There are some spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter, and hence the x-coordinates of start and end of the diameter suffice. The start is always smaller than the end.
An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.
Given an array points where points[i] = [xstart, xend], return the minimum number of arrows that must be shot to burst all balloons.
Example 1:
Input: points = [[10,16],[2,8],[1,6],[7,12]] Output: 2 Explanation: One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).
Example 2:
Input: points = [[1,2],[3,4],[5,6],[7,8]] Output: 4
Example 3:
Input: points = [[1,2],[2,3],[3,4],[4,5]] Output: 2
Example 4:
Input: points = [[1,2]] Output: 1
Example 5:
Input: points = [[2,3],[2,3]] Output: 1
Constraints:
0 <= points.length <= 104points.length == 2-231 <= xstart < xend <= 231 - 1
Similar Questions:
// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int findMinArrowShots(vector<vector<int>>& A) {
if (A.empty()) return 0;
sort(begin(A), end(A));
int ans = 1, arrow = A[0][1];
for (auto &b : A) {
if (b[0] <= arrow) arrow = min(arrow, b[1]);
else {
arrow = b[1];
++ans;
}
}
return ans;
}
};Or
// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int findMinArrowShots(vector<vector<int>>& A) {
sort(begin(A), end(A));
int ans = 0, N = A.size();
for (int i = 0; i < N; ++ans) {
int arrow = INT_MAX;
for (; i < N && A[i][0] <= arrow; ++i) arrow = min(arrow, A[i][1]);
}
return ans;
}
}; // OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int findMinArrowShots(vector<vector<int>>& A) {
sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
long ans = A.size(), e = LONG_MIN;
for (auto &v : A) {
if (v[0] <= e) --ans; // this interval overlaps with another interval. We don't need a separate arrow for it.
else e = v[1];
}
return ans;
}
};