Solved q452 with C++ and Python3.

Algorithm/Greedy/452_minimum-number-of-arrows-to-burst-balloons/q452.cpp

@@ -0,0 +1,26 @@
+class Solution {
+public:
+    int findMinArrowShots(vector<vector<int>>& points) {
+        if (points.empty()) {
+            return 0;
+        }
+        // sort by end points
+        sort(points.begin(), points.end(), [](vector<int>& a, vector<int>& b)->bool{
+        if(a[1] == b[1]){
+            return a[0] < b[0];
+        }
+
+        return a[1] < b[1];
+        });
+        int arrow = 1, prev_end = points[0][1];
+        for (int i = 1; i < points.size(); ++i) {
+            // within range, only one arrow needed
+            if (points[i][0] <= prev_end) {
+                continue;
+            }
+            ++arrow;
+            prev_end = points[i][1];
+        }
+        return arrow;
+    }
+};

Algorithm/Greedy/452_minimum-number-of-arrows-to-burst-balloons/q452.py

@@ -0,0 +1,12 @@
+class Solution:
+    def findMinArrowShots(self, points: List[List[int]]) -> int:
+        if len(points) == 0:
+            return 0
+        points.sort(key = lambda x:x[1])
+        prev_end, arrow = points[0][1], 1
+        for i in range(1, len(points)):
+            if points[i][0] <= prev_end:
+                continue
+            arrow += 1
+            prev_end = points[i][1]
+        return arrow