Hello Everyone,
In daily share trading, a buyer buys shares in the morning and sells them on the same day. If the trader is allowed to make at most 2 transactions in a day, whereas the second transaction can only start after the first one is complete (Sell->buy->sell->buy). Given stock prices throughout the day, find out the maximum profit that a share trader could have made.
Examples:
Input: price[] = {10, 22, 5, 75, 65, 80}
Output: 87
Trader earns 87 as sum of 12, 75
Buy at 10, sell at 22,
Buy at 5 and sell at 80
Input: price[] = {2, 30, 15, 10, 8, 25, 80}
Output: 100
Trader earns 100 as sum of 28 and 72
Buy at price 2, sell at 30, buy at 8 and sell at 80
Input: price[] = {100, 30, 15, 10, 8, 25, 80};
Output: 72
Buy at price 8 and sell at 80.
Input: price[] = {90, 80, 70, 60, 50}
Output: 0
Not possible to earn.
A Simple Solution is to consider every index ‘i’ and do the following
Max profit with at most two transactions =
MAX {max profit with one transaction and subarray price[0…i] +
max profit with one transaction and subarray price[i+1…n-1] }
i varies from 0 to n-1.
Maximum possible using one transaction can be calculated using the following O(n) algorithm
The time complexity of the above simple solution is O(n2).
We can do this O(n) using the following Efficient Solution. The idea is to store the maximum possible profit of every subarray and solve the problem in the following two phases.
1) Create a table profit[0…n-1] and initialize all values in it 0.
2) Traverse price[] from right to left and update profit[i] such that profit[i] stores maximum profit achievable from one transaction in subarray price[i…n-1]
3) Traverse price[] from left to right and update profit[i] such that profit[i] stores maximum profit such that profit[i] contains maximum achievable profit from two transactions in subarray price[0…i].
4) Return profit[n-1]
To do step 2, we need to keep track of the maximum price from right to left side, and to do step 3, we need to keep track of the minimum price from left to right. Why we traverse in reverse directions? The idea is to save space, in the third step, we use the same array for both purposes, maximum with 1 transaction and maximum with 2 transactions. After iteration i, the array profit[0…i] contains the maximum profit with 2 transactions, and profit[i+1…n-1] contains profit with two transactions.
Below are the implementations of the above idea.
// C++ program to find maximum
// possible profit with at most
// two transactions
#include <bits/stdc++.h>
using namespace std;
// Returns maximum profit with
// two transactions on a given
// list of stock prices, price[0..n-1]
int maxProfit( int price[], int n)
{
// Create profit array and
// initialize it as 0
int * profit = new int [n];
for ( int i = 0; i < n; i++)
profit[i] = 0;
/* Get the maximum profit with
only one transaction
allowed. After this loop,
profit[i] contains maximum
profit from price[i..n-1]
using at most one trans. */
int max_price = price[n - 1];
for ( int i = n - 2; i >= 0; i--) {
// max_price has maximum
// of price[i..n-1]
if (price[i] > max_price)
max_price = price[i];
// we can get profit[i] by taking maximum of:
// a) previous maximum, i.e., profit[i+1]
// b) profit by buying at price[i] and selling at
// max_price
profit[i]
= max(profit[i + 1], max_price - price[i]);
}
/* Get the maximum profit with two transactions allowed
After this loop, profit[n-1] contains the result */
int min_price = price[0];
for ( int i = 1; i < n; i++) {
// min_price is minimum price in price[0..i]
if (price[i] < min_price)
min_price = price[i];
// Maximum profit is maximum of:
// a) previous maximum, i.e., profit[i-1]
// b) (Buy, Sell) at (min_price, price[i]) and add
// profit of other trans. stored in profit[i]
profit[i] = max(profit[i - 1],
profit[i] + (price[i] - min_price));
}
int result = profit[n - 1];
delete [] profit; // To avoid memory leak
return result;
}
// Driver code
int main()
{
int price[] = { 2, 30, 15, 10, 8, 25, 80 };
int n = sizeof (price) / sizeof (price[0]);
cout << "Maximum Profit = " << maxProfit(price, n);
return 0;
}
Output
Maximum Profit = 100
The time complexity of the above solution is O(n).
Algorithmic Paradigm: Dynamic Programming
There is one more approach for calculating this problem using the Valley-Peak approach i.e. take a variable profit and initialize it with zero and then traverse through the array of price[] from (i+1)th position whenever the initial position value is greater than the previous value add it to variable profit.
#include <iostream>
using namespace std;
int main()
{
int price[] = { 2, 30, 15, 10, 8, 25, 80 };
int n = 7;
// adding array
int profit = 0;
// Initializing variable
// valley-peak approach
for ( int i = 1; i < n; i++)
{
// traversing through array from (i+1)th
// position
int sub = price[i] - price[i - 1];
if (sub > 0)
profit += sub;
}
cout << "Maximum Profit=" << profit;
return 0;
}
Output
Maximum Profit=100
The time and space complexity is O(n) and O(1) respectively.