Created
August 18, 2020 20:58
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| class Solution { | |
| public: | |
| int maxProfit(vector<int>& prices) { | |
| int n = prices.size(); | |
| if(n==0) | |
| return 0; | |
| vector<int> left(n),right(n); | |
| //Fill 1st transaction (LEFT) | |
| int leftMin = prices[0]; | |
| for(int i=1;i<n;++i) | |
| { | |
| left[i] = max(left[i-1],prices[i]-leftMin); | |
| leftMin = min(leftMin,prices[i]); | |
| } | |
| //Fill 2nd transaction (RIGHT) | |
| int rightMax = prices[n-1]; | |
| for(int i=n-2;i>=0;--i) | |
| { | |
| right[i] = max(right[i+1],rightMax-prices[i]); | |
| rightMax = max(rightMax,prices[i]); | |
| } | |
| //Find the max-profit value | |
| int profit = right[0]; | |
| for(int i=1;i<n;++i) | |
| profit = max(profit,left[i-1]+right[i]); | |
| return profit; | |
| } | |
| }; | |
| //Memoization | |
| class Solution { | |
| /* | |
| pos->current position | |
| t->transactions done | |
| bought->If current stock is bought | |
| */ | |
| vector<vector<vector<int>>> mem; | |
| int recursion(vector<int>& prices,int pos,int t,bool bought) | |
| { | |
| if(pos>=prices.size() || t==0) //Out of bounds case | |
| return 0; | |
| if(mem[bought][t][pos]!=-1) //Return if already calculated | |
| return mem[bought][t][pos]; | |
| //3 choices for a position-->Buy/Sell/Skip | |
| int result = recursion(prices,pos+1,t,bought); //SKIP | |
| if(bought) | |
| result = max(result,recursion(prices,pos+1,t-1,false)+prices[pos]); //SELL | |
| else | |
| result = max(result,recursion(prices,pos+1,t,true)-prices[pos]); //BUY | |
| mem[bought][t][pos] = result; | |
| return result; | |
| } | |
| public: | |
| int maxProfit(vector<int>& prices) { | |
| mem.resize(2,vector<vector<int>>(3,vector<int>(prices.size(),-1))); | |
| int res = recursion(prices,0,2,false); | |
| return res; | |
| } | |
| }; |
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great explanation.