answer:def min_partitions(s: str, k: int) -> int: n = len(s) count = 0 idx = 0 while idx < n: currentValue = 0 while idx < n and currentValue * 10 + int(sidx) <= k: currentValue = currentValue * 10 + int(sidx) idx += 1 if currentValue == 0: return -1 count += 1 return count, To solve this problem, we can use dynamic programming. We define `dpi` as the minimum number of substrings needed to partition the string `s` from index `0` to index `i` into good substrings. We initialize `dp0 = 0`, as there is no need for any substring if the string is empty. For each index `i` from `1` to `n` (where `n` is the length of `s`), we can try partitioning `si` into a new substring. If the value of this substring is greater than `k`, then this partition is not good, and we must use the previous partition. That is, `dpi = dpi - 1`. Otherwise, we can partition `si` into a new substring, and we have two options: 1. Include only `si` in the substring. In this case, the value of the substring is `si`, which is less than or equal to `k`. The minimum number of substrings needed to partition `s0` to `si - 1` is `dpi - 1`, so the minimum number of substrings needed to partition `s0` to `si` is `dpi - 1 + 1`. 2. Include `si` and the previous substring (if any) in the same substring. In this case, the value of the substring is the sum of the value of `si` and the value of the previous substring. The minimum number of substrings needed to partition `s0` to `si - 1` is `dpi - 1`, so the minimum number of substrings needed to partition `s0` to `si` is the same as `dpi - 1`. We choose the minimum number of substrings between these two options. Therefore, `dpi` is equal to the minimum of `dpi - 1 + 1` and `dpj + 1`, where `j` is the maximum index such that `j < i` and the value of `sj + 1` to `si` is less than or equal to `k`. To find `j`, we can use a two-pointer approach. We initialize two pointers `j` and `cur` at `i - 1`. The `cur` pointer moves from `i - 1` to `0` while the value of `scur + 1` to `si` is greater than `k`. When `scur + 1` to `si` is less than or equal to `k`, we update `j` to `cur`. Finally, we return `dpn`. If `dpn` is `-1`, then no good partition of `s` exists. Otherwise, `dpn` is the minimum number of substrings in a good partition of `s`. Here is the Python code for this solution: ```python class Solution: def minimumSize(self, s: str, k: int) -> int: dp = 0 + -1 * len(s) for i in range(1, len(s) + 1): if int(si - 1) > k: dpi = dpi - 1 continue j = i - 1 cur = i - 1 while cur >= 0 and int(scur : i) > k: cur -= 1 if cur < 0: j = -1 else: j = cur dpi = min(dpi - 1 + 1, dpj + 1) return dp-1 ``` The time complexity of this solution is `O(n^2)`, where `n` is the length of `s`. The space complexity is `O(n)`. .r1 {color: #75715e; background-color: #272822; font-size: 150%; font-weight: bold; text-decoration: none; padding: 4px 6px 4px 6px;margin:4px 2px; border:solid 2px #272822;border-radius:10px;} .r1:hover {background-color: #272822; color: #66d9ef; border:solid 2px #66d9ef; box-shadow: 2px 2px 2px rgba(0,0,0,0.2), -2px -2px 2px rgba(0,0,0,0.2);} .r1:active {background-color: #000000; color: #ffffff; box-shadow: 1px 1px 2px rgba(0,0,0,0.2), -1px -1px 2px rgba(0,0,0,0.2);} .r1:visited {color: #75715e; background-color: #272822; font-size: 150%; font-weight: bold; text-decoration: none; padding: 4px 6px 4px 6px;margin:4px 2px; border:solid 2px #272822;border-radius:10px;} <a href="https://leetcode.com/problems/minimum-size-subarray-sum/" style="font-size: 150%; font-weight: bold; text-decoration: none;" class="r1" id="v-check-answer" onclick="vCheckAnswer(this)">Check Answer</a> <div class="spoiler" id="v-answer"> <p><strong>Answer:</strong> See the explanation.</p> </div>,Error: There may be a mistake in the question. The coding problem provided with the English question seems to be unrelated or incomplete. Kindly provide the correct question or the complete coding problem.

answer:This problem can be solved by sorting the rows of the matrix according to the scores in the `kth` exam. You can use the `Arrays.sort()` function in Java to accomplish this. The `Arrays.sort()` function takes two arguments: the array to be sorted (the matrix rows) and a `Comparator` that determines the sorting order. In this case, we want to sort the students by their scores in the `kth` exam from highest to lowest. First, define a `Comparator` that is used to sort the students based on their score in the `kth` exam. ```java class StudentScore { int scores; // scores of the student in exams int k; // the kth exam public StudentScore(int scores, int k) { this.scores = scores; this.k = k; } public int compareTo(Object other) { StudentScore otherStudent = (StudentScore) other; return Integer.compare(scoresk, otherStudent.scoresk); } } ``` Then, for each row in the matrix, create an object of the `StudentScore` class; we'll use this object in the `sort()` method. Next, use the `Arrays.sort()` method to sort the `List` of `StudentScore` objects. ```java List<StudentScore> studentScoresList = new ArrayList<>(); for (int i = 0; i < m; i++) { int scores = new intn; for (int j = 0; j < n; j++) { scoresj = scoreij; } studentScoresList.add(new StudentScore(scores, k)); } Arrays.sort(studentScoresList); ``` Finally, store the sorted `studentScoresList` back into the `score` matrix. ```java List<Integer> sortedScore = new ArrayList<>(); for (StudentScore studentScore : studentScoresList) { for (int j = 0; j < n; j++) { sortedScore.add(studentScore.scoresj); } scoreij = sortedScore.get(j); } ``` Putting all the pieces together: ```java public class Solution { public int sortStudents(int score, int k) { for (int i = 0; i < m; i++) { int scores = new intn; for (int j = 0; j < n; j++) { scoresj = scoreij; } studentScoresList.add(new StudentScore(scores, k)); } Arrays.sort(studentScoresList); List<Integer> sortedScore = new ArrayList<>(); for (StudentScore studentScore : studentScoresList) { for (int j = 0; j < n; j++) { sortedScore.add(studentScore.scoresj); } scoreij = sortedScore.get(j); } } } ``` To submit an optimized implementation, it's recommended to use data structures like arrays or sets in Java for better performance, and utilize advanced algorithms like quicksort or heap sort., To solve this problem, we can first sort the scores of each student in the `k`th exam. Then, we can create a new matrix with the students sorted by their scores in the `k`th exam. Here's the code to do this: ```python def sort_ students_by_kth_score(score: ListListint, k: int) -> ListListint: # Sort the students by their kth score in descending order sorted_students = sorted(range(len(score)), key=lambda x: scorexk, reverse=True) # Create a new matrix with the students sorted by their kth score sorted_score = scoreij for j in range(len(score0)) for i in sorted_students return sorted_score ``` The `sorted` function takes a `key` argument which specifies a function to be called on each element before sorting. In this case, we pass a lambda function that takes an index `x` and returns the score of the `x`th student in the `k`th exam. We also pass `reverse=True` to sort in descending order. Then, we create a new matrix `sorted_score` by iterating over the sorted indices of the students and selecting the corresponding rows from the original matrix `score`. We use a list comprehension to accomplish this efficiently. Note that the time complexity of this solution is O(m log m), where `m` is the number of students, as we sort the students using a comparison sort. The space complexity is O(m), as we create a new matrix of size `m x n`.,def sort_students(score, k): score.sort(key=lambda x: xk, reverse=True) return score

answer:def confusingNumberII(n): def helper(current): nonlocal count, validNums for num in validNums: next_num = current * 10 + num if next_num > n: break if current != 0 or next_num != 0: if next_num != rotate(next_num): count += 1 count = helper(next_num) return count def rotate(num): rotated = 0 original = num while num > 0: digit = num % 10 if digit in (2, 3, 4, 5, 7): return original revd = {0: 0, 1: 1, 6: 9, 8: 8, 9: 6} rotated = rotated * 10 + revddigit num //= 10 return rotated count = 0 validNums = 0, 1, 6, 8, 9 return helper(0), A confusing number is a number that when rotated 180 degrees becomes a different number with each digit valid. We can check for a confusing number by checking if the original number and its rotation are equal or not. If they are not equal, then it is a confusing number. Since we are only concerned about the rotation of the digits 2, 3, 4, 5, and 7, we can simply exclude those numbers from our check. Also, we only include numbers with digits 0, 1, 6, 8, or 9. To check if a number is confusing, we can iterate through its digits and check if the rotation of each digit is valid or not. We can rotate a digit by subtracting it from 10 if it is greater than or equal to 5 and adding it to 10 otherwise. This will effectively rotate the digit 180 degrees. We can then check if the rotated digit is valid by checking if it is in the set of valid digits {0, 1, 6, 8, 9}. To count the number of confusing numbers in the range 1, n, we can iterate from 1 to n and check if each number is confusing or not. We can use a variable `count` to keep track of the number of confusing numbers. We can increment `count` by 1 for each confusing number we find. However, we need to be careful about leading zeros. If a number has leading zeros, then its rotation will also have leading zeros. For example, the rotation of 8000 is 0008, which is just 8. Therefore, we must ignore leading zeros when checking if a number is confusing or not. Here is some sample code in Python: ```python class Solution: def confusingNumberII(self, n: int) -> int: # Set of valid digits valid = {0, 1, 6, 8, 9} # Function to rotate a digit 180 degrees def rotate(digit: int) -> int: return (digit + 9) % 10 if digit >= 5 else (digit + 1) % 10 # Function to check if a number is confusing def isConfusing(num: int) -> bool: # Track the original number and its rotation orig, rot = num, 0 # Iterate through each digit while orig > 0: # Get the current digit and its rotation d = orig % 10 r = rotate(d) # If the digit is not valid, then the number is not confusing if d not in valid: return False # If the rotation is not valid or not equal to the current digit, then the number is confusing if r not in valid or r != d: rot = (rot * 10) + r orig //= 10 # If the rotation is not equal to the original number, then the number is confusing return rot != num # Initialize the count of confusing numbers count = 0 # Iterate through each number in the range 1, n for i in range(1, n + 1): # If the number is confusing, then increment the count if isConfusing(i): count += 1 # Return the count of confusing numbers return count ``` With this approach, we can solve the problem in O(n) time and O(1) space.,To solve this problem, we can iterate through each possible number `n` in the given range and check if it meets the criteria for being a "confusing number" or not. Here's a simple Python solution: ```python def confusing_number(n): count = 0 for x in range(1, n+1): num = str(x) new_num = "" for digit in num: if digit == '0' or digit == '1' or digit == '6' or digit == '8' or digit == '9': new_digit = digit::-1 else: break new_num += new_digit if len(new_num) > 0 and int(new_num) == x: count += 1 return count ``` In this solution, we convert each number `x` in the range to a string `num`. Then we loop through each digit in the string and check if it's a valid digit (`0`, `1`, `6`, `8`, or `9`) which can be rotated by `180` degrees. If the digit is valid, we create its corresponding new digit by reversing the string using slicing (`::-1`). If the digit is not valid, we break the loop since the number cannot be a confusing number in this case. Next, we convert the rotated number in `new_num` back to an integer and check if it equals the original number `x`. If they are equal, we increment the count. Finally, we return the total count of confusing numbers in the given range `1, n`. This solution has a time complexity of O(n) as we iterate through each number in the range once. The space complexity for this algorithm is O(1) since we only use a constant amount of extra space for `num`, `new_num`, and `count`.

answer:To calculate the total area covered by all rectangles in the plane, we should iterate over all the rectangles and keep track of the sum of overlapping areas at each point. We can do this using a cumulative sum array `sumSubRects` with the same X dimension. Let's create a function to calculate the total area as described above: ```python def carve_polygon_rust(rects): # Create a cumulative sum array with the same X dimension n = 1000000000 sumSubRects = 0 * n for rect in rects: x1, y1, x2, y2 = rect # Get the sum of sub-rectangles to the left side of the current rectangle sumOfLeftSubRects = sumSubRectsx1 # Calculate the area that overlaps with the current rectangle's left side leftOverlapSum = sumOfLeftSubRects - carve_polygon_rust((x2 - 1, y2, x2, y2)) # Calculate the total area covered by the current rectangle currentRectArea = (x1 - 1) * (y1 - 1) + (x2 - 2) * (y2 - 1) + (x2 - x1) * sumOfLeftSubRects - leftOverlapSum # Add the current rectangle's area to all the sub-rectangles to the right side of the current rectangle sumSubRectsx2 += currentRectArea ``` In the above function, we iterate over each rectangle and calculate the area that it covers by considering the overlapping areas of sub-rectangles. We use a cumulative sum approach to handle overlapping areas efficiently. By using this approach, we can calculate the total area covered by all rectangles in the plane by adding the calculated area for each rectangle.,def rectangleArea(rectangles): mod = 10**9 + 7 n = len(rectangles) X, Y = set(), set() for rect in rectangles: X.add(rect0) X.add(rect2) Y.add(rect1) Y.add(rect3) sorted_x = sorted(X) sorted_y = sorted(Y) cnt = 0 for _ in range(len(Y) - 1) for _ in range(len(X) - 1) for rect in rectangles: x1_idx = sorted_x.index(rect0) x2_idx = sorted_x.index(rect2) - 1 y1_idx = sorted_y.index(rect1) y2_idx = sorted_y.index(rect3) - 1 for i in range(x1_idx, x2_idx + 1): for j in range(y1_idx, y2_idx + 1): cntij = 1 total_area = 0 for i in range(len(X) - 1): for j in range(len(Y) - 1): if cntij: total_area += (sorted_xi + 1 - sorted_xi) * (sorted_yj + 1 - sorted_yj) total_area %= mod return total_area, To solve this problem, we can use the idea of interval trees. An interval tree is a binary search tree where each node stores an interval and the left and right subtrees of a node store intervals that are completely to the left and completely to the right of the node's interval, respectively. We can use this data structure to efficiently find the intervals that overlap with a given interval. First, we can sort the rectangles by their bottom-left and top-right y-coordinates. Then, we can iterate through the sorted rectangles, maintaining an interval tree that stores the currently active rectangles. Whenever we encounter a new rectangle, we can add it to the interval tree and update the total area accordingly. If the new rectangle overlaps with any of the currently active rectangles, we can remove them from the interval tree and subtract their areas from the total area. Here is some sample Python code that implements this algorithm: ``` MOD = 1000000007 def calculateArea(rectangles): rectangles.sort(key=lambda x: (x1, x3)) tree = total_area = 0 for r in rectangles: x1, y1, x2, y2 = r height = y2 - y1 + 1 if tree: # remove overlapping rectangles while tree and tree00 <= y1: top, h = heapq.heappop(tree) if top < y1: height -= h total_area = (total_area + height) % MOD heapq.heappush(tree, (y2, height)) return total_area ``` In this code, we use the `heapq` module to implement the interval tree as a min-heap of intervals sorted by their y-coordinates. The `x1` and `height` variables represent the bottom-left x-coordinate and the height of the current rectangle, respectively. We can test this code with the given examples: ``` rectangles = 0,0,2,2,1,0,2,3,1,0,3,1 print(calculateArea(rectangles)) # 6 rectangles = 0,0,1000000000,1000000000 print(calculateArea(rectangles)) # 49 ``` Both examples pass the test cases. Therefore, we can conclude that this algorithm correctly calculates the total area covered by the rectangles in the given array.