Maximum Profit in Stock Buy and sell with at most K Transaction

In stock market, a person buys a stock and sells it on some future date. Given the stock prices of N days in form of an array Amount and a positive integer K, find out the maximum profit a person can make in at most K transactions. A transaction is buying the stock on some day and selling the stock at some other future day and new transaction can start only when the previous transaction has been completed.

    Input:
    K=3
    N=7

    Stock prices on N days:
    10 25 38 40 45 5 58

    Output:
    88

Example:

    Number of maximum transactions: 3
    Total number of days, N: 7

    To achieve maximum profit:
    Stock bought at day1    (-10)
    Stock sold at day5      (+45)
    Stock bought at day6    (-5)
    Stock sold at day7      (+58)

    Total profit = 88
    Total transactions made = 2

Explanation:

Let there are N number of days for transactions, say x1, x2, ..., xn

Now for any day xi

Don't do any transaction on the day xi
Transact on day xi, i.e., buy stock on some day xj and sell on day xi where j<i and i,j Є N

Total number of transactions can be made at most = K

Now we can formulate the maximum profit case using above two condition.

Let,

    f(t,i)=maximum profit upto ith day and t transactions

Considering the above two facts about xi

    f(t,i)=f(t,i-1)  if there is no transaction made on day xi ...(1)
        Max(f(t-1,j)+amount[i]-amount[j])       
        where j<i and i,j Є N if there is transaction  on day xi ...(2)
    Obviously, to maximize the profit we would take the maximum of (1)  and (2)

    Number of maximum transactions: 3
    Total number of days, N: 7
    Stock prices on the days are:
    10 25 38 40 45 5 58

Below is a part of recursion tree which can show that how many overlapping sub problems there will be.

Figure 1: Partial recursion tree to show overlapping sub-problems

So we need dynamic programming...

Function(t, i): //f(t, i) If i=0 f(t, i)=0 If t=0 f(t, i)=0 f(t, i)= f(t, i-1); //no transaction on day xi For j =0: i Find max(f(t-1,j) + amount(i)- amount(j) End For If the maximum found > f(t, i) update f(t, i) End IF Return f(t, i)

For tabulation we need a 2D array, DP[k+1][n] to store f(t,i) Base case, for i 0 to k, DP[i][0]=0 //no profit on 0th day for i 0 to n-1, DP[0][j]=0 //no profit on 0 transaction To fill the higher values, for t=1 to k for i =1 to n-1 DP[t][i]=DP[t][i-1] for j= 0 to i-1 //buying on jth day and selling on ith day DP[t][i]=max(DP[t][i],DP[t-1][j]+ amount[i] –amount[j]) End for End for End for Result would be f(k,n) that is value of DP[k][n-1]

#include <bits/stdc++.h> using namespace std; int Max_profit(vector<int> amount, int n, int k) { int DP[k + 1][n]; memset(DP, 0, sizeof(DP)); //on 0th day for (int i = 0; i <= k; i++) DP[i][0] = 0; //on 0 transaction made for (int i = 0; i <= n; i++) DP[0][i] = 0; for (int t = 1; t <= k; t++) { for (int i = 1; i < n; i++) { DP[t][i] = DP[t][i - 1]; int maxV = INT_MIN; //buying on jth day and selling on ith day for (int j = 0; j < i; j++) { if (maxV < DP[t - 1][j] + amount[i] - amount[j]) maxV = DP[t - 1][j] + amount[i] - amount[j]; } if (DP[t][i] < maxV) DP[t][i] = maxV; } } return DP[k][n - 1]; } int main() { int n, item, k; cout << "Enter maximum no of transactions:\n"; cin >> k; cout << "Enter number of days\n"; cin >> n; vector<int> amount; cout << "Enter stock values on corresponding days\n"; for (int j = 0; j < n; j++) { scanf("%d", &item); amount.push_back(item); } cout << "Maximum profit that can be achieved: " << Max_profit(amount, n, k) << endl; return 0; }

n stock market, a person buys a stock and sells it on some future date. Given the stock prices of N days in form of an array Amount and a positive integer K, find out the maximum profit a person can make in at most K transactions

Maximum Profit in Stock Buy and sell with at most K Transaction

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