Introduction

In this blog post, we will explore how to solve the LeetCode problem "1203" 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 are n items each belonging to zero or one of m groups where group[i] is the group that the i-th item belongs to and it's equal to -1 if the i-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it. Return a sorted list of the items such that: The items that belong to the same group are next to each other in the sorted list. There are some relations between these items where beforeItems[i] is a list containing all the items that should come before the i-th item in the sorted array (to the left of the i-th item). Return any solution if there is more than one solution and return an empty list if there is no solution. Example 1: Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]] Output: [6,3,4,1,5,2,0,7] Example 2: Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]] Output: [] Explanation: This is the same as example 1 except that 4 needs to be before 6 in the sorted list. Constraints: 1 <= m <= n <= 3 * 104 group.length == beforeItems.length == n -1 <= group[i] <= m - 1 0 <= beforeItems[i].length <= n - 1 0 <= beforeItems[i][j] <= n - 1 i != beforeItems[i][j] beforeItems[i] does not contain duplicates elements.

Explanation

Here's a breakdown of the approach and the Python code solution:

• High-Level Approach:

  • Topological Sort: The problem fundamentally involves topological sorting. We need to sort items while respecting dependencies (beforeItems) and group affiliations.
  • Two-Level Sort: Perform topological sort at two levels: First, sort the groups themselves based on inter-group dependencies. Second, sort the items within each group.
  • Dependency Graph: Build dependency graphs for both inter-group and intra-group dependencies. Use these graphs to detect cycles, which would indicate no solution.

• Complexity:

  • Runtime Complexity: O(n + m + E), where n is the number of items, m is the number of groups, and E is the total number of dependencies in beforeItems.
  • Storage Complexity: O(n + m), primarily due to storing the dependency graphs, group assignments, and topological sort results.

Code

    from collections import defaultdict, deque

def sortItems(n: int, m: int, group: list[int], beforeItems: list[list[int]]) -> list[int]:
    """
    Sorts items based on group affiliation and dependencies.

    Args:
        n: The number of items.
        m: The number of groups.
        group: A list where group[i] is the group the i-th item belongs to (-1 if no group).
        beforeItems: A list of lists where beforeItems[i] are items that must come before item i.

    Returns:
        A sorted list of items, or an empty list if no solution exists.
    """

    # Assign unique group IDs to items in no group
    for i in range(n):
        if group[i] == -1:
            group[i] = m
            m += 1

    # 1. Group-level topological sort

    group_graph = defaultdict(list)
    group_indegree = [0] * m

    item_graph = defaultdict(list)
    item_indegree = [0] * n

    for i in range(n):
        for pre in beforeItems[i]:
            if group[i] != group[pre]:
                if group[pre] not in group_graph[group[i]]:
                    group_graph[group[pre]].append(group[i])
                    group_indegree[group[i]] += 1
            else:
                item_graph[pre].append(i)
                item_indegree[i] += 1

    def topological_sort(graph, indegree, nodes):
        queue = deque([node for node in nodes if indegree[node] == 0])
        result = []
        while queue:
            node = queue.popleft()
            result.append(node)
            for neighbor in graph[node]:
                indegree[neighbor] -= 1
                if indegree[neighbor] == 0:
                    queue.append(neighbor)

        if len(result) != len(nodes):
            return []
        return result


    sorted_groups = topological_sort(group_graph, group_indegree, list(range(m)))
    if not sorted_groups:
        return []

    # 2. Item-level topological sort within each group

    group_items = defaultdict(list)
    for i in range(n):
        group_items[group[i]].append(i)

    result = []
    for group_id in sorted_groups:
        items_in_group = group_items[group_id]

        item_graph_ingroup = defaultdict(list)
        item_indegree_ingroup = {item: 0 for item in items_in_group}

        for u in items_in_group:
          for v in beforeItems[u]:
            if v in items_in_group:
              item_graph_ingroup[v].append(u)
              item_indegree_ingroup[u]+=1

        sorted_items = topological_sort(item_graph_ingroup, item_indegree_ingroup, items_in_group)
        if not sorted_items:
            return []
        result.extend(sorted_items)

    return result