**Count Contained Intervals - Amazon Top Interview Questions**

### Problem Statement :

You are given a two-dimensional list of integers intervals. Each value is unique and contains an inclusive interval [start, end]. Return the number of intervals are contained by another interval. An interval that's contained by multiple other intervals should only be counted once. An interval [s0, e0] contains another interval [s1, e1] if s0 ≤ s1 and e0 ≥ e1. Constraints 0 ≤ n ≤ 100,000 where n is the length of intervals Example 1 Input intervals = [ [1, 5], [2, 3], [3, 6], [4, 4] ] Output 2 Explanation [2, 3] and [4, 4] are contained by [1, 5]. [4, 4] is also contained by [3, 6] but we don't count it again. Example 2 Input intervals = [ [1, 2], [2, 3], [1, 5] ] Output 2 Explanation Both [1, 2] and [2, 3] are contained by [1, 5]

### Solution :

` ````
Solution in C++ :
int solve(vector<vector<int>>& nums) {
sort(nums.begin(), nums.end(), [](vector<int>& a, vector<int>& b) {
if (a[0] == b[0]) return a[1] > b[1];
return a[0] < b[0];
});
int count = 0, pe = INT_MIN;
for (auto& n : nums) {
if (n[1] <= pe) {
count += 1;
} else {
pe = n[1];
}
}
return count;
}
```

` ````
Solution in Python :
class Solution:
def solve(self, intervals):
events = []
OPEN, CLOSE = 0, 1
for i, (start, end) in enumerate(intervals):
events.append((start, OPEN, end, i))
events.append((end, CLOSE, end, i))
events.sort(key=lambda x: (x[0], -x[2]))
ans = 0
balance = 0
max_end = float("-inf")
for time, event_type, end, idx in events:
if event_type == OPEN and end <= max_end:
ans += 1
if event_type == OPEN:
max_end = max(max_end, end)
return ans
```

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