1288. Remove Covered Intervals

Problem statement

Given an array intervals where intervals[i] = [li, ri] represent the interval [li, ri), remove all intervals that are covered by another interval in the list.

The interval [a, b) is covered by the interval [c, d) if and only if c <= a and b <= d.

Return the number of remaining intervals.

Example 1

Input: intervals = [[1,4],[3,6],[2,8]]
Output: 2
Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.

Example 2

Input: intervals = [[1,4],[2,3]]
Output: 1

Constraints

• 1 <= intervals.length <= 1000.

• intervals[i].length == 2.

• 0 <= li <= ri <= 10^5.

• All the given intervals are unique.

Solution 1: Bruteforce

For each interval i, find if any other interval j such that j covers i or i covers j then remove the smaller one from intervals.

Example 1

For intervals = [[1,4],[3,6],[2,8]].

• With interval i = [1,4], there is no other interval j such that covers i or j covers i.

• With interval i = [3,6], there is interval j = [2,8] convering i. Remove [3,6] from intervals.

Final intervals = [[1,4],[2,8]].

Code

#include <vector>
#include <iostream>
using namespace std;
inline bool isCovered(vector<int>& i, vector<int>& j) {
    return j[0] <= i[0] && i[1] <= j[1];
}
int removeCoveredIntervals(vector<vector<int>>& intervals) {
    int i = 0;
    while (i < intervals.size() - 1) {
        int j = i + 1;
        bool erase_i = false;
        while (j < intervals.size()) {
            if (isCovered(intervals[i], intervals[j])) {
                intervals.erase(intervals.begin() + i);
                erase_i = true;
                break;
            } else if (isCovered(intervals[j], intervals[i])) {
                intervals.erase(intervals.begin() + j);
            } else {
                j++;
            }
        }
        if (!erase_i) {
            i++;
        }
    }
    return intervals.size();
}
int main() {
    vector<vector<int>> intervals{{1,4},{3,6},{2,8}};
    cout << removeCoveredIntervals(intervals) << endl;
    intervals = {{1,4},{2,3}};
    cout << removeCoveredIntervals(intervals) << endl;
}

Output:
2
1

Complexity

• Runtime: O(N^3), where N = intervals.length.

• Extra space: O(1).

Solution 2: Using dictionary order

You might know how to look up words in a dictionary.

The word apple appears before candy in the dictionary because the starting letter a of apple appears before c of candy in the English alphabet.

And apple appears after animal since the next letter p appears after n.

The C++ Standard Library uses that dictionary order to compare two std::vectors.

Example 1

Rewriting intervals = [[1,4],[3,6],[2,8]] in dictionary order you get intervals = [[1,4],[2,8],[3,6]]. In this order, the left bounds of the intervals are sorted first.

If intervals is sorted like that, you can avoid bruteforce in Solution 1 by a simpler algorithm.

Check if each interval i covers or is covered by some of the previous ones.

Remember that the left bound of interval i is always bigger than or equal to all left bounds of the previous ones. So,

1. i is covered by some previous interval if the right bound of i is less than some of the right bounds before.

2. Otherwise i can only cover its exact previous one that has the same left bound.

Code

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int removeCoveredIntervals(vector<vector<int>>& intervals) {
    sort(intervals.begin(), intervals.end());
    int count = 0;
    int maxRight = -1;
    int preLeft = -1;
    for (auto& i : intervals) {
        if (i[1] <= maxRight) { // i is covered by some previous interval
            count++;
        } else if (i[0] == preLeft) { // i covers its exact previous one
            count++;
        } else {
            preLeft = i[0];
        }
        maxRight = max(maxRight, i[1]);
    }
    return intervals.size() - count;
}
int main() {
    vector<vector<int>> intervals{{1,4},{3,6},{2,8}};
    cout << removeCoveredIntervals(intervals) << endl;
    intervals = {{1,4},{2,3}};
    cout << removeCoveredIntervals(intervals) << endl;
}

Output:
2
1

Complexity

• Runtime: O(NlogN), where N = intervals.length.

• Extra space: O(1).

Key takeaway

• Two std::vectors can be compared using the dictionary order.

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