LeetCode //C - 452. Minimum Number of Arrows to Burst Balloons

452. Minimum Number of Arrows to Burst Balloons

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [$x_{start},x_{end}$] denotes a balloon whose horizontal diameter stretches between $x_{start}$ and $x_{end}$. You do not know the exact y-coordinates of the balloons.

Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with $x_{start}$ and $x_{end}$ is burst by an arrow shot at x if $x_{start}<=x<=x_{end}$. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.

Given the array points, 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: The balloons can be burst by 2 arrows:
Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].

Example 2:

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.

Example 3:

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].

Constraints:

• $1<=points.length<=10^5$
• points[i].length == 2
• $-2^{31}<=xstart<xend<=2^{31}-1$

From: LeetCode
Link: 452. Minimum Number of Arrows to Burst Balloons

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Solution:

Ideas：

1. Problem Analysis:
The problem is essentially asking how many arrows are needed such that each arrow hits at least one balloon, and each balloon is hit by at least one arrow. An important observation here is that if an arrow is shot at some point x, it will burst all balloons whose range covers x.

2. Sorting the Balloons by End Point:
The first key idea in the solution is to sort the balloons by their ending points (i.e., $x_{end}$). The reasoning behind this is that if we shoot an arrow at the smallest available end point, we ensure that we burst as many balloons as possible that started before this end point.
The compare function helps the qsort function in sorting the balloons based on their end points.

3. Counting Arrows:
After sorting, we initialize our arrow count and set the position of the first arrow to be the end point of the first balloon.

4. Iterating Over the Balloons:
We then iterate over the rest of the balloons. For each balloon, we check its start point:

• If the start point is less than or equal to the current arrow’s position, it means this balloon can be burst by the current arrow, and we move to the next balloon.
• If the start point is greater than the current arrow’s position, it means we need a new arrow. We then increment our arrow count and set the new arrow’s position to be the end point of the current balloon.

5. Return the Total Number of Arrows:
After iterating over all balloons, the arrows variable will hold the minimum number of arrows needed to burst all balloons. We return this value.

6. Handling Integer Overflow:
The initial solution had a subtraction in the compare function, which led to integer overflow for large values. We then changed the comparison logic to avoid subtraction, thereby preventing the overflow.

Code:

int compare(const void* a, const void* b) {
    int end1 = (*(int**)a)[1];
    int end2 = (*(int**)b)[1];
    if (end1 < end2) return -1;
    if (end1 > end2) return 1;
    return 0;
}

int findMinArrowShots(int** points, int pointsSize, int* pointsColSize) {
    if(pointsSize == 0) {
        return 0;
    }

    // Sort the points based on the end values
    qsort(points, pointsSize, sizeof(int*), compare);

    int arrows = 1;
    int arrowPos = points[0][1];

    for(int i = 1; i < pointsSize; i++) {
        // If the start of the balloon is greater than the arrowPos, it means the current arrow can't burst this balloon
        if(points[i][0] > arrowPos) {
            arrows++;
            arrowPos = points[i][1];
        }
    }

    return arrows;
}