LeetCode #57: Insert Interval (C/C++).

medium

source: https://leetcode.com/problems/insert-interval/
C/C++ Solution to LeetCode problem 57. Insert Interval.

Problem

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You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

Examples

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Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Constraints

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• 0 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 105
• intervals is sorted by start_i in ascending order.
• newInterval.length == 2
• 0 <= start <= end <= 105

Solution

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• For this problem, we need to know if the newInterval array has been merged.
• If has not been merged, we select what is the next array to merge:
  • min(newInterval[0], intervals[i][0])
• If our result array is empty, we just add the current array to it.
• Else:
  • We determine if the current array fits inside the last one added to the result.
    • If so, we do nothing with it and move to the next one.
    • If not, we determine if the lower index fits inside the last one added.
      • If yes, then we just set the upper index of the last one added to max(lastAdded[1], current[1]).
    • If not, then we push the current array to the result.
  • We finish once we added all the arrays and the new interval.

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

    vector<vector<int>> result;
    bool newMerged = false;
    bool isNewNext = false;

    int i=0;
    while (i < intervals.size() || !newMerged) {
      isNewNext = false;
      vector<int> *next = &intervals[i];
      if ((!newMerged && i == intervals.size()) || (!newMerged && newInterval[0] <= intervals[i][0])) {
        isNewNext = true;
        next = &newInterval;
      }

      if (result.size() == 0)
        result.push_back(*next);
      else if ((*next)[0] >= result[result.size()-1][0] &&
          (*next)[0] <= result[result.size()-1][1])
            result[result.size()-1][1] = max((*next)[1], result[result.size()-1][1]);
      else
        result.push_back(*next);

      if (isNewNext)
        newMerged = true;      
      else
        i++;
    }

    return result;
  }
};