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question:# Heap Sort Enhancement and Application You are part of a team developing educational tools for algorithm visualization. Your task is to create a new feature that extends the functionality of the provided Heap Sort implementations, and then build an advanced usage case that simulates real-world scenarios. # Part 1: Multi-Purpose Median Finder Build a function that utilizes both Max and Min Heap Sort to find the median of an unsorted list of integers, `arr`. If the list length is even, the median is the average of the two middle numbers after sorting. **Function Signature** ```python def find_median(arr: List[int]) -> float: pass ``` **Input** - `arr`: List of integers. **Output** - Returns a floating point number representing the median of the array. **Constraints** - The list can contain up to (10^6) numbers. - The numbers in the list follow standard integer constraints. Example ```python >>> find_median([7, 1, 3, 5]) 4.0 >>> find_median([1, 2, 3, 4, 5]) 3.0 >>> find_median([5]) 5.0 ``` # Part 2: Educational Visualization Mode Enhance both the Max Heap Sort and Min Heap Sort functions with a `step_by_step` parameter. When `step_by_step` is set to `True`, each internal step of the heapification process must be printed, showing the state of the array after each significant change. **Function Signatures** ```python def max_heap_sort(arr: List[int], step_by_step: bool = False) -> List[int]: pass def min_heap_sort(arr: List[int], step_by_step: bool = False) -> List[int]: pass ``` **Input/Output Specifications** - **Input**: - `arr`: List of integers to be sorted. - `step_by_step`: Boolean flag. Default value is False. - **Output**: - Sorted List of integers in ascending order for both functions. Example ```python >>> max_heap_sort([3, 5, 1, 2, 4], step_by_step=True) iteration 0 : 3 5 1 2 4 iteration 1 : 3 5 1 2 4 iteration 2 : 5 3 1 2 4 iteration 3 : 5 3 1 2 4 iteration 4 : 5 3 4 2 1 iteration 5 : 4 3 1 2 5 iteration 6 : 4 3 1 2 5 # (continues printing until the process is complete) # Final sorted array [1, 2, 3, 4, 5] ``` Apply the enhanced functions to demonstrate step-by-step visualization.

answer:from typing import List def find_median(arr: List[int]) -> float: if not arr: raise ValueError("The array must not be empty") arr.sort() n = len(arr) if n % 2 == 1: return float(arr[n // 2]) else: return (arr[(n // 2) - 1] + arr[n // 2]) / 2.0 def max_heapify(arr, n, i, step_by_step): largest = i l = 2 * i + 1 r = 2 * i + 2 if l < n and arr[l] > arr[largest]: largest = l if r < n and arr[r] > arr[largest]: largest = r if largest != i: arr[i], arr[largest] = arr[largest], arr[i] if step_by_step: print(f'Heapify: {arr}') max_heapify(arr, n, largest, step_by_step) def max_heap_sort(arr: List[int], step_by_step: bool = False) -> List[int]: n = len(arr) for i in range(n // 2 - 1, -1, -1): max_heapify(arr, n, i, step_by_step) if step_by_step: print(f'Build Max Heap: {arr}') for i in range(n - 1, 0, -1): arr[i], arr[0] = arr[0], arr[i] if step_by_step: print(f'Swap {arr[0]} with {arr[i]}: {arr}') max_heapify(arr, i, 0, step_by_step) return arr def min_heapify(arr, n, i, step_by_step): smallest = i l = 2 * i + 1 r = 2 * i + 2 if l < n and arr[l] < arr[smallest]: smallest = l if r < n and arr[r] < arr[smallest]: smallest = r if smallest != i: arr[i], arr[smallest] = arr[smallest], arr[i] if step_by_step: print(f'Heapify: {arr}') min_heapify(arr, n, smallest, step_by_step) def min_heap_sort(arr: List[int], step_by_step: bool = False) -> List[int]: n = len(arr) for i in range(n // 2 - 1, -1, -1): min_heapify(arr, n, i, step_by_step) if step_by_step: print(f'Build Min Heap: {arr}') for i in range(n - 1, 0, -1): arr[i], arr[0] = arr[0], arr[i] if step_by_step: print(f'Swap {arr[0]} with {arr[i]}: {arr}') min_heapify(arr, i, 0, step_by_step) return arr[::-1]

question:# Combination Sum with Constraints Description Given an integer array `nums` containing all positive numbers with no duplicates, write a function that finds the number of possible combinations that add up to a given positive integer target. Different sequences of numbers leading to the same sum are counted as different combinations. Implementation Implement the following two functions: 1. `combination_sum_topdown(nums, target)` - Input: - `nums`: List of unique positive integers. - `target`: A positive integer target sum. - Output: The number of possible combinations as an integer. - Approach: Use a top-down dynamic programming approach with memoization. 2. `combination_sum_bottom_up(nums, target)` - Input: - `nums`: List of unique positive integers. - `target`: A positive integer target sum. - Output: The number of possible combinations as an integer. - Approach: Use a bottom-up dynamic programming approach. Examples - Example 1: - Input: `nums = [1, 2, 3]`, `target = 4` - Output: `7` Explanation: ``` The possible combination ways are: (1, 1, 1, 1) (1, 1, 2) (1, 2, 1) (1, 3) (2, 1, 1) (2, 2) (3, 1) Therefore, there are 7 ways to achieve the target 4. ``` - Example 2: - Input: `nums = [4]`, `target = 4` - Output: `1` Explanation: ``` The only possible combination way is: (4) Therefore, there is 1 way to achieve the target 4. ``` Follow-up What if negative numbers are allowed in the given array `nums`? How does it impact the problem, and what constraints should be added?

answer:def combination_sum_topdown(nums, target): Top-down dynamic programming approach with memoization. memo = {} def dp(rem): if rem == 0: return 1 if rem < 0: return 0 if rem in memo: return memo[rem] count = 0 for num in nums: count += dp(rem - num) memo[rem] = count return count return dp(target) def combination_sum_bottom_up(nums, target): Bottom-up dynamic programming approach. dp = [0] * (target + 1) dp[0] = 1 for i in range(1, target + 1): for num in nums: if i >= num: dp[i] += dp[i - num] return dp[target]

question:<|Analysis Begin|> # Core Identification * **Algorithm/Data Structure**: Union-Find (or Disjoint Set) * **Type**: Data Structure * **Main Purpose**: Efficiently manage and merge dynamic sets of elements. # Complexity * **Time Complexity**: * `root` operation: O(α(n)), where α is the inverse Ackermann function. * `unite` operation: O(α(n)), where α is the inverse Ackermann function. * `add` operation: O(1) * **Space Complexity**: O(n), where `n` is the number of elements. # Principles * **Core Operational Steps**: * **Find** (or `root`): Determine the representative or "root" of the set an element belongs to. * **Union** (or `unite`): Merge two sets. * **Path Compression**: Flatten the structure whenever `root` is called. * **Union by Size**: Attach the smaller tree to the root of the larger tree during the union. # Characteristics & Applications * **Properties**: * Elements within the same set share the same root. * The `root` operation uses path compression for efficiency. * The `unite` operation uses union by size to balance trees and keep them shallow. * **Common Use Cases**: * Network connectivity. * Image processing (finding connected components). * Kruskal’s Minimum Spanning Tree algorithm. * Detecting cycles in a graph. * **Strengths**: * Very efficient for union and find operations. * Supports dynamic connectivity queries. * **Limitations**: * Poor efficiency for non-tree-based applications. * Requires initialization overhead for large sets. # Implementation Challenges * **Edge Cases**: * **Reuniting elements**: Ensuring that attempting to union the same elements repeatedly doesn’t break the structure. * **Large Inputs**: Handling very large number of elements efficiently. * **Performance Bottlenecks**: * **Deep Recursion**: In paths where many nodes point directly or indirectly to a distant root without path compression. * **Error Scenarios**: * **Invalid Elements**: Trying to `root` or `unite` an element that hasn't been added. * **Optimization Points**: * **Path Compression**: Ensure it effectively reduces heights of trees. * **Union by Size**: Keeps trees balanced by always attaching the smaller tree under the root of the larger tree. <|Analysis End|> <|Question Begin|> # Problem Statement You are given a 2D grid map of `m x n` containing integers where each cell represents water (0) or land (1). Initially, the grid is all water. We operate on this grid by adding one land cell at a time and counting the number of islands after each operation. An island is defined as a group of adjacent lands connected vertically or horizontally. Write a function `num_islands(positions: List[Tuple[int, int]], m: int, n: int) -> List[int]` that takes a list of positions where land will be added and returns a list where the i-th element is the number of islands after we add the i-th land position. # Input and Output Format * **Input**: * `positions`: A list of tuples where each tuple has two integers representing the row and column of the cell in the grid to convert into land. * `m` and `n`: Two integers representing the number of rows and columns of the grid respectively. * **Output**: * A list of integers where the i-th integer represents the number of islands after addition of the i-th position. # Constraints * 1 <= m, n <= 10^4 * 1 <= positions.length <= 10^4 * positions(i) will have distinct values. # Example ```python num_islands([(0, 0), (0, 1), (1, 2), (2, 1)], 3, 3) # Output: [1, 1, 2, 3] ``` # Explanation 1. The grid starts as water: ``` 0 0 0 0 0 0 0 0 0 ``` 2. Adding land at (0, 0): ``` 1 0 0 0 0 0 Number of islands = 1 0 0 0 ``` 3. Adding land at (0, 1): ``` 1 1 0 0 0 0 Number of islands = 1 0 0 0 ``` 4. Adding land at (1, 2): ``` 1 1 0 0 0 1 Number of islands = 2 0 0 0 ``` 5. Adding land at (2, 1): ``` 1 1 0 0 0 1 Number of islands = 3 0 1 0 ``` # Notes * Be sure to handle any edge cases, such as lands added that are already connected to existing islands. * Optimize your solution for efficiency.

answer:from typing import List, Tuple class UnionFind: def __init__(self, size): self.parent = list(range(size)) self.rank = [0] * size self.count = 0 def find(self, p): if self.parent[p] != p: self.parent[p] = self.find(self.parent[p]) return self.parent[p] def union(self, p, q): rootP = self.find(p) rootQ = self.find(q) if rootP != rootQ: if self.rank[rootP] > self.rank[rootQ]: self.parent[rootQ] = rootP elif self.rank[rootP] < self.rank[rootQ]: self.parent[rootP] = rootQ else: self.parent[rootQ] = rootP self.rank[rootP] += 1 self.count -= 1 # Count decreases by 1 each time we union two disjoint sets def set_count(self, count): self.count = count def get_count(self): return self.count def num_islands(positions: List[Tuple[int, int]], m: int, n: int) -> List[int]: def index(x, y): return x * n + y uf = UnionFind(m * n) added = set() results = [] directions = [(0, 1), (1, 0), (0, -1), (-1, 0)] for (x, y) in positions: if (x, y) not in added: added.add((x, y)) uf.set_count(uf.get_count() + 1) for dx, dy in directions: nx, ny = x + dx, y + dy if 0 <= nx < m and 0 <= ny < n and (nx, ny) in added: uf.union(index(x, y), index(nx, ny)) results.append(uf.get_count()) return results

question:# Right Triangle Side Calculation You are tasked with implementing a function that calculates the unknown side of a right-angled triangle using the Pythagorean theorem. The triangle's sides are referred to as `opposite`, `adjacent`, and `hypotenuse`. # Function Definition You should write a function `calculate_third_side(opposite, adjacent, hypotenuse)` that returns the length of the unknown side. # Input Format Your function signature should be: ```python def calculate_third_side(opposite: Union[float, str], adjacent: Union[float, str], hypotenuse: Union[float, str]) -> str: ``` - `opposite`, `adjacent`, and `hypotenuse` should be either a floating point number indicating the length or the string `"?"` indicating the unknown side. # Output Format The function should return a string formatted as follows based on the side being calculated: - `"Opposite = <calculated_length>"` if the `opposite` side is unknown. - `"Adjacent = <calculated_length>"` if the `adjacent` side is unknown. - `"Hypotenuse = <calculated_length>"` if the `hypotenuse` side is unknown. # Constraints - You may assume that exactly one side will be `"?"` at any given call. - All given side lengths will be positive numbers. - If the provided inputs do not form a valid right-angled triangle configuration, you should raise a `ValueError` with an appropriate message. # Example Usage ```python # Example 1 print(calculate_third_side(3, 4, "?")) # Should output: Hypotenuse = 5.0 # Example 2 print(calculate_third_side(3, "?", 5)) # Should output: Adjacent = 4.0 # Example 3 print(calculate_third_side("?", 4, 5)) # Should output: Opposite = 3.0 ``` Note: Implementers should ensure to handle potential edge cases correctly, such as invalid side lengths and maintaining numerical precision based on floating-point operations.

answer:import math from typing import Union def calculate_third_side(opposite: Union[float, str], adjacent: Union[float, str], hypotenuse: Union[float, str]) -> str: Calculate the unknown side of a right-angled triangle using the Pythagorean theorem. At least one side must be provided as a string "?" which indicates the unknown side. if hypotenuse == "?": opposite = float(opposite) adjacent = float(adjacent) if opposite <= 0 or adjacent <= 0: raise ValueError("Both opposite and adjacent sides must be positive numbers.") hypotenuse_squared = opposite ** 2 + adjacent ** 2 return f"Hypotenuse = {math.sqrt(hypotenuse_squared)}" elif opposite == "?": hypotenuse = float(hypotenuse) adjacent = float(adjacent) if hypotenuse <= 0 or adjacent <= 0 or hypotenuse <= adjacent: raise ValueError("Invalid side lengths: hypotenuse must be greater than adjacent and both must be positive numbers.") opposite_squared = hypotenuse ** 2 - adjacent ** 2 if opposite_squared < 0: raise ValueError("Invalid right triangle configuration") return f"Opposite = {math.sqrt(opposite_squared)}" elif adjacent == "?": hypotenuse = float(hypotenuse) opposite = float(opposite) if hypotenuse <= 0 or opposite <= 0 or hypotenuse <= opposite: raise ValueError("Invalid side lengths: hypotenuse must be greater than opposite and both must be positive numbers.") adjacent_squared = hypotenuse ** 2 - opposite ** 2 if adjacent_squared < 0: raise ValueError("Invalid right triangle configuration") return f"Adjacent = {math.sqrt(adjacent_squared)}" else: raise ValueError("Exactly one side must be unknown (provided as '?').")