This card builds on the Interval Scheduling card. That card sorts by end and counts the intervals to remove. This card sorts by end too, but it counts groups. A group is a set of intervals that all contain one common point.

In Minimum Number of Arrows to Burst Balloons (LC 452), each balloon is a range [start, end] on the x axis. One arrow at position x bursts every balloon with start <= x <= end. The question asks for the fewest arrows, so it asks for the fewest groups.

Both cards use the same sort and the same loop. Two things change. The loop does not remove anything: it shoots a new arrow instead. And the touch rule is reversed. In Interval Scheduling, two intervals that only touch do not overlap. Here, a balloon that only touches the arrow is hit.

Look at the balloon that ends first. Some arrow must hit it, so that arrow sits at a position x with start <= x <= end for this balloon. Move the arrow right, up to this balloon's end. It still hits this balloon. It now hits every balloon that the old arrow hit, because every other balloon ends at or after this end, and it already contains the old position x. So an arrow at end is never worse.

After that arrow, a balloon is hit when start <= arrow_x. Since the list is sorted by end, every balloon that comes later ends at or after arrow_x. The only question is where it starts.

The loop does this with arrow_x. A balloon with start > arrow_x starts after the arrow, so no earlier arrow reaches it. The loop adds 1 to arrows and sets arrow_x = end. Every other balloon is already hit, so nothing changes. The starting value is None, so the first balloon always gets an arrow.

Example: in LC 452 with points = [[10, 16], [2, 8], [1, 6], [7, 12]], the sorted order is [1, 6], [2, 8], [7, 12], [10, 16]. The first arrow goes to 6 and hits [1, 6] and [2, 8]. [7, 12] starts at 7, which is greater than 6, so the second arrow goes to 12. It hits [7, 12] and [10, 16]. The answer is 2.

The board replays points = [[1, 6], [2, 3], [4, 5], [5, 8], [7, 9]]. The rows are in sorted order by end, so the first row is [2, 3].

The first arrow goes to 3 and bursts [2, 3]. [4, 5] starts at 4, which is greater than 3, so a second arrow goes to 5. [1, 6] contains [4, 5]. It starts at 1, which is not greater than 5, so the same arrow hits it.

[5, 8] starts at 5, and 5 > 5 is false. The balloon only touches the arrow at 5, and the arrow still hits it. This is the step marked as the trap on the board. Then [7, 9] starts at 7, which is greater than 5, so a third arrow goes to 9. The answer is 3.

The test start >= arrow_x would shoot a new arrow for [5, 8]. For the four balloons [[1, 2], [2, 3], [3, 4], [4, 5]] the code would then return 4 instead of 2, because every pair of neighbours only touches.

Keep the same balloons and sort by start. The order is [1, 6], [2, 3], [4, 5], [5, 8], [7, 9]. The first arrow goes to 6, the end of the long balloon [1, 6]. The code then treats [2, 3] as hit, but 6 is not inside [2, 3]. The code returns 2 instead of 3.

The reason is that the end of the first balloon by start can lie to the right of the short balloons inside it. With the sort by end, arrow_x is always the smallest end among the balloons that are not hit yet, so every later balloon that starts at or before it really contains it.

The sort takes O(N log N) time and the loop takes O(N) time. The loop keeps only arrow_x and arrows, so the extra space is O(1) besides the sort. Start arrow_x at None and not at 0, because coordinates can be negative.

The same idea solves Maximum Length of Pair Chain (LC 646), but there the touch rule is the opposite: a pair can follow another only when its first number is strictly greater than the previous second number. Check the rule of each question before you choose between > and >=.