Dec18

Author: aliasvishnu

README.md

@@ -3,7 +3,7 @@
 Hints for ~250 leetcode problems. Mostly medium, some hard.
 Good practice list for internships/job hunt.
 
-If this helps you, please star the repo (I commit a lot, so I suggest you unwatch this repo).
+If this helps you, please star the repo (I commit a lot, so I suggest you unwatch the repo).
 
 Solutions <a href = "https://github.com/aliasvishnu/leetcode/tree/master/Solutions" > here </a>
 
@@ -79,6 +79,7 @@ Solutions <a href = "https://github.com/aliasvishnu/leetcode/tree/master/Solutio
 | 118 | BF | Generate new row from old row. | O(n^2) |
 | 119 | BF | Generate new row from old row. | O(n^2) |
 | 128 | DP | DP problem, find recurrence relation. | |
+| 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) |
 
 ## Second hint list
 

Solutions/413-ArithmeticSlices.cpp

@@ -0,0 +1,31 @@
+class Solution {
+public:
+
+    // n = number of values sequences are different for, 1, 2, 3, 4, 5 becomes 1, 1, 1, 1 i.e. n = 4.
+    int f(int n){
+        return (n*n+n)/2;
+    }
+
+    int numberOfArithmeticSlices(vector<int>& A) {
+        int len = A.size();
+        if(len <= 2) return 0;
+
+
+        int curdiff = A[1] - A[0], curcount = 0, answer = 0;
+
+        for(int i = 2; i < len; i++){
+            if(A[i]-A[i-1] == curdiff) curcount += 1;
+            else{
+                answer += f(curcount);
+                curcount = 0;
+                curdiff = A[i]-A[i-1];
+            }
+
+        }
+
+        answer += f(curcount);
+
+
+        return answer;
+    }
+};