O(N)

Author: sjain700

.DS_Store

@@ -0,0 +1,56 @@
+class Solution {
+public:
+    string tictactoe(vector<vector<int>>& moves) {
+      /* Read second approach directly from solution */
+        vector<int> a(3,0);
+        vector<int> b(3,0);
+        int dia = 0;
+        int anti = 0;
+        
+        int count = 0;
+        
+        for(int i = 0 ; i < moves.size(); i++)
+        {
+            if(count % 2 == 0)
+            {
+                a[moves[i][0]]++;
+                b[moves[i][1]]++;
+                if(moves[i][0] + moves[i][1] == 2)
+                {
+                    anti++;
+                }
+                if(moves[i][0] == moves[i][1])
+                {
+                    dia++;
+                }
+            }
+            else
+            {
+                a[moves[i][0]]--;
+                b[moves[i][1]]--;
+                 if(moves[i][0] + moves[i][1] == 2)
+                {
+                    anti--;
+                }
+                if(moves[i][0] == moves[i][1])
+                {
+                    dia--;
+                }
+            }
+            
+            if(a[0] == 3 || a[1] == 3 || a[2] == 3 || dia == 3 || anti == 3 || b[0] == 3 || b[1] == 3 || b[2] == 3)
+                return "A";
+            else if(a[0] == -3 || a[1] == -3 || a[2] == -3 || dia == -3 || anti == -3 || b[0] == -3 || b[1] == -3 || b[2] == -3)
+                return "B";
+            
+           count++; 
+        }
+        if(moves.size()<9)
+            return "Pending";
+        else
+            return "Draw";
+        
+        
+        
+    }
+};