Suppose we have a list of intervals. In this list interval[i] has [start, end] values. We have to find the number of intervals are contained by another interval. If there is an interval that is contained by multiple other intervals that should only be counted once. An interval [s0, e0] is inside another interval [s1, e1] when s0 ≤ s1 and e0 ≥ e1.
So, if the input is like intervals = [[2, 6],[3, 4],[4, 7],[5, 5]], then the output will be 2, because [3, 4] and [5, 5] are inside [2, 6] and [4, 7] respectively.
To solve this, we will follow these steps −
- if intervals list is empty, then
- return 0
- sort the intervals list based on start time, when start times are same, sort in decreasing order of end time
- end_mx := -infinity
- ans := 0
- for each (start, end) pair in intervals, do
- if end <= end_mx, then
- ans := ans + 1
- end_mx := maximum of end_mx and end
- if end <= end_mx, then
- return ans
Example
Let us see the following implementation to get better understanding −
def solve(intervals): if not intervals: return 0 intervals.sort(key=lambda x: (x[0], -x[1])) end_mx = float("-inf") ans = 0 for start, end in intervals: if end <= end_mx: ans += 1 end_mx = max(end_mx, end) return ans intervals = [[2, 6],[3, 4],[4, 7],[5, 5]] print(solve(intervals))
Input
[[2, 6],[3, 4],[4, 7],[5, 5]]
Output
2