Minimum Number of Arrows to Burst Balloons - LeetCode

Description

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] = [xstart, xend] denotes a balloon whose horizontal diameter stretches between xstart and xend. 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 xstart and xend is burst by an arrow shot at x if xstart <= x <= xend. 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 <= 105
• points[i].length == 2
• -231 <= xstart < xend <= 231 - 1

Code

類似 Insert Interval，不同之處在於如何處理重疊的部分。在這裏，重疊的部分是越來越小的，在前者，merge 後的 interval 會是越來越大的。

和 Maximum Length of Pair Chain 是同一種題目。

maintain points 的重疊部分（left, right），對於下一個 point 而言，考慮其左邊屆是否在重疊部分中，藉此判斷是否需要一支新的 arrow。

class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& points) {
        sort(points.begin(), points.end());
        int s_max = points[0][0], e_min = points[0][1];
        int shot = 1;
        for(int i = 1; i < points.size(); i++) {
            if(points[i][0] > e_min) {
                s_max = points[i][0];
                e_min = points[i][1];
                shot++;
            } else {
                s_max = max(s_max, points[i][0]);
                e_min = min(e_min, points[i][1]);
            }
        }
        return shot;
    }
};

class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& points) {
        sort(points.begin(), points.end());
        int n = points.size();
        int left = points[0][0], right = points[0][1];
        int arrow = 1;
        for(int i = 1; i < n; i++) {
            if(points[i][0] > right) {
                left = points[i][0];
                right = points[i][1];
                arrow++;
            } else if (points[i][0] < right) {
                left = points[i][0];
                right = min(points[i][1], right);
            } else {
                left = points[i][0];
                right = points[i][0];
            }
        }
        return arrow;
    }
};

Source

• Minimum Number of Arrows to Burst Balloons - LeetCode