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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 <= 104
• points.length == 2
• -231 <= xstart < xend <= 231 - 1

Related Topics:
Greedy, Sort

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;
for (auto &b : A) {
if (b <= arrow) arrow = min(arrow, b);
else {
arrow = b;
++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] <= arrow; ++i) arrow = min(arrow, A[i]);
}
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 < b; });
long ans = A.size(), e = LONG_MIN;
for (auto &v : A) {
if (v <= e) --ans; // this interval overlaps with another interval. We don't need a separate arrow for it.
else e = v;
}
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 != point2)
return point1 - point2;
else
return point1 - point2;
}
});
int arrows = 0;
int left = points, right = points;
int length = points.length;
for (int i = 1; i < length; i++) {
int[] point = points[i];
if (point > right) {
arrows++;
left = point;
right = point;
} else {
left = Math.max(left, point);
right = Math.min(right, point);
}
}
arrows++;
return arrows;
}
}

############

class Solution {
public int findMinArrowShots(int[][] points) {
Arrays.sort(points, (a, b) -> a < b ? -1 : 1);
int ans = 1;
int x = points;
for (int[] v : points) {
int a = v, b = v;
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;
for (auto &b : A) {
if (b <= arrow) arrow = min(arrow, b);
else {
arrow = b;
++ans;
}
}
return ans;
}
};

• class Solution:
def findMinArrowShots(self, points: List[List[int]]) -> int:
points.sort(key=lambda x: x)
ans = 1
x = points
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)
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] < points[j] })
ans := 1
x := points
for _, v := range points {
a, b := v, v
if a > x {
ans++
x = b
}
}
return ans
}

• function findMinArrowShots(points: number[][]): number {
points.sort((a, b) => a - b);
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 < b ? -1 : a > b ? 1 : 0);
int ans = 0;
long last = long.MinValue;
foreach (var point in points) {
if (point > last) {
++ans;
last = point;
}
}
return ans;
}
}