• # 452. Minimum Number of Arrows to Burst Balloons

## Notation

For convenience, we say the kth ballon as bk, first element of bk as sk, the second element of bk as ek, the total number of ballons as n.

## Algorithm Description

(* If n = 0, directly return 0. The following is talking about the other case, when there are ballons. )

1. First sort the array according to their sk

2. Then we define Sk and Ek as follows (Intuitively, if the intersection of the bk and the [Sk-1 , Ek-1] exsits then [Sk, Ek] is that intersection, else [Sk, Ek] is bk):

1. Sk is always same as sk

• If k = 0, Ek=ek.
• Else if k > 0:
• If sk ≤ Ek-1 and ek > Ek-1 (It's c1, short for case 1):
• Ek = Ek-1
• If sk > Ek-1 (also means ek > Ek-1) (It's c2, short for case 2):
• Ek = ek
• If ek ≤ Ek-1 (also means Sk-1 < sk < Ek-1) (It's c3, short for case 3):
• Ek = ek

Since sk is non-decreasing with k increasing, the above definition is complete (c1, c2 and c3 cover all possible cases) and unique.

3. Now initially, set result r as 1, get S1 and E1 from b1, do the following iteration:

• For every k (from 2 to n):
• if c1 or c3:
• r remain same
• if c2:
• r increase by 1
• anyway get Sk and Ek according to definition
• Return r

## Proof

For convenience, we say the result is rt after the tth iterateration.

(1). This algorithm will terminate in at most O(nlogn) time if n is finite.

(1) is proved because if we use merge sort, it takes O(nlogn) time to finish the sorting part. This is followed by n iterations where each iteration takes constant time.

(2.1)For the first ballon, this algorithm result r 1 is correct and the r1th arrow could be placed anywhere between [S1, E1]

(2.1) is obviously correct.

(2.2) If for the first tth ballons, rt is correct and the rtth arrow could be anywhere between [St, Et], then for the first t+1th ballons, rt+1 is correct and the rt+1th arrow could be anywhere between [St+1, Et+1]. (t = 1, 2, 3...)

For different cases of bt+1

• If c1 and c3:
• Since there is intersection it+1 between [st+1, et+1] and [St, Et] and the rtth arrow could be anywhere between [St, Et], let the arrow in it+1, which is both inside [St, Et] and bt+1, then r remains same, that is rt+1 = rt so r is correct according to the algorithm.

In this case, [St+1, Et+1] is exactly it + 1, so according to above paragraph, the rt+1th arrow could be anywhere between [St+1, Et+1]

• If c2:
• Since st+1 > Et and Et is at most et, so there is no intersection between any of the ballons with bt+1. Thus r need to be increased by 1 to cover bt+1, that is rt+1 = rt + 1, according to the algorithm rt+1 is correct and the new arrow, a.k.a the rt+1th arrow could obviously be anywhere in [St+1, Et+1], which is also bt+1

Thus, (2.2) is proved.

According to (2.1) and (2.2), this algorithm returns correct result for any finite valid input.

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