Create 1937. Maximum Number of Points with Cost

1937. Maximum Number of Points with Cost

@@ -0,0 +1,30 @@
+class Solution {
+public:
+    long long maxPoints(vector<vector<int>>& points) {
+        //dynamic programming
+        int m = points.size(), n = points[0].size();
+        long long currMax;
+        vector<long long> maxPoints(n), rightRow(n);
+        for(auto row: points){
+            currMax = 0;
+
+            //Calculate maximum points from the right
+            //We do this as we need to change the value of maxPoints[j]
+            for(int j = n-1; j >= 0; j--){
+                currMax = max(currMax, maxPoints[j]);
+                rightRow[j] = currMax--;
+            }
+
+            currMax = 0; //Maximum points from the left
+            for(int j = 0; j < n; j++){
+                currMax = max(currMax, maxPoints[j]);
+                //Consider maximum points possible if we pick 
+                //the cell of index j of current row
+                maxPoints[j] = max(currMax--, rightRow[j]) + row[j];
+            }
+        }
+
+        // return maximum possible amount of points
+        return *max_element(maxPoints.begin(), maxPoints.end());
+    }
+};