LeetCode #56: Merge Intervals (C/C++).

medium

source: https://leetcode.com/problems/merge-intervals
C/C++ Solution to LeetCode problem 56. Merge Intervals.

Problem

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Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Examples

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Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints

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• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104

Solution

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We need to sort the initial vector, so that every range will have a starti value >= than the previous range.

• Once sorted, we push the first range to the result.
• We iterate through all the next elements.
  • If starti >= starti-1
    • If endi >= endi-1 we have a new end for the last element in the results.
  • If not, we have a new range to add to the results.

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class Solution {
public:
  vector<vector<int>> merge(vector<vector<int>>& intervals) {
    if (intervals.size() == 0) return { {} };
    vector<vector<int>> result;

    sort(intervals.begin(), intervals.end());

    result.push_back(intervals[0]);
    for (int i=1; i<intervals.size(); i+=1) {
      if (intervals[i][0] >= result[result.size()-1][0] &&
          intervals[i][0] <= result[result.size()-1][1]) {
        if(intervals[i][1] >= result[result.size()-1][1])
          result[result.size()-1][1] = intervals[i][1];
      } else {
        result.push_back(intervals[i]);
      }
    }

    return result;
  }
};