LeetCode 2508: Add Edges to Make Degrees of All Nodes Even Solution

Master LeetCode problem 2508 (Add Edges to Make Degrees of All Nodes Even), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2508. Add Edges to Make Degrees of All Nodes Even

Hard

Hash Table Graph

Problem Explanation

Explanation

To solve this problem, we need to check if it is possible to add at most two additional edges to the given graph such that each node in the graph has an even degree. If a node has an odd degree, we need to add edges such that it becomes even.

We can determine if it is possible by checking the total number of nodes with odd degrees in the graph. If there are 0 or 2 nodes with odd degrees, then it is possible to add edges to make all nodes have even degrees.

We can count the degrees of each node and store them in a map. Then, we iterate through the map to count the number of nodes with odd degrees. If there are more than 2 nodes with odd degrees, it is not possible to make all nodes have even degrees.

The time complexity of this approach is O(n) where n is the number of edges since we need to iterate through all edges to count the degrees of nodes. The space complexity is also O(n) to store the degrees of each node.

Solution Code

class Solution {
    public boolean canMakeAllNodesEven(int n, int[][] edges) {
        Map<Integer, Integer> degrees = new HashMap<>();
        
        for (int i = 1; i <= n; i++) {
            degrees.put(i, 0);
        }
        
        for (int[] edge : edges) {
            degrees.put(edge[0], degrees.get(edge[0]) + 1);
            degrees.put(edge[1], degrees.get(edge[1]) + 1);
        }
        
        int oddDegreeCount = 0;
        for (int degree : degrees.values()) {
            if (degree % 2 != 0) {
                oddDegreeCount++;
            }
        }
        
        return oddDegreeCount == 0 || oddDegreeCount == 2;
    }
}

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