Given a list containing future prediction of share prices, find maximum profit that can be earned by buying and selling shares any number of times with constraint that a new transaction can only start after previous transaction is complete. i.e. we can only hold at-most one share at a time.

**rate[]:** {1, 5, 2, 3, 7, 6, 4, 5}

Total profit earned is 10

Buy on day 1 and sell on day 2

Buy on day 3 and sell on day 5

Buy on day 7 and sell on day 8

**rate[]:** {10, 8, 6, 5, 4, 2}

Total profit earned is 0

The idea is very simple. We traverse the given list of prices and find local minimum of every increasing sequence. For example, in the array {1, 5, 2, 3, 7, 6, 4, 5}, below are three increasing sequences of length 2 or more.

{1, 5}

{2, 3, 7}

{4, 5}

The local minimum of each sequence is 1, 2 and 4 respectively. We can gain maximum profit if we buy the shares at the starting of every increasing sequence (local minimum) and sell them at the end of the increasing sequence (local maximum).

**C++ implementation –**

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#include <bits/stdc++.h> using namespace std; // Function to find maximum profit that can be earned by buying and // selling shares any number of times int maxProfit(int rate[], int n) { // store maximum profit gained int profit = 0; // initialize local minimum to first element's index int j = 0; // start from second element for (int i = 1; i < n; i++) { // update local minimum if decreasing sequence is found if (rate[i - 1] > rate[i]) j = i; // sell shares if current element is peak // i.e. (previous < current > next) if (rate[i - 1] < rate[i] && (i + 1 == n || rate[i] > rate[i + 1])) { profit += (rate[i] - rate[j]); printf("Buy on day %d and sell on day %d\n", j + 1, i + 1); } } return profit; } // main function int main() { int rate[] = { 1, 5, 2, 3, 7, 6, 4, 5 }; int n = sizeof(rate) / sizeof(rate[0]); cout << "\nTotal profit earned is " << maxProfit(rate, n); return 0; } |

`Output:`

Buy on day 1 and sell on day 2

Buy on day 3 and sell on day 5

Buy on day 7 and sell on day 8

Total profit earned is 10

The time complexity of above solution is O(n) and auxiliary space used by the program is O(1).

**Thanks for reading.**

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Java implementation http://ideone.com/IfgpAW