LeetCode 2508: Add Edges to Make Degrees of All Nodes Even
LeetCode 2508 Solution 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.
LeetCode 2508 Solutions in Java, C++, Python
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;
}
}Interactive Code Editor for LeetCode 2508
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