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Problem solution · C++

Merge2

Merge2: a C++ solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Sorting and greedy selection
Source
Kamyu LeetCode Solutions
Length
29 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sorting and greedy selection

For Merge2, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 29 lines of C++ from the credited upstream file merge2.cpp.
  • The implementation visibly relies on sequence storage.
  • 2 loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMerge2 · C++C++
Use this to learn the idea, then write your own version.
// Time Complexity: O(nlogn)// Space Complexity: O(1) /** * Definition for an interval. * struct Interval { *     int start; *     int end; *     Interval() : start(0), end(0) {} *     Interval(int s, int e) : start(s), end(e) {} * }; */class Solution {    public:        vector<Interval> merge(vector<Interval> &intervals) {            vector<Interval> ans;            sort(intervals.begin(), intervals.end(),                     [](const Interval &v1, const Interval &v2) { return v1.start < v2.start; }                );            const int n = intervals.size();            for(int i = 0; i < n;) {                ans.push_back(intervals[i++]);                while(i < n && intervals[i].start <= ans.back().end)                    ans.back().end = max(ans.back().end, intervals[i++].end);            }            return ans;        }};

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