new soln

Author: shreymehul

198.Rob.cs

@@ -49,4 +49,24 @@ public int Rob(int[] nums) {
         }
         return Math.Max(nums[0],nums[1]);
     }
+}
+
+public class Solution {
+    public int Rob(int[] nums) {
+        int[] dp = new int[nums.Length];
+        Array.Fill(dp,-1);
+        if(nums.Length == 1)
+            return nums[0];
+        if(nums.Length == 2)
+            return Math.Max(nums[1],nums[0]);
+        if(nums.Length == 3)
+            return Math.Max(nums[1],nums[0]+nums[2]);
+        dp[0] = nums[0];
+        dp[1] = nums[1];
+        dp[2] = nums[0]+nums[2];
+        for(int i = 3; i<nums.Length;i++){
+            dp[i] = nums[i] + Math.Max(dp[i-2],dp[i-3]);
+        }
+        return Math.Max(dp[nums.Length-1],dp[nums.Length-2]);
+    }
 }

213.Rob.cs

@@ -0,0 +1,53 @@
+// 213. House Robber II
+// You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.
+// Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.
+// Example 1:
+// Input: nums = [2,3,2]
+// Output: 3
+// Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
+// Example 2:
+// Input: nums = [1,2,3,1]
+// Output: 4
+// Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
+// Total amount you can rob = 1 + 3 = 4.
+// Example 3:
+// Input: nums = [1,2,3]
+// Output: 3
+// Constraints:
+// 1 <= nums.length <= 100
+// 0 <= nums[i] <= 1000
+
+
+//Approch: same as 198, but here first and last cant be include at once.
+public class Solution {
+    public int Rob(int[] nums) {
+        int[] dp = new int[nums.Length];
+        if(nums.Length == 1)
+            return nums[0];
+        if(nums.Length == 2)
+            return Math.Max(nums[1],nums[0]);
+        if(nums.Length == 3)
+            return Math.Max(nums[1],Math.Max(nums[0],nums[2]));
+
+        //Assume array without last element
+        dp[0] = nums[0];
+        dp[1] = nums[1];
+        dp[2] = nums[0]+nums[2];
+        for(int i = 3; i<nums.Length-1;i++){
+            dp[i] = nums[i] + Math.Max(dp[i-2],dp[i-3]);
+        }
+        int profi1 = Math.Max(dp[nums.Length-2],dp[nums.Length-3]);
+        Array.Fill(dp,0);
+
+        //Assume array without first element
+        dp[0] = 0;
+        dp[1] = nums[1];
+        dp[2] = nums[2];
+        for(int i = 3; i<nums.Length;i++){
+            dp[i] = nums[i] + Math.Max(dp[i-2],dp[i-3]);
+        }
+        int profi2 = Math.Max(dp[nums.Length-1],dp[nums.Length-2]);
+
+        return Math.Max(profi1,profi2);
+    }
+}

452.FindMinArrowShots.cs

@@ -0,0 +1,40 @@
+// 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] = [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
+
+
+public class Solution {
+    public int FindMinArrowShots(int[][] points) {
+        points = points.OrderBy(x => x[1]).ThenBy(x => x[0]).ToArray();
+        int count = 0;
+        for(int i = 0; i < points.Length; i++){
+            count++;
+            int j = i + 1;
+            while(j < points.Length && points[i][1] >= points[j][0])
+                j++;
+            i = j-1;
+        }
+        return count;
+    }
+}