We are given an array Arr[] of length n. It represents the price of a stock on ‘n’ days. The following guidelines need to be followed:
- We can buy and sell the stock any number of times.
- In order to sell the stock, we need to first buy it on the same or any previous day.
- We can’t buy a stock again after buying it once. In other words, we first buy a stock and then sell it. After selling we can buy and sell again. But we can’t sell before buying and can’t buy before selling any previously bought stock.
- After every transaction, there is a transaction fee (‘fee’) associated with it.
Examples
Practice:
Disclaimer: Don’t jump directly to the solution, try it out yourself first.
Memoization Approach
Algorithm / Intuition
Intuition:
Every day, we will have two choices, either to do nothing and move to the next day or to buy/sell (based on the last transaction) and find out the profit. Therefore we need to generate all the choices in order to compare the profit. As we need to try out all the possible choices, we will use recursion.
Steps to form the recursive solution:
We will first form the recursive solution by the three points mentioned in the Dynamic Programming Introduction.
Step 1: Express the problem in terms of indexes.
We need to think in the terms of the number of days, therefore one variable will be the array index( say ind). Next, we need to respect the condition that we can’t buy a stock again, that is we need to first sell a stock, and then we can buy that again. Therefore we need a second variable ‘buy’ which tells us on a particular day whether we can buy or sell the stock. We can generalize the function as :
Step 2: Try out all possible choices at a given index.
Every day, we have two choices:
- To either buy/sell the stock(based on the buy variable’s value).
- To do nothing and move on to the next day.
We need to generate all the choices. We will use the pick/non-pick technique as discussed in this video “Recursion on Subsequences”.
Case 1: When buy == 0, we can buy the stock.
If we can buy the stock on a particular day, we have two options:
- Option 1: To do no transaction and move to the next day. In this case, the net profit earned will be 0 from the current transaction, and to calculate the maximum profit starting from the next day, we will recursively call f(ind+1,0). As we have not bought the stock, the ‘buy’ variable value will still remain 0, indicating that we can buy the stock the next day.
- Option 2: The other option is to buy the stock on the current day. In this case, the net profit earned from the current transaction will be -Arr[i]. As we are buying the stock, we are giving money out of our pocket, therefore the profit we earn is negative. To calculate the maximum profit starting from the next day, we will recursively call f(ind+1,1). As we have bought the stock, the ‘buy’ variable value will change to 1, indicating that we can’t buy and only sell the stock the next day.
Case 2: When buy == 1, we can sell the stock.
If we can buy the stock on a particular day, we have two options:
- Option 1: To do no transaction and move to the next day. In this case, the net profit earned will be 0 from the current transaction, and to calculate the maximum profit starting from the next day, we will recursively call f(ind+1,1). As we have not bought the stock, the ‘buy’ variable value will still remain at 1, indicating that we can’t buy and only sell the stock the next day.
- Option 2: The other option is to sell the stock on the current day. In this case, the net profit earned from the current transaction will be +Arr[i]. As we are selling the stock, we are putting the money into our pocket, therefore the profit we earn is positive. Next, as we have bought and sold the stock, we have made one complete transaction. Hence we will incur the transaction fees. So we subtract that from the profit earned. To calculate the maximum profit starting from the next day, we will recursively call f(ind+1,0). As we have sold the stock, the ‘buy’ variable value will change to 0, indicating that we can buy the stock the next day.
Note: Buying and selling together count as one complete transaction. Therefore we will account for the fee only once ( after selling the stock).
The figure below gives us the summary:
Step 3: Return the maximum
As we are looking to maximize the profit earned, we will return the maximum value in both cases.
The final pseudocode after steps 1, 2, and 3:
Base Cases:
- If ind==n, it means we have finished trading on all days, and there is no more money that we can get, therefore we simply return 0.
Steps to memoize a recursive solution:
If we draw the recursion tree, we will see that there are overlapping subproblems. In order to convert a recursive solution the following steps will be taken:
- Create a dp array of size [n][2]. The size of the input array is ‘n’, so the index will always lie between ‘0’ and ‘n-1’. The ‘buy’ variable can take only two values: 0 and 1. Therefore we take the dp array as dp[n][2]
- We initialize the dp array to -1.
- Whenever we want to find the answer to particular parameters (say f(ind, buy)), we first check whether the answer is already calculated using the dp array(i.e dp[ind][buy]!= -1 ). If yes, simply return the value from the dp array.
- If not, then we are finding the answer for the given value for the first time, we will use the recursive relation as usual but before returning from the function, we will set dp[ind][buy] to the solution we get.
Code
#include <bits/stdc++.h>
using namespace std;
int getAns(vector<int> &Arr, int ind, int buy, int n, int fee, vector<vector<int>> &dp) {
// Base case: If we've reached the end of the array, return 0 profit.
if (ind == n) return 0;
// Check if the result is already computed
if (dp[ind][buy] != -1)
return dp[ind][buy];
int profit;
if (buy == 0) { // We can buy the stock
profit = max(0 + getAns(Arr, ind + 1, 0, n, fee, dp), -Arr[ind] + getAns(Arr, ind + 1, 1, n, fee, dp));
}
if (buy == 1) { // We can sell the stock
profit = max(0 + getAns(Arr, ind + 1, 1, n, fee, dp), Arr[ind] - fee + getAns(Arr, ind + 1, 0, n, fee, dp));
}
// Store the computed profit in the DP array
return dp[ind][buy] = profit;
}
int maximumProfit(int n, int fee, vector<int> &Arr) {
vector<vector<int>> dp(n, vector<int>(2, -1));
if (n == 0) return 0; // Edge case: No stocks to trade.
int ans = getAns(Arr, 0, 0, n, fee, dp);
return ans;
}
int main() {
vector<int> prices = {1, 3, 2, 8, 4, 9};
int n = prices.size();
int fee = 2;
cout << "The maximum profit that can be generated is " << maximumProfit(n, fee, prices) << endl;
return 0;
}
import java.util.*;
class TUF {
// Function to calculate the maximum profit
static int getAns(int[] Arr, int ind, int buy, int n, int fee, int[][] dp) {
// Base case
if (ind == n) {
return 0;
}
// If the result is already calculated, return it
if (dp[ind][buy] != -1) {
return dp[ind][buy];
}
int profit = 0;
if (buy == 0) { // We can buy the stock
profit = Math.max(0 + getAns(Arr, ind + 1, 0, n, fee, dp), -Arr[ind] + getAns(Arr, ind + 1, 1, n, fee, dp));
}
if (buy == 1) { // We can sell the stock
profit = Math.max(0 + getAns(Arr, ind + 1, 1, n, fee, dp), Arr[ind] - fee + getAns(Arr, ind + 1, 0, n, fee, dp));
}
// Store the result in dp and return it
dp[ind][buy] = profit;
return profit;
}
// Function to calculate the maximum profit with a transaction fee
static int maximumProfit(int n, int fee, int[] Arr) {
int dp[][] = new int[n][2];
// Initialize dp array with -1 to mark states as not calculated yet
for (int row[] : dp) {
Arrays.fill(row, -1);
}
if (n == 0) {
return 0;
}
int ans = getAns(Arr, 0, 0, n, fee, dp);
return ans;
}
public static void main(String args[]) {
int prices[] = {1, 3, 2, 8, 4, 9};
int n = prices.length;
int fee = 2;
System.out.println("The maximum profit that can be generated is " + maximumProfit(n, fee, prices));
}
}
def get_max_profit(prices, ind, buy, n, fee, dp):
# Base case: if we reach the end of the array
if ind == n:
return 0
# Check if the result is already calculated
if dp[ind][buy] != -1:
return dp[ind][buy]
profit = 0
if buy == 0: # We can buy the stock
profit = max(
0 + get_max_profit(prices, ind + 1, 0, n, fee, dp),
-prices[ind] + get_max_profit(prices, ind + 1, 1, n, fee, dp)
)
if buy == 1: # We can sell the stock
profit = max(
0 + get_max_profit(prices, ind + 1, 1, n, fee, dp),
prices[ind] - fee + get_max_profit(prices, ind + 1, 0, n, fee, dp)
)
dp[ind][buy] = profit
return profit
def maximum_profit(n, fee, prices):
dp = [[-1 for _ in range(2)] for _ in range(n)]
if n == 0:
return 0
ans = get_max_profit(prices, 0, 0, n, fee, dp)
return ans
if __name__ == "__main__":
prices = [1, 3, 2, 8, 4, 9]
n = len(prices)
fee = 2
result = maximum_profit(n, fee, prices)
print(f"The maximum profit that can be generated is {result}")
function getAns(Arr, ind, buy, n, fee, dp) {
// Base case: if the index reaches n, return 0
if (ind === n) return 0;
// Check if the result is already computed and stored in dp
if (dp[ind][buy] !== -1) return dp[ind][buy];
let profit;
if (buy === 0) { // We can buy the stock
profit = Math.max(
0 + getAns(Arr, ind + 1, 0, n, fee, dp),
-Arr[ind] + getAns(Arr, ind + 1, 1, n, fee, dp)
);
}
if (buy === 1) { // We can sell the stock
profit = Math.max(
0 + getAns(Arr, ind + 1, 1, n, fee, dp),
Arr[ind] - fee + getAns(Arr, ind + 1, 0, n, fee, dp)
);
}
// Store the result in dp and return it
dp[ind][buy] = profit;
return profit;
}
function maximumProfit(n, fee, Arr) {
// Create a 2D dp array filled with -1 to store computed results
const dp = new Array(n).fill().map(() => new Array(2).fill(-1));
// Check if n is 0 and return 0 if so
if (n === 0) return 0;
// Call the recursive function to compute the maximum profit
const ans = getAns(Arr, 0, 0, n, fee, dp);
return ans;
}
// Main function
function main() {
const prices = [1, 3, 2, 8, 4, 9];
const n = prices.length;
const fee = 2;
const result = maximumProfit(n, fee, prices);
console.log("The maximum profit that can be generated is", result);
}
// Call the main function
main();
The maximum profit that can be generated is 8
Complexity Analysis
Time Complexity: O(N*2)
Reason: There are N*2 states therefore at max ‘N*2’ new problems will be solved and we are running a for loop for ‘N’ times to calculate the total sum
Space Complexity: O(N*2) + O(N)
Reason: We are using a recursion stack space(O(N)) and a 2D array ( O(N*2)).
Tabulation Approach
Algorithm / Intuition
To convert the memoization approach to a tabulation one, create a dp array with the same size as done in memoization. We can set its type as bool and initialize it as false.
- First, we declare the dp array of size [n+1][2] as zero.
- Next, we set the base condition, we set dp[n][0] = dp[n][1] = 0(the case when we had exhausted the number of days of the stock market).
- Next, we set two nested for loops, the outer loop (for variable ind) moves from n-1 to 0 and the inner loop (for variable buy) moves from 0 to 1.
- In every iteration, we calculate the value according to the memoization logic.
- At last, we will print dp[0][0] as our answer.
Code
#include <bits/stdc++.h>
using namespace std;
int maximumProfit(int n, int fee, vector<int>& Arr) {
if (n == 0) return 0; // Edge case: No stocks to trade.
// Create a 2D DP array with dimensions (n+1) x 2, initialized to 0
vector<vector<int>> dp(n + 1, vector<int>(2, 0));
// Loop through the stock prices from the end to the beginning
for (int ind = n - 1; ind >= 0; ind--) {
for (int buy = 0; buy <= 1; buy++) {
int profit;
if (buy == 0) { // We can buy the stock
profit = max(0 + dp[ind + 1][0], -Arr[ind] + dp[ind + 1][1]);
}
if (buy == 1) { // We can sell the stock
profit = max(0 + dp[ind + 1][1], Arr[ind] - fee + dp[ind + 1][0]);
}
dp[ind][buy] = profit;
}
}
return dp[0][0]; // Return the maximum profit for buying.
}
int main() {
vector<int> prices = {1, 3, 2, 8, 4, 9};
int n = prices.size();
int fee = 2;
cout << "The maximum profit that can be generated is " << maximumProfit(n, fee, prices) << endl;
return 0;
}
import java.util.*;
class TUF {
// Function to calculate the maximum profit
static int maximumProfit(int n, int fee, int[] Arr) {
// Handle the base case when n is 0
if (n == 0) {
return 0;
}
int dp[][] = new int[n + 1][2];
// Iterate through the array backwards
for (int ind = n - 1; ind >= 0; ind--) {
for (int buy = 0; buy <= 1; buy++) {
int profit = 0;
if (buy == 0) { // We can buy the stock
profit = Math.max(0 + dp[ind + 1][0], -Arr[ind] + dp[ind + 1][1]);
}
if (buy == 1) { // We can sell the stock
profit = Math.max(0 + dp[ind + 1][1], Arr[ind] - fee + dp[ind + 1][0]);
}
dp[ind][buy] = profit;
}
}
// The maximum profit is stored in dp[0][0]
return dp[0][0];
}
public static void main(String args[]) {
int prices[] = {1, 3, 2, 8, 4, 9};
int n = prices.length;
int fee = 2;
System.out.println("The maximum profit that can be generated is " + maximumProfit(n, fee, prices));
}
}
def maximum_profit(n, fee, prices):
if n == 0:
return 0
# Create a 2D dp array of size [n+1][2] initialized to 0
dp = [[0 for _ in range(2)] for _ in range(n + 1)]
# Loop through the array from right to left
for ind in range(n - 1, -1, -1):
for buy in range(2):
profit = 0
if buy == 0: # We can buy the stock
profit = max(
0 + dp[ind + 1][0],
-prices[ind] + dp[ind + 1][1]
)
if buy == 1: # We can sell the stock
profit = max(
0 + dp[ind + 1][1],
prices[ind] - fee + dp[ind + 1][0]
)
dp[ind][buy] = profit
return dp[0][0]
if __name__ == "__main__":
prices = [1, 3, 2, 8, 4, 9]
n = len(prices)
fee = 2
result = maximum_profit(n, fee, prices)
print(f"The maximum profit that can be generated is {result}")
function maximumProfit(n, fee, Arr) {
// Check if n is 0 and return 0 if so
if (n === 0) return 0;
// Create a 2D dp array of size (n+1) x 2, initialized with 0s
const dp = new Array(n + 1).fill().map(() => new Array(2).fill(0));
// The base condition is handled by initializing the dp array as 0
// Loop through the prices in reverse order
for (let ind = n - 1; ind >= 0; ind--) {
for (let buy = 0; buy <= 1; buy++) {
let profit;
if (buy === 0) { // We can buy the stock
profit = Math.max(
0 + dp[ind + 1][0],
-Arr[ind] + dp[ind + 1][1]
);
}
if (buy === 1) { // We can sell the stock
profit = Math.max(
0 + dp[ind + 1][1],
Arr[ind] - fee + dp[ind + 1][0]
);
}
dp[ind][buy] = profit;
}
}
// Return the maximum profit that can be generated
return dp[0][0];
}
// Main function
function main() {
const prices = [1, 3, 2, 8, 4, 9];
const n = prices.length;
const fee = 2;
const result = maximumProfit(n, fee, prices);
console.log("The maximum profit that can be generated is", result);
}
// Call the main function
main();
The maximum profit that can be generated is 8
Complexity Analysis
Time Complexity: O(N*2)
Reason: There are two nested loops that account for O(N*2) complexity.
Space Complexity: O(N*2)
Reason: We are using an external array of size ‘N*2’. Stack Space is eliminated.
Space Optimization Approach
Algorithm / Intuition
If we closely look the relation,
dp[ind][buy] = max( dp[ind+1][buy] , max( dp[ind+1][!buy])
We see that to calculate a value of a cell of the dp array, we need only the next column values(say ahead). So, we don’t need to store an entire 2-D array. Hence we can space optimize it.
- We set the ahead column as 0 (base condition)
- Then we set the nested loops to calculate the cur column values.
- We replace dp[ind] with cur and dp[ind+1] with ahead in our memoization code.
- After the inner loop execution, we set ahead as cur for the next outer loop iteration.
- At last, we return cur[0] as our answer.
Code
#include <bits/stdc++.h>
using namespace std;
int maximumProfit(int n, int fee, vector<int>& Arr) {
if (n == 0) return 0; // Edge case: No stocks to trade.
vector<long> ahead(2, 0); // To track maximum profit one step ahead
vector<long> cur(2, 0); // To track current maximum profit
// Initialize both ahead[0] and ahead[1] to 0 as the base condition
ahead[0] = ahead[1] = 0;
long profit;
for (int ind = n - 1; ind >= 0; ind--) {
for (int buy = 0; buy <= 1; buy++) {
if (buy == 0) { // We can buy the stock
profit = max(0 + ahead[0], -Arr[ind] + ahead[1]);
}
if (buy == 1) { // We can sell the stock
profit = max(0 + ahead[1], Arr[ind] - fee + ahead[0]);
}
cur[buy] = profit;
}
ahead = cur;
}
return cur[0]; // Return the maximum profit for buying.
}
int main() {
vector<int> prices = {1, 3, 2, 8, 4, 9};
int n = prices.size();
int fee = 2;
cout << "The maximum profit that can be generated is " << maximumProfit(n, fee, prices) << endl;
return 0;
}
import java.util.*;
class TUF {
// Function to calculate the maximum profit
static long maximumProfit(int n, int fee, int[] Arr) {
// Handle the base case when n is 0
if (n == 0) {
return 0;
}
long ahead[] = new long[2];
long cur[] = new long[2];
// Initialize base conditions
ahead[0] = ahead[1] = 0;
long profit = 0;
for (int ind = n - 1; ind >= 0; ind--) {
for (int buy = 0; buy <= 1; buy++) {
if (buy == 0) { // We can buy the stock
profit = Math.max(0 + ahead[0], -Arr[ind] + ahead[1]);
}
if (buy == 1) { // We can sell the stock
profit = Math.max(0 + ahead[1], Arr[ind] - fee + ahead[0]);
}
cur[buy] = profit;
}
ahead = (long[]) (cur.clone());
}
return cur[0];
}
public static void main(String args[]) {
int prices[] = {1, 3, 2, 8, 4, 9};
int n = prices.length;
int fee = 2;
System.out.println("The maximum profit that can be generated is " + maximumProfit(n, fee, prices));
}
}
def maximum_profit(n, fee, prices):
if n == 0:
return 0
# Initialize two lists to track the profit states
ahead = [0, 0]
cur = [0, 0]
# Base condition
ahead[0] = ahead[1] = 0
for ind in range(n - 1, -1, -1):
for buy in range(2):
profit = 0
if buy == 0: # We can buy the stock
profit = max(
0 + ahead[0],
-prices[ind] + ahead[1]
)
if buy == 1: # We can sell the stock
profit = max(
0 + ahead[1],
prices[ind] - fee + ahead[0]
)
cur[buy] = profit
# Update the 'ahead' list for the next iteration
ahead = cur.copy()
return cur[0]
if __name__ == "__main__":
prices = [1, 3, 2, 8, 4, 9]
n = len(prices)
fee = 2
result = maximum_profit(n, fee, prices)
print(f"The maximum profit that can be generated is {result}")
function maximumProfit(n, fee, Arr) {
// Check if n is 0 and return 0 if so
if (n === 0) return 0;
// Create arrays 'ahead' and 'cur', each initialized with two zeros
let ahead = [0, 0];
let cur = [0, 0];
// Base condition: Initialize 'ahead' with zeros
ahead[0] = ahead[1] = 0;
let profit;
// Loop through the prices in reverse order
for (let ind = n - 1; ind >= 0; ind--) {
for (let buy = 0; buy <= 1; buy++) {
if (buy === 0) { // We can buy the stock
profit = Math.max(
0 + ahead[0],
-Arr[ind] + ahead[1]
);
}
if (buy === 1) { // We can sell the stock
profit = Math.max(
0 + ahead[1],
Arr[ind] - fee + ahead[0]
);
}
cur[buy] = profit;
}
// Update 'ahead' with 'cur' for the next iteration
ahead = [...cur];
}
// Return the maximum profit that can be generated
return cur[0];
}
// Main function
function main() {
const prices = [1, 3, 2, 8, 4, 9];
const n = prices.length;
const fee = 2;
const result = maximumProfit(n, fee, prices);
console.log("The maximum profit that can be generated is", result);
}
// Call the main function
main();
The maximum profit that can be generated is 8
Complexity Analysis
Time Complexity: O(N*2)
Reason: There are two nested loops that account for O(N*2) complexity
Space Complexity: O(1)
Reason: We are using an external array of size ‘2’.
Video Explanation
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