Stock Buy Sell to Maximize Profit - GeeksforGeeks
Stock Buy Sell to Maximize Profit
- Difficulty Level : Medium
- Last Updated : 23 Aug, 2021
The cost of a stock on each day is given in an array, find the max profit that you can make by buying and selling in those days. For example, if the given array is {100, 180, 260, 310, 40, 535, 695}, the maximum profit can earned by buying on day 0, selling on day 3. Again buy on day 4 and sell on day 6. If the given array of prices is sorted in decreasing order, then profit cannot be earned at all.
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
Naive approach: A simple approach is to try buying the stocks and selling them on every single day when profitable and keep updating the maximum profit so far.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the maximum profit
// that can be made after buying and
// selling the given stocks
int maxProfit(int price[], int start, int end)
{
// If the stocks can't be bought
if (end <= start)
return 0;
// Initialise the profit
int profit = 0;
// The day at which the stock
// must be bought
for (int i = start; i < end; i++) {
// The day at which the
// stock must be sold
for (int j = i + 1; j <= end; j++) {
// If buying the stock at ith day and
// selling it at jth day is profitable
if (price[j] > price[i]) {
// Update the current profit
int curr_profit = price[j] - price[i]
+ maxProfit(price, start, i - 1)
+ maxProfit(price, j + 1, end);
// Update the maximum profit so far
profit = max(profit, curr_profit);
}
}
}
return profit;
}
// Driver code
int main()
{
int price[] = { 100, 180, 260, 310,
40, 535, 695 };
int n = sizeof(price) / sizeof(price[0]);
cout << maxProfit(price, 0, n - 1);
return 0;
}
Output
865
Efficient approach: If we are allowed to buy and sell only once, then we can use following algorithm. Maximum difference between two elements. Here we are allowed to buy and sell multiple times.
Following is the algorithm for this problem.
- Find the local minima and store it as starting index. If not exists, return.
- Find the local maxima. and store it as an ending index. If we reach the end, set the end as the ending index.
- Update the solution (Increment count of buy-sell pairs)
- Repeat the above steps if the end is not reached.
- C++
- C
- Java
- Python3
- C#
- Javascript
// C++ Program to find best buying and selling days
#include <bits/stdc++.h>
using namespace std;
// This function finds the buy sell
// schedule for maximum profit
void stockBuySell(int price[], int n)
{
// Prices must be given for at least two days
if (n == 1)
return;
// Traverse through given price array
int i = 0;
while (i < n - 1) {
// Find Local Minima
// Note that the limit is (n-2) as we are
// comparing present element to the next element
while ((i < n - 1) && (price[i + 1] <= price[i]))
i++;
// If we reached the end, break
// as no further solution possible
if (i == n - 1)
break;
// Store the index of minima
int buy = i++;
// Find Local Maxima
// Note that the limit is (n-1) as we are
// comparing to previous element
while ((i < n) && (price[i] >= price[i - 1]))
i++;
// Store the index of maxima
int sell = i - 1;
cout << "Buy on day: " << buy
<< "\t Sell on day: " << sell << endl;
}
}
// Driver code
int main()
{
// Stock prices on consecutive days
int price[] = { 100, 180, 260, 310, 40, 535, 695 };
int n = sizeof(price) / sizeof(price[0]);
// Function call
stockBuySell(price, n);
return 0;
}
// This is code is contributed by rathbhupendra
Output
Buy on day: 0 Sell on day: 3Buy on day: 4 Sell on day: 6
Time Complexity: The outer loop runs till I become n-1. The inner two loops increment value of I in every iteration. So overall time complexity is O(n)
Valley Peak Approach:
In this approach, we just need to find the next greater element and subtract it from the current element so that the difference keeps increasing until we reach a minimum. If the sequence is a decreasing sequence so the maximum profit possible is 0.
- C++
- Java
#include <iostream>
using namespace std;
// Preprocessing helps the code run faster
#define fl(i, a, b) for (int i = a; i < b; i++)
// Function that return
int maxProfit(int* prices, int size)
{
// maxProfit adds up the difference between
// adjacent elements if they are in increaisng order
int maxProfit = 0;
// The loop starts from 1
// as its comparing with the previous
fl(i, 1, size) if (prices[i] > prices[i - 1]) maxProfit
+= prices[i] - prices[i - 1];
return maxProfit;
}
// Driver Function
int main()
{
int prices[] = { 100, 180, 260, 310, 40, 535, 695 };
int N = sizeof(prices) / sizeof(prices[0]);
cout << maxProfit(prices, N) << endl;
return 0;
}
// This code is contributed by Kingshuk Deb
Output
865
Time Complexity: O(n)
Auxiliary Space: O(1)
This article is compiled by Ashish Anand and reviewed by the GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
