# Set Intersection Size At Least Two Problem

## Description

LeetCode Problem 757.

You are given a 2D integer array intervals where intervals[i] = [start_i, end_i] represents all the integers from starti to endi inclusively.

A containing set is an array nums where each interval from intervals has at least two integers in nums.

For example, if intervals = [[1,3], [3,7], [8,9]], then [1,2,4,7,8,9] and [2,3,4,8,9] are containing sets.

Return the minimum possible size of a containing set.

Example 1:

``````1
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4
Input: intervals = [[1,3],[3,7],[8,9]]
Output: 5
Explanation: let nums = [2, 3, 4, 8, 9].
It can be shown that there cannot be any containing array of size 4.
``````

Example 2:

``````1
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4
Input: intervals = [[1,3],[1,4],[2,5],[3,5]]
Output: 3
Explanation: let nums = [2, 3, 4].
It can be shown that there cannot be any containing array of size 2.
``````

Example 3:

``````1
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4
Input: intervals = [[1,2],[2,3],[2,4],[4,5]]
Output: 5
Explanation: let nums = [1, 2, 3, 4, 5].
It can be shown that there cannot be any containing array of size 4.
``````

Constraints:

• 1 <= intervals.length <= 3000
• intervals[i].length == 2
• 0 <= start_i < end_i <= 10^8

## Sample C++ Code

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class Solution {
public:
int intersectionSizeTwo(vector<vector<int>>& intervals) {
sort(intervals.begin(), intervals.end(), [](vector<int>& a, vector<int>& b) {
return a[1] < b[1] || (a[1] == b[1] && a[0] > b[0]);
});

int n = intervals.size(), ans = 0, p1 = -1, p2 = -1;

for (int i = 0; i < n; i++) {
// current p1, p2 works for intervals[i]
if (intervals[i][0] <= p1)
continue;

// Neither of p1, p2 works for intervals[i]
// replace p1, p2 by ending numbers
if (intervals[i][0] > p2) {
ans += 2;
p2 = intervals[i][1];
p1 = p2-1;
}

// only p2 works;
else {
ans++;
p1 = p2;
p2 = intervals[i][1];
}
}

return ans;
}
};
``````