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Greedy Algorithm - Question 1

 Leetcode 435 Non-overlapping Intervals

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Constraints:

1 <= intervals.length <= 10^5

intervals[i].length == 2

-5 * 10^4 <= starti < endi <= 5 * 10^4


Analysis:

This question is equivalent to a classic question: interval scheduling, which searches for the maximum number of un-overlapping intervals. Once this number is found, then it is straightforward to calculate the minimum number of intervals to be removed.

When all the intervals are sorted with the ending time, then we can always pick up the first interval as the first step. Why does this always give the optimal result?

Because this interval ends the earliest!

Just because it ends the earliest, choosing it cannot be worse than choosing other intervals. Thus we can greedily choose the first one.

After the first one is chosen, we can again pick up the first interval with a starting time after the ending time of the first interval picked, The argument of the correctness is similar to the first one.

Therefore, for each step, we can always greedily pick up the first valid interval (no overlapping with the previous interval). (Just because it ends the earliest for all the rest intervals).


See the code below:


class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        int n = intervals.size(), ct = 0, front = -1e5;
        sort(intervals.begin(), intervals.end(), [](auto &a, auto &b) {
            // if(a[1] == b[1]) return a[0] < b[0];
            return a[1] < b[1];
        });
        for(int i=0; i<n; ++i) {
            if(intervals[i][0] < front) continue;
            ++ct;
            front = intervals[i][1];
        }
        return n - ct;
    }
};


Upper Layer

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