October leetcoding challenge, week 2, Q 3

Author: tans105

ds-algorithms-competitive-programming/src/main/java/competitiveProgramming/leetcode/thirtyDaysLeetcodingChallenge/october/week2/MinimumNumberOfArrowsToBurst.java

@@ -0,0 +1,91 @@
+package competitiveProgramming.leetcode.thirtyDaysLeetcodingChallenge.october.week2;
+
+import utils.ArrayUtils;
+
+import java.util.Arrays;
+import java.util.Comparator;
+import java.util.Stack;
+
+/*
+https://leetcode.com/explore/challenge/card/october-leetcoding-challenge/560/week-2-october-8th-october-14th/3490/
+
+Minimum Number of Arrows to Burst Balloons
+
+There are some spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter, and hence the x-coordinates of start and end of the diameter suffice. The start is always smaller than the end.
+
+An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.
+
+Given an array points where points[i] = [xstart, xend], 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: One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).
+
+Example 2:
+
+Input: points = [[1,2],[3,4],[5,6],[7,8]]
+Output: 4
+
+Example 3:
+
+Input: points = [[1,2],[2,3],[3,4],[4,5]]
+Output: 2
+
+Example 4:
+
+Input: points = [[1,2]]
+Output: 1
+
+Example 5:
+
+Input: points = [[2,3],[2,3]]
+Output: 1
+
+
+
+Constraints:
+
+    0 <= points.length <= 104
+    points.length == 2
+    -231 <= xstart < xend <= 231 - 1
+
+ */
+public class MinimumNumberOfArrowsToBurst {
+    public static void main(String[] args) {
+        System.out.println(findMinArrowShots(ArrayUtils.parse2DArray("[[10,16],[2,8],[1,6],[7,12]]")));
+        System.out.println(findMinArrowShots(ArrayUtils.parse2DArray("[[1,2],[3,4],[5,6],[7,8]]")));
+        System.out.println(findMinArrowShots(ArrayUtils.parse2DArray("[[1,2],[2,3],[3,4],[4,5]]")));
+        System.out.println(findMinArrowShots(ArrayUtils.parse2DArray("[[1,2]")));
+        System.out.println(findMinArrowShots(ArrayUtils.parse2DArray("[[2,3],[2,3]]")));
+    }
+
+    public static int findMinArrowShots(int[][] points) {
+        if(points == null || points.length == 0) return 0;
+
+        Stack<int[]> stack = new Stack<>();
+        int count = 1;
+        int i = 1;
+
+        Arrays.sort(points, Comparator.comparingInt(a -> a[1]));
+        stack.push(points[0]);
+
+        while (i < points.length) {
+            int[] prev = stack.peek();
+            int[] current = points[i];
+
+            if(prev[1] >= current[0]) { //intersection => continue
+                stack.push(new int[]{current[0], prev[1]});
+            } else {
+                stack.push(current);
+                count++;
+            }
+            i++;
+        }
+
+        return count;
+    }
+}