Introduction

In this blog post, we will explore how to solve the LeetCode problem "2846" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There is an undirected tree with n nodes labeled from 0 to n - 1. You are given the integer n and a 2D integer array edges of length n - 1, where edges[i] = [ui, vi, wi] indicates that there is an edge between nodes ui and vi with weight wi in the tree. You are also given a 2D integer array queries of length m, where queries[i] = [ai, bi]. For each query, find the minimum number of operations required to make the weight of every edge on the path from ai to bi equal. In one operation, you can choose any edge of the tree and change its weight to any value. Note that: Queries are independent of each other, meaning that the tree returns to its initial state on each new query. The path from ai to bi is a sequence of distinct nodes starting with node ai and ending with node bi such that every two adjacent nodes in the sequence share an edge in the tree. Return an array answer of length m where answer[i] is the answer to the ith query. Example 1: Input: n = 7, edges = [[0,1,1],[1,2,1],[2,3,1],[3,4,2],[4,5,2],[5,6,2]], queries = [[0,3],[3,6],[2,6],[0,6]] Output: [0,0,1,3] Explanation: In the first query, all the edges in the path from 0 to 3 have a weight of 1. Hence, the answer is 0. In the second query, all the edges in the path from 3 to 6 have a weight of 2. Hence, the answer is 0. In the third query, we change the weight of edge [2,3] to 2. After this operation, all the edges in the path from 2 to 6 have a weight of 2. Hence, the answer is 1. In the fourth query, we change the weights of edges [0,1], [1,2] and [2,3] to 2. After these operations, all the edges in the path from 0 to 6 have a weight of 2. Hence, the answer is 3. For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from ai to bi. Example 2: Input: n = 8, edges = [[1,2,6],[1,3,4],[2,4,6],[2,5,3],[3,6,6],[3,0,8],[7,0,2]], queries = [[4,6],[0,4],[6,5],[7,4]] Output: [1,2,2,3] Explanation: In the first query, we change the weight of edge [1,3] to 6. After this operation, all the edges in the path from 4 to 6 have a weight of 6. Hence, the answer is 1. In the second query, we change the weight of edges [0,3] and [3,1] to 6. After these operations, all the edges in the path from 0 to 4 have a weight of 6. Hence, the answer is 2. In the third query, we change the weight of edges [1,3] and [5,2] to 6. After these operations, all the edges in the path from 6 to 5 have a weight of 6. Hence, the answer is 2. In the fourth query, we change the weights of edges [0,7], [0,3] and [1,3] to 6. After these operations, all the edges in the path from 7 to 4 have a weight of 6. Hence, the answer is 3. For each queries[i], it can be shown that answer[i] is the minimum number of operations needed to equalize all the edge weights in the path from ai to bi. Constraints: 1 <= n <= 104 edges.length == n - 1 edges[i].length == 3 0 <= ui, vi < n 1 <= wi <= 26 The input is generated such that edges represents a valid tree. 1 <= queries.length == m <= 2 * 104 queries[i].length == 2 0 <= ai, bi < n

Explanation

Here's the breakdown of the solution:

• Build Adjacency List and Perform DFS: Represent the tree as an adjacency list where each node maps to its neighbors and the edge weights. Use Depth-First Search (DFS) to find the path between the two nodes given in each query.
• Count Edge Weight Frequencies: Once the path is found, iterate through the edges of the path and count the occurrences of each edge weight.

• Calculate Minimum Operations: The minimum number of operations needed is equal to the total number of edges in the path minus the frequency of the most frequent edge weight.

• Runtime Complexity: O(m * n), where 'm' is the number of queries and 'n' is the number of nodes. This is because, in the worst case, DFS might visit all nodes for each query.

• Storage Complexity: O(n) to store the adjacency list and path.

Code

    from collections import defaultdict

def min_operations_to_equalize_path(n: int, edges: list[list[int]], queries: list[list[int]]) -> list[int]:
    """
    Calculates the minimum number of operations required to make the weight of every edge on the path from ai to bi equal.

    Args:
        n: The number of nodes in the tree.
        edges: A list of edges, where each edge is a list [ui, vi, wi].
        queries: A list of queries, where each query is a list [ai, bi].

    Returns:
        A list of answers, where each answer is the minimum number of operations for the corresponding query.
    """

    adj = defaultdict(list)
    for u, v, w in edges:
        adj[u].append((v, w))
        adj[v].append((u, w))

    def find_path(start: int, end: int) -> list[int]:
        """
        Finds the path between two nodes using Depth-First Search (DFS).

        Args:
            start: The starting node.
            end: The ending node.

        Returns:
            A list of edges representing the path, or None if no path is found.
        """

        path = []
        visited = set()

        def dfs(node: int, current_path: list[tuple[int, int]]) -> bool:
            nonlocal path
            visited.add(node)

            if node == end:
                path = current_path
                return True

            for neighbor, weight in adj[node]:
                if neighbor not in visited:
                    if dfs(neighbor, current_path + [(node, neighbor, weight)]):
                        return True
            return False

        dfs(start, [])
        return path

    answers = []
    for a, b in queries:
        path = find_path(a, b)

        if not path:
            answers.append(0)  # No path found (shouldn't happen in a tree)
            continue

        weights = [w for _, _, w in path]
        weight_counts = defaultdict(int)
        for weight in weights:
            weight_counts[weight] += 1

        max_frequency = 0
        for weight in weight_counts:
            max_frequency = max(max_frequency, weight_counts[weight])

        answers.append(len(path) - max_frequency)

    return answers