Merge Intervals : A unique Graph-based approach

So I was solving this leetcode question titled Merge Intervals and after trying for 30-40 mins I got an intuition that it could be solved by sorting but anyway, I was not able to implement it in one go. I missed some edge cases.

But when I opened the editorial on this problem, I was shocked. Two approaches were discussed on that page, one was the sorting-based approach which solved the problem in O(NlogN) time and O(1) space. Indeed that one was the optimised one.

The other approach got me. It was based on the concept of Connected Components in Graphs and trust me this is the first topic in my current Data Structures course of Sem 4 in my college. I feel terrible that I could not apply what I have learned from that college course😭 .

Anyway here is the implementation of that approach:

class Solution {
public:
    map<vector<int>, vector<vector<int>>> graph;
    map<int, vector<vector<int>>> nodes_in_comp;
    set<vector<int>> visited;

    bool overlap(vector<int>& a, vector<int>& b) {
        return a[0] <= b[1] and b[0] <= a[1];
    }

    // build a graph where an undirected edge between intervals u and v exists
    // iff u and v overlap.
    void buildGraph(vector<vector<int>>& intervals) {
        for (auto interval1 : intervals) {
            for (auto interval2 : intervals) {
                if (overlap(interval1, interval2)) {
                    graph[interval1].push_back(interval2);
                    graph[interval2].push_back(interval1);
                }
            }
        }
    }

    // merges all of the nodes in this connected component into one interval.
    vector<int> mergeNodes(vector<vector<int>>& nodes) {
        int min_start = nodes[0][0];
        for (auto node : nodes) {
            min_start = min(min_start, node[0]);
        }

        int max_end = nodes[0][1];
        for (auto node : nodes) {
            max_end = max(max_end, node[1]);
        }

        return {min_start, max_end};
    }

    // use depth-first search to mark all nodes in the same connected component
    // with the same integer.
    void markComponentDFS(vector<int>& start, int comp_number) {
        stack<vector<int>> stk;
        stk.push(start);

        while (!stk.empty()) {
            vector<int> node = stk.top();
            stk.pop();

            // not found
            if (visited.find(node) == visited.end()) {
                visited.insert(node);

                nodes_in_comp[comp_number].push_back(node);

                for (auto child : graph[node]) {
                    stk.push(child);
                }
            }
        }
    }

    // gets the connected components of the interval overlap graph.
    void buildComponents(vector<vector<int>>& intervals) {
        int comp_number = 0;

        for (auto interval : intervals) {
            if (visited.find(interval) == visited.end()) {
                markComponentDFS(interval, comp_number);
                comp_number++;
            }
        }
    }

    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        buildGraph(intervals);
        buildComponents(intervals);

        // for each component, merge all intervals into one interval.
        vector<vector<int>> merged;
        for (size_t comp = 0; comp < nodes_in_comp.size(); comp++) {
            merged.push_back(mergeNodes(nodes_in_comp[comp]));
        }

        return merged;
    }
};

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Merge Intervals : A unique Graph-based approach

Atharv Mishra

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