Introduction

In this blog post, we will explore how to solve the LeetCode problem "2508" 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 graph consisting of n nodes numbered from 1 to n. You are given the integer n and a 2D array edges where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi. The graph can be disconnected. You can add at most two additional edges (possibly none) to this graph so that there are no repeated edges and no self-loops. Return true if it is possible to make the degree of each node in the graph even, otherwise return false. The degree of a node is the number of edges connected to it. Example 1: Input: n = 5, edges = [[1,2],[2,3],[3,4],[4,2],[1,4],[2,5]] Output: true Explanation: The above diagram shows a valid way of adding an edge. Every node in the resulting graph is connected to an even number of edges. Example 2: Input: n = 4, edges = [[1,2],[3,4]] Output: true Explanation: The above diagram shows a valid way of adding two edges. Example 3: Input: n = 4, edges = [[1,2],[1,3],[1,4]] Output: false Explanation: It is not possible to obtain a valid graph with adding at most 2 edges. Constraints: 3 <= n <= 105 2 <= edges.length <= 105 edges[i].length == 2 1 <= ai, bi <= n ai != bi There are no repeated edges.

Explanation

Here's the approach:

• Identify Odd Degree Nodes: Find all nodes with an odd degree. The goal is to make all these degrees even.
• Case Analysis: Analyze different cases based on the number of odd-degree nodes. If there are zero odd-degree nodes, the graph is already valid. If there are two, add an edge between them. If there are four, consider different pairings to add two edges. If the number of odd-degree nodes is neither 0, 2, or 4, it's impossible to make all degrees even with at most two edge additions. If the number of odd-degree nodes is greater than 4, it is impossible to make all degrees even with at most two edge additions.

• Connectivity Check: If there are no odd degree nodes after adding at most two edges, the problem is solved.

• Time & Space Complexity: O(n + m), where n is the number of nodes and m is the number of edges. Space complexity is O(n).

Code

    def solve():
    n, edges = map(int, input().split())
    adj = [[] for _ in range(n + 1)]
    for _ in range(edges):
        u, v = map(int, input().split())
        adj[u].append(v)
        adj[v].append(u)

    odd_degree_nodes = []
    for i in range(1, n + 1):
        if len(adj[i]) % 2 != 0:
            odd_degree_nodes.append(i)

    if not odd_degree_nodes:
        print("true")
        return

    if len(odd_degree_nodes) == 2:
        print("true")
        return

    if len(odd_degree_nodes) == 4:
        print("true")
        return

    if len(odd_degree_nodes) == 0:
         print("true")
         return

    print("false")



n, edges_count = map(int, input().split())
edges = []
for _ in range(edges_count):
    edges.append(list(map(int, input().split())))

def is_possible(n, edges):
    adj = [[] for _ in range(n + 1)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    odd_degree_nodes = []
    for i in range(1, n + 1):
        if len(adj[i]) % 2 != 0:
            odd_degree_nodes.append(i)

    if not odd_degree_nodes:
        return True

    if len(odd_degree_nodes) == 2:
        return True

    if len(odd_degree_nodes) == 4:
        return True

    if len(odd_degree_nodes) == 0:
        return True

    return False

print(is_possible(n, edges))