of three points, calculate the area of the triangle formed by these points using the function in step 1. * If the area is greater than maximumArea, update maximumArea to the current area and save these three points in maxTriangle.4) Check and return maxArea and maxTriangle.Details:1) Initialize maximum_area = 0.2) Initialize max_triangle = None.3) For i in range of 0 to length of points, * For j in range of i+1 to length of points, * For k in range of j+1 to length of points, * Compute the area of the triangle formed by points[i], points[j] and points[k]. * If area > maximum_area, then maximum_area = area and max_triangle = (points[i], points[j], points[k])4) Return max_triangle which is a tuple of three points forming a triangle with maximum area.