## Monday, November 7, 2016

### LettCode 452 - Minimum Number of Arrows to Burst Balloons

https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/
There are a number of 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. Start is always smaller than end. There will be at most 104 balloons.
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 travelling up infinitely. The problem is to find the minimum number of arrows that must be shot to burst all balloons.
Example:
```Input:
[[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).
```
https://discuss.leetcode.com/topic/66709/c-easy-understood-solution-sort
First, we sort balloons by increasing points.end (if ends are the same, then by increasing of points.start). Every time arrow shot points.end, say, points[i].second. If next balloon.start <= points[i].second, it is also shot, thus we continue.
https://discuss.leetcode.com/topic/73981/share-my-explained-greedy-solution-as-the-highest-voted-java-solution-right-now-is-not-ideal
No offense but the currently highest voted java solution is not ideal, the sorting can be adjusted so that there's no need to check again in the for loop.
Idea:
We know that eventually we have to shoot down every balloon, so for each ballon there must be an arrow whose position is between balloon[0] and balloon[1]. Given that, we can sort the array of balloons by their ending position. Then we make sure that while we take care of each balloon from the beginning, we can shoot as many following balloons as possible.
So what position should we pick? We should shoot as right as possible, because all balloons' end position is to the right of the current one. Therefore the position should be currentBalloon[1], because we still need to shoot down the current one.
This is exactly what I do in the for loop: check how many balloons I can shoot down with one shot aiming at the ending position of the current balloon. Then I skip all these balloons and start again from the next one (or the leftmost remaining one) that needs another arrow.
Example:
``````balloons = [[7,10], [1,5], [3,6], [2,4], [1,4]]
``````
After sorting, it becomes:
``````balloons = [[2,4], [1,4], [1,5], [3,6], [7,10]]
``````
So first of all, we shoot at position 4, we go through the array and see that all first 4 balloons can be taken care of by this single shot. Then we need another shot for one last balloon. So the result should be 2.
``````public int findMinArrowShots(int[][] points) {
if (points.length == 0) {
return 0;
}
Arrays.sort(points, (a, b) -> a[1] - b[1]);
int arrowPos = points[0][1];
int arrowCnt = 1;
for (int i = 1; i < points.length; i++) {
if (arrowPos >= points[i][0]) {
continue;
}
arrowCnt++;
arrowPos = points[i][1];
}
return arrowCnt;
}``````
Almost the same with 435. Non-overlapping Intervals
``````public int findMinArrowShots(int[][] points) {
int count = 0;
long end = Long.MIN_VALUE;
Arrays.sort(points, (a, b) -> (a[1]-b[1]));

for(int[] p: points){
if(p[0] > end){
end = p[1];
count++;
}
}

return count;
}``````
https://discuss.leetcode.com/topic/66579/java-greedy-soution

``````    int findMinArrowShots(vector<pair<int, int>>& points) {
int count = 0, arrow = INT_MIN;
sort(points.begin(), points.end(), mysort);
for(int i = 0; i<points.size(); i++){
if(arrow!=INT_MIN && points[i].first<=arrow){continue;} //former arrow shot points[i]
arrow = points[i].second; // new arrow shot the end of points[i]
count++;
}
return count;
}
static bool mysort(pair<int, int>& a, pair<int, int>& b){
return a.second==b.second?a.first<b.first:a.second<b.second;
}``````
http://blog.jerkybible.com/2016/11/11/LeetCode-452-Minimum-Number-of-Arrows-to-Burst-Balloons/

```
public int findMinArrowShots(int[][] points) {

if (points == null || points.length == 0 || points[0].length == 0) {

return 0;

}

Arrays.sort(points, new Comparator<int[]>() {

public int compare(int[] a, int[] b) {

return a[1] - b[1];

}

});

long lastEnd = Long.MIN_VALUE;

int minArrows = 0;

for (int i = 0; i < points.length; i++) {

if (lastEnd < points[i][0]) {

lastEnd = points[i][1];

minArrows++;

}

}

return minArrows;

}
```
https://discuss.leetcode.com/topic/66548/concise-java-solution-tracking-the-end-of-overlapping-intervals
``````    public int findMinArrowShots(int[][] points) {
if(points == null || points.length < 1) return 0;
Arrays.sort(points, (a, b)->(a[0]-b[0]));
int result = 1;
int end = points[0][1];
for(int i = 1; i < points.length; i ++) {
if(points[i][0] > end) {
result ++;
end = points[i][1];
} else {
end = Math.min(end, points[i][1]);
}
}
return result;
}``````
https://discuss.leetcode.com/topic/66579/java-greedy-soution
He is actually sorting by start in ascending order, note that `if(a[0]==b[0]) return a[1]-b[1];` means only when starts of a and b are equal we then look at ends.

when sorted by end value, you will always need a shot for the first balloon(at its end) because if it's a stand alone balloon you obviously have to otherwise it has other balloons overlapping, in which case you also have to use an arrow at its end otherwise you will miss this balloon. 2, If you have overlapping balloon, it's good, your arrow will destroy them, and this way, you are creating a sub-problem, which start with a new set of balloons with these characteristics. You keep doing until no balloon is left.
``````public int findMinArrowShots(int[][] points) {
if(points==null || points.length==0 || points[0].length==0) return 0;
Arrays.sort(points, new Comparator<int[]>() {
public int compare(int[] a, int[] b) {
if(a[0]==b[0]) return a[1]-b[1];
else return a[0]-b[0];
}
});

int minArrows = 1;
int arrowLimit = points[0][1];
for(int i=1;i<points.length;i++) {
int[] baloon = points[i];
if(baloon[0]<=arrowLimit) {
arrowLimit=Math.min(arrowLimit, baloon[1]);
} else {
minArrows++;
arrowLimit=baloon[1];
}
}
return minArrows;
}``````

```按照气球的起点排序

def findMinArrowShots(self, points): """ :type points: List[List[int]] :rtype: int """ ans = 0 emin = MAXINT = 0x7FFFFFFF for s, e in sorted(points): if emin < s: ans += 1 emin = MAXINT emin = min(emin, e) return ans + bool(points)

http://www.cnblogs.com/grandyang/p/6050562.html
http://blog.csdn.net/mebiuw/article/details/53096708
http://brookebian.blogspot.com/2016/11/452-minimum-number-of-arrows-to-burst.html

`````` public int findMinArrowShots(int[][] points) {
if(points.length == 0) return 0;
int n = points.length;
Arrays.sort(points, (a, b)->{
if(a[0] == b[0]) return Integer.compare(a[1],b[1]);
return Integer.compare(a[0],b[0]);
});
int p1 = 0;
int ans = 1;
// point 0's end
int end = points[0][1];
for(int i = 1; i < n; i++){
//when end is smaller than next start, we need another arrow
//we need to update end, if end is smaller than previous ones
end = Math.min(end,points[i][1]);
if( end < points[i][0]){
end = points[i][1];
ans++;
}
}
return ans;
}  ``````