Merge pull request neetcode-gh#1541 from elcabalero/patch-10

Updated code as in the video adapting to C++

cpp/neetcode_150/14_2-d_dynamic_programming/best_time_to_buy_and_sell_stock_with_cooldown.cpp

@@ -7,7 +7,7 @@
  sell rest buy
 
  Time: O(n)
- Space: O(1) -> optimized from O(n) since only need i - 1 prev state
+ Space: O(n)
 */
 
 // class Solution {
@@ -28,7 +28,9 @@
 // return max(s0[n - 1], s2[n - 1]);
 // }
 // };
-
+//
+// Optimized solution with O(1) space follows
+/*
 class Solution {
 public:
  int maxProfit(vector<int>& prices) {
@@ -46,3 +48,37 @@ class Solution {
  return max(sold, rest);
  }
 };
+*/
+
+// NeetCode's solution
+// Uses an array of arrays of size two (i.e. a n x 2 matrix)
+// representing at position 0 of each row the "selling" value
+// whereas at position 1 the "buying" value.
+// Could've used a map<pair<int,int>> instead of the bidimensional
+// array at the expense of logn search.
+class Solution {
+public:
+ int maxProfit(vector<int>& prices) {
+ vector<vector<int>> DP(prices.size(), vector<int>(2, -1));
+ return dfs(prices, DP, 0, 1);
+ }
+
+private:
+ int dfs(vector<int>& prices, vector<vector<int>>& DP, int i, int buying){
+ if (i >= prices.size())
+ return 0;
+ if (DP[i][buying] != -1)
+ return DP[i][buying];
+
+ int cooldown = dfs(prices, DP, i+1, buying);
+ if (buying){
+ int buy = dfs(prices, DP, i+1, 0) - prices[i];
+ DP[i][buying] = max(buy, cooldown);
+ } else {
+ int sell = dfs(prices, DP, i+2, 1) + prices[i];
+ DP[i][buying] = max(sell, cooldown);
+ }
+
+ return DP[i][buying];
+ }
+};