# Leetcode 2655 - Find Maximal Uncover Ranges [Medium] (Premium)
## Description
You are given an integer `n` which is the length of a 0-indexed array `nums`, and a 0-indexed 2D-array `ranges`, which is a list of sub-ranges of `nums` (sub-ranges may overlap).

Each row `ranges[i]` has exactly 2 cells:

* `ranges[i][0]`, which shows the start of the ith range (inclusive)
* `ranges[i][1]`, which shows the end of the ith range (inclusive)

These ranges cover some cells of `nums` and leave some cells uncovered. Your task is to find all of the uncovered ranges with maximal length.

Return a 2D-array `answer` of the uncovered `ranges`, sorted by the starting point in ascending order.

By all of the uncovered ranges with maximal length, we mean satisfying two conditions:

Each uncovered cell should belong to exactly one sub-range
There should not exist two ranges (l1, r1) and (l2, r2) such that r1 + 1 = l2

### Example 1:

Input: n = 10, ranges = [[3,5],[7,8]]
Output: [[0,2],[6,6],[9,9]]
Explanation: The ranges (3, 5) and (7, 8) are covered, so if we simplify the array nums to a binary array where 0 shows an uncovered cell and 1 shows a covered cell, the array becomes [0,0,0,1,1,1,0,1,1,0] in which we can observe that the ranges (0, 2), (6, 6) and (9, 9) aren't covered.

### Example 2:

Input: n = 3, ranges = [[0,2]]
Output: []
Explanation: In this example, the whole of the array nums is covered and there are no uncovered cells so the output is an empty array.

### Example 3:

Input: n = 7, ranges = [[2,4],[0,3]]
Output: [[5,6]]
Explanation: The ranges (0, 3) and (2, 4) are covered, so if we simplify the array nums to a binary array where 0 shows an uncovered cell and 1 shows a covered cell, the array becomes [1,1,1,1,1,0,0] in which we can observe that the range (5, 6) is uncovered.

### Constraints:

* `1 <= n <= 109`
* `0 <= ranges.length <= 106`
* `ranges[i].length = 2`
* `0 <= ranges[i][j] <= n - 1`
* `ranges[i][0] <= ranges[i][1]`

## Solution
* Sort the `ranges`
* init a pointer to retrieve the `0` ~ `n`
* When the pointer is smaller than the range's left, it means there is an uncover range `p` ~ `range's left - 1`
* When the pointer is smaller than the range's right, it means the number here is covered, and should update `p` to `range's right + 1` to find the next uncover range.
* At the end, should check the pointer is reach the end `n`. if not, add the uncover range `p` ~ `n-1`
### C++

vector<vector<int>> findMaximalUncoveredRanges(int n, vector<vector<int>>& ranges) {
    sort(ranges.begin(), ranges.end());
    
    vector<vector<int>> ans;
    int p = 0;
    
    for(int i=0; i<ranges.size(); i++){
        if(p < ranges[i][0]) ans.push_back({p, ranges[i][0]-1});
        if(p <= ranges[i][1]) p = ranges[i][1]+1;
    }
    
    if(p!=n) ans.push_back({p, n-1});
    return ans;
}