Easy
Points That Intersect With Cars — C++
Full explanation · Time O(nlogn) · Space O(1)
// Time: O(nlogn)
// Space: O(1)
// sort, line sweep
class Solution {
public:
int numberOfPoints(vector<vector<int>>& nums) {
sort(begin(nums), end(nums));
int result = 0;
vector<int> curr = nums[0];
for (int i = 1; i < size(nums); ++i) {
if (nums[i][0] <= curr[1]) {
curr[1] = max(curr[1], nums[i][1]);
} else {
result += curr[1] - curr[0] + 1;
curr = nums[i];
}
}
result += curr[1] - curr[0] + 1;
return result;
}
};