Create cherry_pickup_ii.cpp

Author: avidLearnerInProgress

Diff for: cherry_pickup_ii.cpp

@@ -0,0 +1,55 @@
+class Solution {
+    int dp[71][71][71];
+    int delta[3] = {0, -1, 1};
+    
+    bool isNotInRange(int &c1, int &c2, int &m, int &n) {
+        return c1 < 0 || c2 < 0 || c1 >= n || c2 >= n; 
+    }
+        
+    int cherryPickupHelper(vector<vector<int>>& grid, int r, int c1, int c2) {
+        int m = grid.size(), n = grid[0].size();
+        
+        if(r == m) {
+            return 0;
+        }
+        
+        if(isNotInRange(c1, c2, m, n)) {
+            return INT_MIN;
+        }
+        
+        if(dp[r][c1][c2] != -1) { //memoized call
+            return dp[r][c1][c2];
+        }
+        
+        int currMaxCherryCount = 0;
+        
+        for(int i = 0; i < 3; ++i) {
+            for(int j = 0; j < 3; ++j) {
+                currMaxCherryCount = max(currMaxCherryCount, cherryPickupHelper(grid, r + 1, c1 + delta[i], c2 + delta[j]));
+            }
+        }
+        
+        currMaxCherryCount += (c1 == c2) ? grid[r][c1] : grid[r][c1] + grid[r][c2];
+        
+        return dp[r][c1][c2] = currMaxCherryCount;     //store computation
+    }
+    
+public:
+    int cherryPickup(vector<vector<int>>& grid) {
+         
+        //too much of work in this problem!! DP
+        //both robots move simultaneously (this reduces a dimension in terms of storage for the dp matrix)
+        //three directions .. max_cherries: 3 * r1 + 3 * r2 => 9 different states 
+        //one edgecase - r1 and r2 at same position, add value of grid only once
+        //Top Down approach
+        
+        int m = grid.size();
+        if(!m) return 0;
+        
+        int n = grid[0].size();
+    
+        memset(dp, -1, sizeof dp); 
+        return cherryPickupHelper(grid, 0, 0, n - 1);
+    }
+};
+