Dec18

Author: aliasvishnu

README.md

@@ -80,7 +80,8 @@ Solutions <a href = "https://github.com/aliasvishnu/leetcode/tree/master/Solutio
 | 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) |
-| 738 | Array | Find the first time Xi > Xi+1, Xi -= 1 and turn all Xi+k = 9, For eg, 321 becomes 299. However if there are repeating digits like 33332, then subtract the first occurrence then rest become 9, 33332 becomes 29999. | 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) |
 
 ## Second hint list
 

Solutions/694-NumberOfDistinctIslands.cpp

@@ -0,0 +1,43 @@
+class Solution {
+public:
+    unordered_set<string> st;
+    int rows, cols, count = 0;
+
+    void DFS(vector<vector<int>> &grid, int r, int c, string &s, string code){
+        if(r < 0 || r >= rows) { s += ","; return;}
+        if(c < 0 || c >= cols) { s += ","; return;}
+        if(grid[r][c] == 0) { s += ","; return;}
+        else{
+            grid[r][c] = 0;
+            s += code;
+        }
+
+        DFS(grid, r+1, c, s, "d");
+        DFS(grid, r, c+1, s, "r");
+        DFS(grid, r-1, c, s, "l");
+        DFS(grid, r, c-1, s, "u");
+    }
+
+    int numDistinctIslands(vector<vector<int>>& grid) {
+        rows = grid.size();
+        if(rows == 0) return 0;
+        cols = grid[0].size();
+        if(cols == 0) return 0;
+
+        for(int i = 0; i < rows; i++){
+            for(int j = 0; j < cols; j++){
+                if(grid[i][j] == 1){
+                    string order = "";
+                    DFS(grid, i, j, order, "s");
+                    if(st.find(order) == st.end()){
+                        count += 1;
+                        st.insert(order);
+                        // cout << order << endl;
+                    }
+                }
+            }
+        }
+
+        return count;
+    }
+};