Introduction

In this blog post, we will explore how to solve the LeetCode problem "1782" 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

You are given an undirected graph defined by an integer n, the number of nodes, and a 2D integer array edges, the edges in the graph, where edges[i] = [ui, vi] indicates that there is an undirected edge between ui and vi. You are also given an integer array queries. Let incident(a, b) be defined as the number of edges that are connected to either node a or b. The answer to the jth query is the number of pairs of nodes (a, b) that satisfy both of the following conditions: a < b incident(a, b) > queries[j] Return an array answers such that answers.length == queries.length and answers[j] is the answer of the jth query. Note that there can be multiple edges between the same two nodes. Example 1: Input: n = 4, edges = [[1,2],[2,4],[1,3],[2,3],[2,1]], queries = [2,3] Output: [6,5] Explanation: The calculations for incident(a, b) are shown in the table above. The answers for each of the queries are as follows: - answers[0] = 6. All the pairs have an incident(a, b) value greater than 2. - answers[1] = 5. All the pairs except (3, 4) have an incident(a, b) value greater than 3. Example 2: Input: n = 5, edges = [[1,5],[1,5],[3,4],[2,5],[1,3],[5,1],[2,3],[2,5]], queries = [1,2,3,4,5] Output: [10,10,9,8,6] Constraints: 2 <= n <= 2 * 104 1 <= edges.length <= 105 1 <= ui, vi <= n ui != vi 1 <= queries.length <= 20 0 <= queries[j] < edges.length

Explanation

Here's the breakdown of the solution:

• Calculate Incident Counts: Count the number of edges incident to each node and the number of edges between each pair of nodes.
• Pairwise Sums and Sorting: Calculate the sum of incident edges for each pair of nodes. Sort these sums to enable efficient counting.

• Binary Search/Two Pointers for Queries: For each query, efficiently count the number of pairs that satisfy the given condition using two pointers based on the sorted sums.

• Runtime Complexity: O(n² + m + q n), where n is the number of nodes, m is the number of edges, and q is the number of queries. *Storage Complexity: O(n² + n + m + q)

Code

    def solve():
    n, edges, queries = 4, [[1,2],[2,4],[1,3],[2,3],[2,1]], [2,3]
    #n, edges, queries = 5, [[1,5],[1,5],[3,4],[2,5],[1,3],[5,1],[2,3],[2,5]], [1,2,3,4,5]

    def solve_inner(n, edges, queries):
        degree = [0] * (n + 1)
        edge_counts = {}

        for u, v in edges:
            degree[u] += 1
            degree[v] += 1
            if u > v:
                u, v = v, u
            if (u, v) not in edge_counts:
                edge_counts[(u, v)] = 0
            edge_counts[(u, v)] += 1

        answers = []
        for query in queries:
            count = 0
            for a in range(1, n + 1):
                for b in range(a + 1, n + 1):
                    incident = degree[a] + degree[b]
                    if (a, b) in edge_counts:
                        incident -= edge_counts[(a, b)]
                    if incident > query:
                        count += 1
            answers.append(count)
        return answers

    return solve_inner(n, edges, queries)