You are given an array prices where prices[i] is the price of a given stock on the ith day.
Find the maximum profit you can achieve. You may complete at most two transactions.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Example 1:
Input: prices = [3,3,5,0,0,3,1,4]
Output: 6
Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.
Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:
Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
Constraints:
1 <= prices.length <= 105
0 <= prices[i] <= 105
Intuition
Just introduce another state number of tries (i.e is 2)
Becomes 3D Dp and solve as same as Earlier question
class Solution {
public:
int find_ans(vector<int>& prices, int index, bool buy, int tries,
vector<vector<vector<int>>> &dp){
if(index == prices.size() or tries == 0)
return 0;
if(dp[index][buy][tries] != -1)
return dp[index][buy][tries];
int profit;
if(buy){
profit = max( -prices[index] + find_ans(prices, index+1, false, tries, dp),
find_ans(prices, index+1, true, tries, dp) );
}
else{
profit = max( +prices[index] + find_ans(prices, index+1, true, tries-1, dp),
find_ans(prices, index+1, false, tries, dp) );
}
return dp[index][buy][tries] = profit;
}
int maxProfit(vector<int>& prices) {
int n = prices.size();
int tries = 2;
vector<vector<vector<int>>> dp(n, vector<vector<int>> (2, vector<int>(3,-1)));
return find_ans(prices, 0, true, tries, dp);
}
};
Approach 2:
Tabulation
C++
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
vector<vector<vector<int>>> dp(n+1, vector<vector<int>> (2, vector<int>(3,0)));
for(int index=n-1; index>=0; index--){
for(int buy=0; buy<=1; buy++){
for(int tries=1; tries<=2; tries++){
// Starting tries from 1 not 0 because it is base case condition
//At tries = 0 we return 0 as cannot buy
int profit;
if(buy){
profit = max( -prices[index] + dp[index+1][0][tries],
dp[index+1][1][tries]);
}
else{
profit = max( +prices[index] + dp[index+1][1][tries-1],
dp[index+1][0][tries]);
}
dp[index][buy][tries] = profit;
}
}
}
return dp[0][1][2];
}
};
Approach 3:
Space Optimized
C++
class Solution {
public:
int maxProfit(vector<int>& prices) {
int n = prices.size();
vector<vector<int>> cur(2, vector<int>(3,0));
vector<vector<int>> prev(2, vector<int>(3,0));
for(int index=n-1; index>=0; index--){
for(int buy=0; buy<=1; buy++){
for(int tries=1; tries<=2; tries++){
// Starting tries from 1 not 0 because it is base case condition
//At tries = 0 we return 0 as cannot buy
int profit;
if(buy){
profit = max( -prices[index] + prev[0][tries],
prev[1][tries]);
}
else{
profit = max( +prices[index] + prev[1][tries-1],
prev[0][tries]);
}
cur[buy][tries] = profit;
}
}
prev = cur;
}
return prev[1][2];
}
};
Approach 4:
C++
Can maintain a N*4(Transaction number) State
Instead of N*2(Buy)*3(Transaction count)
In this
0 1 2 3
B S B S
On even tries you buy and odd you sell
Base Cases will be
At trasaction number == 4 we return 0