Introduction

In this blog post, we will explore how to solve the LeetCode problem "2322" 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 connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges. You are given a 0-indexed integer array nums of length n where nums[i] represents the value of the ith node. You are also given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. Remove two distinct edges of the tree to form three connected components. For a pair of removed edges, the following steps are defined: Get the XOR of all the values of the nodes for each of the three components respectively. The difference between the largest XOR value and the smallest XOR value is the score of the pair. For example, say the three components have the node values: [4,5,7], [1,9], and [3,3,3]. The three XOR values are 4 ^ 5 ^ 7 = 6, 1 ^ 9 = 8, and 3 ^ 3 ^ 3 = 3. The largest XOR value is 8 and the smallest XOR value is 3. The score is then 8 - 3 = 5. Return the minimum score of any possible pair of edge removals on the given tree. Example 1: Input: nums = [1,5,5,4,11], edges = [[0,1],[1,2],[1,3],[3,4]] Output: 9 Explanation: The diagram above shows a way to make a pair of removals. - The 1st component has nodes [1,3,4] with values [5,4,11]. Its XOR value is 5 ^ 4 ^ 11 = 10. - The 2nd component has node [0] with value [1]. Its XOR value is 1 = 1. - The 3rd component has node [2] with value [5]. Its XOR value is 5 = 5. The score is the difference between the largest and smallest XOR value which is 10 - 1 = 9. It can be shown that no other pair of removals will obtain a smaller score than 9. Example 2: Input: nums = [5,5,2,4,4,2], edges = [[0,1],[1,2],[5,2],[4,3],[1,3]] Output: 0 Explanation: The diagram above shows a way to make a pair of removals. - The 1st component has nodes [3,4] with values [4,4]. Its XOR value is 4 ^ 4 = 0. - The 2nd component has nodes [1,0] with values [5,5]. Its XOR value is 5 ^ 5 = 0. - The 3rd component has nodes [2,5] with values [2,2]. Its XOR value is 2 ^ 2 = 0. The score is the difference between the largest and smallest XOR value which is 0 - 0 = 0. We cannot obtain a smaller score than 0. Constraints: n == nums.length 3 <= n <= 1000 1 <= nums[i] <= 108 edges.length == n - 1 edges[i].length == 2 0 <= ai, bi < n ai != bi edges represents a valid tree.

Explanation

Here's the breakdown of the solution:

• Calculate XOR Subtrees: Perform a Depth-First Search (DFS) to calculate the XOR value of each subtree rooted at each node.
• Iterate and Evaluate Edge Removals: Iterate through all possible pairs of edges to remove. For each pair, determine the three resulting connected components and their corresponding XOR values.

• Minimize Score: Calculate the score (max XOR - min XOR) for each edge removal pair and maintain the minimum score found so far.

• Runtime Complexity: O(n^3), where n is the number of nodes. Storage Complexity: O(n)

Code

    def minimumScore(nums, edges):
    n = len(nums)
    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    total_xor = 0
    for num in nums:
        total_xor ^= num

    xor_values = [0] * n

    def dfs(node, parent):
        xor_values[node] = nums[node]
        for neighbor in adj[node]:
            if neighbor != parent:
                dfs(neighbor, node)
                xor_values[node] ^= xor_values[neighbor]

    dfs(0, -1)

    min_score = float('inf')

    for i in range(n - 1):
        for j in range(i + 1, n - 1):
            u1, v1 = edges[i]
            u2, v2 = edges[j]

            # Find which nodes are connected when removing edge i
            def find_component(start_node, blocked_edge):
                visited = {start_node}
                stack = [start_node]
                component = [start_node]

                while stack:
                    node = stack.pop()
                    for neighbor in adj[node]:
                        if (node == blocked_edge[0] and neighbor == blocked_edge[1]) or \
                           (node == blocked_edge[1] and neighbor == blocked_edge[0]):
                            continue
                        if neighbor not in visited:
                            visited.add(neighbor)
                            stack.append(neighbor)
                            component.append(neighbor)
                return component

            # Remove edge i, form component 1
            component1_nodes = find_component(0, (u1, v1))

            # Remove edge j, form component 2 starting from a node NOT in component1
            start_node2 = -1
            for k in range(n):
                if k not in component1_nodes:
                    start_node2 = k
                    break

            if start_node2 == -1:
                continue  # All nodes are in the same component after removing first edge

            component2_nodes = find_component(start_node2, (u2, v2))

            # The remaining nodes form component 3
            component3_nodes = []
            for k in range(n):
                if k not in component1_nodes and k not in component2_nodes:
                    component3_nodes.append(k)

            # Calculate XOR values of each component
            xor1 = 0
            for node in component1_nodes:
                xor1 ^= nums[node]

            xor2 = 0
            for node in component2_nodes:
                xor2 ^= nums[node]

            xor3 = 0
            for node in component3_nodes:
                xor3 ^= nums[node]

            if not component3_nodes: # if component 3 is empty, calculate it from total_xor - xor1 - xor2
                 xor3 = total_xor ^ xor1 ^ xor2

            xor_values_components = [xor1, xor2, xor3]
            min_xor = min(xor_values_components)
            max_xor = max(xor_values_components)
            score = max_xor - min_xor
            min_score = min(min_score, score)

    return min_score