Added hints Dec 28

Author: aliasvishnu

README.md

@@ -95,6 +95,13 @@ Solutions <a href = "https://github.com/aliasvishnu/leetcode/tree/master/Solutio
 | 187 | Hash | Store all 10 letter substrings in a hashtable. | O(n) |
 | 190 | BF | Remember to store answer in unsigned long. Construct answer from bits. | O(log(n)) |
 | 191 | BF | Straighforward. | O(log(n)) |
+| 198 | DP | DP[i] = max(DP[i-2]+nums[i], DP[i-1]). | O(n) |
+| 200 | DFS | Do DFS whenever you encounter a 1; #islands = # of times the DFS routine is called from main(). | O(n*m) |
+| 202 | Set | Use a set to keep track of all visited numbers. | Don't know. |
+| 205 | Hash | Make sure you have a 1-1 mapping. | O(n) |
+| 206 | BF | Use variables prevNode, curNode, nextNode. You need to do it this step by step, decide in how to and in what order to set the said variables. | O(n) |
+| 207 | DFS | You're looking for a loop. So set the nodes whom you're visiting in the current loop as gray, and if you end up visiting a gray node then there is a loop. | O(|V|+|E|) |
+| 208 | Tree | Multiway tree. | O(n) for both search and insert, n = # letters |
 | 413 | Array | [1, 2, 3, 4, 5, 7, 9, 11]. can be written as [5, 3] i.e. 5 sequence of difference 1 and 3 sequence of difference 2, you need to figure out how many parts you can split 5 and 3 into. | O(n) |
 | 694 | DFS | Keep track of the directions in which DFS proceeds in some form, maybe a string like ddr for down down right. | O(rows*cols) |
 | 738 | Array | Find the first time Xi > Xi+1, Xi -= 1 and turn all Xi+k = 9, For eg, 321 becomes 299. Figure out cases like 33332. | O(n) |

Solutions/198-HouseRobber.cpp

@@ -11,20 +11,15 @@ class Solution {
         if(len == 0) return 0;
         if(len == 1) return nums[0];
         if(len == 2) return max(nums[1], nums[0]);
-        if(len == 3) return max(nums[1], nums[2]+nums[0]);
 
         dp[0] = nums[0];
         dp[1] = max(nums[0], nums[1]);
-        dp[2] = max(nums[1], nums[2]+nums[0]);
 
-        for(int i = 3; i < len; i++){
-            dp[i] = max(dp[i-3], dp[i-2]) + nums[i];
-        }
-        int maxval = 0;
-        for(int i = 0; i < len; i++){
-            if(maxval < dp[i]) maxval = dp[i];
+        for(int i = 2; i < len; i++){
+            dp[i] = max(dp[i-2] + nums[i], dp[i-1]);
         }
 
-        return maxval;
+
+        return dp[len-1];
     }
 };