452 - Minimum Number Of Arrows To Burst Balloons

#medium

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.

class Solution {
public:

    static bool cmp( vector<int> & a, vector<int> & b ) {
        if ( a[0] < b[0] ) return true; 
        return false;
    }

    int findMinArrowShots(vector<vector<int>>& points) {
        sort( points.begin(), points.end(), cmp );
        int result = 1;
        int max_value = points[0][1]; // 右邊界
        for ( int i = 1; i < points.size(); i++ ) {
            if ( points[i][0] > max_value ) {
                result++;
                max_value = points[i][1];
            }
            // key point: 第二個區間可能較小，需更新右邊界
            else {
                max_value = min( points[i][1], max_value );
            }

        }

        return result;
    }   
};

435 - Non-overlapping Intervals

#medium

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:
Input: intervals = [ [1,2],[2,3],[3,4],[1,3] ]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.

Example 2:
Input: intervals = [ [1,2],[1,2],[1,2] ]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.

Example 3:
Input: intervals = [ [1,2],[2,3] ]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

class Solution {
public:
    static bool cmp( vector<int> & a, vector<int> & b ) {
        if ( a[0] == b[0] ) return b[1] > a[1];
        return b[0] > a[0];
    }

    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        // 先排序左區間小到大
        // 思考 [1,2] [1,3] => 取前面
        // 思考 [1,6] [2,4] => 取後面
        // 皆取右區間最小的。
        int result = 0;
        sort(intervals.begin(), intervals.end(), cmp );
        int max_value = intervals[0][1];
        for ( int i = 1; i < intervals.size(); i++ ) {
            if ( intervals[i][0] < max_value ) {
                result++;
                max_value = min( intervals[i][1], max_value );
            }
            else {
                max_value = intervals[i][1]; // 更新右邊界
            }

        }
        return result;
    }