answer:```jsfunction getRandomNumber(min, max) { return Math.floor(Math.random() * (max - min + 1) + min);}function hasDuplicateNameKey(array, newObj) { for (let i = 0; i < array.length; i++) { if (array[i].name === newObj.name) { return true; } } return false;}function createArray(size) { const array = []; for (let i = 0; i < size; i++) { let newObj; do { newObj = { name: `name{getRandomNumber(1, 100)}`, value: getRandomNumber(1, 100) }; } while (hasDuplicateNameKey(array, newObj)); array.push(newObj); } return array;}const myArray = createArray(10);console.log(myArray);```Here is a JavaScript code snippet that does the following:1. Generates a random number between two specified numbers.2. Checks if an array has a duplicate name key.3. Creates an array of objects with unique name keys and unique values.The function `createArray(size)` accepts a size parameter which determines the number of objects in the resulting array. You can change the argument passed to the `createArray` function to create an array of a different size. Currently, the argument 10 has been passed to create an array of 10 objects.

answer:Here is a Python class named `StringModifier` which has a `prepend` method to add given characters to the start of the string until it reaches a specified minimum length. It also has a `remove_added` method to remove the added characters.```pythonclass StringModifier: def __init__(self, string: str): self.string = string self.original = string def prepend(self, element, min_len): if not isinstance(element, str) or not isinstance(min_len, int): raise TypeError("element must be str, min_len must be int") if min_len <= len(self.string): raise ValueError("min_len should be more than length of the string") self.string = (element * (min_len - len(self.string))) + self.string return self.string def remove_added(self): self.string = self.original return self.string```You can instantiate this class and call its methods as follows:```pythonmod = StringModifier("example")print(mod.prepend('-', 12)) # prints: ----exampleprint(mod.remove_added()) # prints: example```In the above code, error handling has been incorporated using Python's `raise` statement. If `element` is not a string or `min_len` is not an integer, a `TypeError` is raised. If `min_len` is less than or equal to the length of the initial string, a `ValueError` is raised.

answer:This problem is a typical dynamic programming problem. We can use a 2D DP matrix to keep track of the minimum painting cost at each point.The main idea is that for each house, we want to find the minimum cost of painting that house, considering that it may not have the same color as its previous house. We do that by looking at the minimum cost of painting the previous house with a different color and adding it to the cost of painting the current house.Here is a Python solution to the problem:```pythondef minCostII(costs): if not costs: return 0 n, k = len(costs), len(costs[0]) dp = [[0] * k for _ in range(n)] dp[0] = costs[0] for i in range(1, n): min1 = min(dp[i-1]) idx = dp[i-1].index(min1) min2 = min(dp[i-1][:idx]+dp[i-1][idx+1:]) for j in range(k): if j == idx: dp[i][j] = min2 + costs[i][j] else: dp[i][j] = min1 + costs[i][j] return min(dp[-1])```This function builds up the `dp` matrix by iterating over each row (house) and each column (color) in the `costs` matrix. It keeps track of the minimum cost for the previous houses and the second minimum cost in case the color is the same as the minimum.Eventually, it returns the minimum cost of the last row of the `dp` matrix, which gives the minimum cost of painting all the houses with the condition that no two adjacent houses have the same color.

answer:Just as the previous problem did, this one involves lengths and triangles. In this case, you are asked to find the number of quadrilaterals with certain specific properties. But unlike last time, you are now dealing with a quadrilateral and a line, not an equilateral triangle. To solve the problem, we can make use of the fact that the quadrilateral is inscribed in a circle. This allows us to use some of the properties of the circles, tangents and the bisectors. From property of circle, we know that the opposite angles of quadrilateral are supplemental i.e., ∠B + ∠D = 180. Now, ∠BCD = ∠BAD = ∠A and ∠BDC = ∠BAC = ∠C. Due to angular bisector, ∠BCD = ∠CDP i.e., ∠A = ∠CDP.The triangles ADC and PDC are similar by AA rule of similarity (ANGLE-ANGLE rule). Therefore, ratios of their corresponding sides are equal. We can write AD/DP = CD/CDSince AB = p and CD = r, AD = k*p and DP = k*r for some k belongs to set of positive real numbers.Since AD <= DA (given), we get that 0 < k <= 1Finally, AD + DP = AE = k*p + k*r = k*(p+r) Since AE has integral length, k*(p+r) should be integral. This means p+r should be divide AE which has integral length. To find the number of quadrilaterals, we can sum over all valid perimeters, and for each perimeter we find all valid lengths of DE by dividing the perimeter by the number of sides and rounding down. This is done by taking perimeter // 4, where // is the integer division operator in Python. We then subtract 1 because the length of DE cannot equal the perimeter. We add one to the cumulative count for each valid length of DE. We can write a python code based on the above reasoning as follows:```pythondef quadrilaterals(): limit = 100000 count = 0 for perimeter in range(4, limit + 1): count += perimeter // 4 - 1 return countprint(quadrilaterals())```Please note, we have to consider all the possible perimeter values from 4 to 100,000 to get the count of quadrilaterals. For each perimeter, we are calculating all the valid lengths of DE by performing integer division of the perimeter by 4 and then subtracting 1. The reason we subtract 1 is because the length of DE cannot be the same as the perimeter. This calculation is done inside a loop that iterates over all possible perimeter values, so we can simply add the result of each iteration to the count to get the total number of quadrilaterals. Also, perimeter of quadrilateral starts from 4 because quadrilateral is a polygon with 4 sides and minimum possible length of each side can be 1 since side lengths are integers as per the given question. So, minimum perimeter of quadrilateral will be 1+1+1+1 = 4. That's why range for perimeter is from 4 to 100000 in the code snippet.

answer:The code provided seems to be already optimized with loop usage to calculate the sum of the numbers from 0 till 9. As repetition is already handled by the for loop, there's no need for further optimization. In JavaScript, a for loop is one of the fastest ways to perform iteration so without additional context, this example couldn't be optimized further.

answer:["VIPeR", "DukeMTMC-reID"]