answer:To eliminate every single occurrence of white spaces present within a string, you can use the replace method like so: ```pythondef remove_spaces(string): return string.replace(' ', '')```This will replace each space character in the string with an empty string, effectively removing it. Here, string.replace is a Python built-in method that replaces all occurrences of a specified character. However, you have mentioned not to use trimming or built-in function which makes this method non-applicable.Without using trim or any other built-in functions, we would need an approach that manually checks and manipulates each character in the string. Here's such a method:```pythondef manual_remove_spaces(string): result = [] for character in string: if character != ' ': result.append(character) return ''.join(result)```The function manual_remove_spaces loops through each character in the string. If the character is not a space, it adds it to the results list. Finally, it joins the characters in the result list together into a single string, with no spaces between. However, to replace multiple occurrences of white spaces with a single white space, we need to slightly modify the function:```pythondef replace_spaces(string): result = [] previous_space = False # Flag to check if previous character checked was a "space"; for character in string: if character != ' ': result.append(character) previous_space = False elif character == ' ' and not previous_space: # If current character is "space" AND the previous character was not a "space". result.append(character) previous_space = True return ''.join(result)```The additional logic in the function replace_spaces checks if the previous character was a space. If it was, it skips any additional spaces it comes across, until it finds a non-space character. The previous_space variable is used to track if the last character was a space.Note that the efficiency of these methods is O(n), where n is the length of the string. This is because both require a loop that goes through each character in the string once. They are therefore quite efficient and suitable for handling strings of reasonable length.

answer:Here is an example of how you can achieve this in Python:```pythondef count_word_frequency(text): words = text.split(" ") frequency_dict = {} for word in words: if word not in frequency_dict: frequency_dict[word] = 1 else: frequency_dict[word] += 1 return frequency_dictuser_input_text = "Hello world!"print(count_word_frequency(user_input_text))```This program will output:```python{'Hello': 1, 'world!': 1}```The `count_word_frequency()` function splits the input text into words using the `split(" ")` string method and then loops over each word. If it's the first time it sees a word, it adds the word to the dictionary with a value of 1. If it sees a word it has seen before, it just increments the count of the word in the dictionary.The `print(count_word_frequency(user_input_text))` will take the user text as input and print the resulting dictionary.

answer:To provide a solution to the mentioned problem, I would require specific data along with the criteria to identify anomalies. However, I can give you a generalized solution using the openpyxl and pandas package to read data from the Excel file and identify common anomalies like missing values, outliers, etc.This is a simple example of how this could be done:```python#importing required modulesimport pandas as pdfrom openpyxl import load_workbook# load workbookwb = load_workbook(filename='filename.xlsx')# define which sheet you want to work onsheet_ranges = wb['NameOfYourSheet']# read the data from excel using pandasdata = pd.read_excel('filename.xlsx')# identify missing values in datamissing_values = data.isnull().sum() # identify outliers in data (using Z-score method here for example, other methods like IQR can also be used)# calcualte meanmean = data.mean() # calculate standard deviationstd_dev = data.std() # calculate z scoresz_scores = (data - mean) / std_dev # identify rows with z score greater than 3 (commonly used threshold for identifying outliers)outliers = data[(z_scores > 3).any(axis=1)]```In the above code, replace 'filename.xlsx' with the name of your Excel file and 'NameOfYourSheet' with the sheet name of your excel work book. This code will first load the excel and identify missing values. Then it will calculate the Z-scores for each data point. A Z-score indicates how many standard deviations an element is from the mean, and can be used to identify outliers in the data. In this case, those rows where any column has a Z-score greater than 3, which is the commonly used threshold for identifying outliers, are considered as anomalies. The subset of data containing these outliers is stored in 'outliers'. Please tailor this script to your specific needs as identifying anomalies can be a complex task depending on the nature of the data and the specific definition of what an anomaly is in the context of that data, which is not mentioned here.

answer:Here is the Swift translation of your Java code:```swiftimport Foundationclass Solution { func isBored(_ s: String?) -> Int { //Ensure string is not empty guard let s = s, !s.isEmpty else { return 0 } //Split the string using punctuation marks let sentences = s.components(separatedBy: CharacterSet(charactersIn: ".!?")) var boredomCount = 0 for sentence in sentences { // Remove leading and trailing spaces let trimmedSentence = sentence.trimmingCharacters(in: .whitespaces) if !trimmedSentence.isEmpty, trimmedSentence.first == "I" { // Increment boredom count boredomCount += 1 } } // Return count of boredom sentences return boredomCount }}```This Swift code does the exact same thing as your original Java code. It splits the text on each occurrence of ".", "!", or "?" to separate sentences, then trims off any white-space characters from the beginning and end of each sentence, and checks if the first character is "I". If it is, it increments the boredom count by one. It finally returns this count.

answer:Below is a Python solution for the given problem.```pythonfactor = 12_sum = 26x, y = 0, 0for i in range (1, _sum): j = _sum - i if i * j == factor: x, y = i, j breakprint(f"The two numbers that meet both conditions are {x} and {y}")```This program scans over a range of possible pairs of numbers (i, j), checking if the product of i and j equals the given factor while the sum equals the given sum. If it finds such a pair, it records them in x and y, then immediately ends the loop. It then prints out the two numbers.

answer:This solution involves building a responsive and accessible webpage that satisfies the outlined requirements using HTML, CSS and JavaScript.Due to the complexity of the problem, a simplified educational example is given below. However, for a fully functional, professional-grade webpage, more sophisticated libraries and techniques would need to be employed by an experienced programmer.Please replace YOUR_API_KEY with your own Google Maps API key.```html<!DOCTYPE html><html><head> <title>Interactive Map</title> <style> #map { height: 400px; width: 100%; } </style></head><body> <div id="info">Click the map to get the location's latitude and longitude.</div> <div id="map" tabindex="0" role="application" aria-label="interactive map"></div> <script> let map, infoWindow, geocoder; function initMap() { map = new google.maps.Map(document.getElementById('map'), { center: {lat: -34.397, lng: 150.644}, zoom: 6 }); infoWindow = new google.maps.InfoWindow; geocoder = new google.maps.Geocoder; if (navigator.geolocation) { navigator.geolocation.getCurrentPosition(function(position) { let pos = { lat: position.coords.latitude, lng: position.coords.longitude }; infoWindow.setPosition(pos); infoWindow.setContent('Location found.'); infoWindow.open(map); map.setCenter(pos); }, function() { handleLocationError(true, infoWindow, map.getCenter()); }); } else { handleLocationError(false, infoWindow, map.getCenter()); } map.addListener('click', function(e) { geocoder.geocode({'location': e.latLng}, function(results, status) { if (status === 'OK') { if (results[0]) { let pos = { lat: e.latLng.lat(), lng: e.latLng.lng() }; infoWindow.setPosition(pos); infoWindow.setContent(results[0].formatted_address); infoWindow.open(map); } else { window.alert('No results found'); } } else { window.alert('Geocoder failed due to: ' + status); } }); }); } function handleLocationError(browserHasGeolocation, infoWindow, pos) { infoWindow.setPosition(pos); infoWindow.setContent(browserHasGeolocation ? 'Error: The Geolocation service failed.' : 'Error: Your browser doesn't support geolocation.'); infoWindow.open(map); } </script> <script async defer src="https://maps.googleapis.com/maps/api/js?key=YOUR_API_KEY&callback=initMap"> </script></body></html>```The code creates an interactive webpage that displays a map powered by Google Maps. The map responds to browser window size changes due to the width: 100% in the CSS rules. Furthermore, if the browser supports Geolocation, the map will center on the user's current location when loaded. Otherwise, it will show an error message. With Google Maps Geocoding, users can click anywhere on the map to get the latitude and longitude coordinates of the clicked location, these coordinates are then displayed in the infoWindow. This code runs equally well on both desktop and mobile platforms and it's made accessible using ARIA attributes on the div containing the map.