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Formatted question description: https://leetcode.ca/all/452.html
452. Minimum Number of Arrows to Burst Balloons (Medium)
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:
Solution 1. Greedy
// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/
// 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/
// 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;
}
};
Solution 2. Interval Scheduling Maximization (ISM)
// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/
// 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;
}
};
-
class Solution { public int findMinArrowShots(int[][] points) { if (points == null || points.length == 0) return 0; Arrays.sort(points, new Comparator<int[]>() { public int compare(int[] point1, int[] point2) { if (point1[0] != point2[0]) return point1[0] - point2[0]; else return point1[1] - point2[1]; } }); int arrows = 0; int left = points[0][0], right = points[0][1]; int length = points.length; for (int i = 1; i < length; i++) { int[] point = points[i]; if (point[0] > right) { arrows++; left = point[0]; right = point[1]; } else { left = Math.max(left, point[0]); right = Math.min(right, point[1]); } } arrows++; return arrows; } } ############ class Solution { public int findMinArrowShots(int[][] points) { Arrays.sort(points, (a, b) -> a[1] < b[1] ? -1 : 1); int ans = 1; int x = points[0][1]; for (int[] v : points) { int a = v[0], b = v[1]; if (a > x) { ++ans; x = b; } } return ans; } } -
// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/ // 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; } }; -
class Solution: def findMinArrowShots(self, points: List[List[int]]) -> int: points.sort(key=lambda x: x[1]) ans = 1 x = points[0][1] for a, b in points: if a > x: ans += 1 x = b return ans ############ class Solution(object): def findMinArrowShots(self, points): """ :type points: List[List[int]] :rtype: int """ ans = 0 points.sort(key=lambda p: p[1]) end = float("-inf") for s, e in points: if s > end: ans += 1 end = e return ans -
func findMinArrowShots(points [][]int) int { sort.Slice(points, func(i, j int) bool { return points[i][1] < points[j][1] }) ans := 1 x := points[0][1] for _, v := range points { a, b := v[0], v[1] if a > x { ans++ x = b } } return ans } -
function findMinArrowShots(points: number[][]): number { points.sort((a, b) => a[1] - b[1]); let ans = 0; let last = -Infinity; for (const [a, b] of points) { if (last < a) { ans++; last = b; } } return ans; } -
public class Solution { public int FindMinArrowShots(int[][] points) { Array.Sort(points, (a, b) => a[1] < b[1] ? -1 : a[1] > b[1] ? 1 : 0); int ans = 0; long last = long.MinValue; foreach (var point in points) { if (point[0] > last) { ++ans; last = point[1]; } } return ans; } }