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# Stock Buy Sell to Maximize Profit

• Difficulty Level : Medium
• Last Updated : 29 Jul, 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.

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++

 `// C++ implementation of the approach``#include ``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);` `    ``cout << maxProfit(price, 0, n - 1);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the maximum profit``// that can be made after buying and``// selling the given stocks``static` `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 = Math.max(profit, curr_profit);``            ``}``        ``}``    ``}``    ``return` `profit;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `price[] = { ``100``, ``180``, ``260``, ``310``,``                    ``40``, ``535``, ``695` `};``    ``int` `n = price.length;` `    ``System.out.print(maxProfit(price, ``0``, n - ``1``));``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 implementation of the approach` `# Function to return the maximum profit``# that can be made after buying and``# selling the given stocks``def` `maxProfit(price, start, end):` `    ``# If the stocks can't be bought``    ``if` `(end <``=` `start):``        ``return` `0``;` `    ``# Initialise the profit``    ``profit ``=` `0``;` `    ``# The day at which the stock``    ``# must be bought``    ``for` `i ``in` `range``(start, end, ``1``):` `        ``# The day at which the``        ``# stock must be sold``        ``for` `j ``in` `range``(i``+``1``, end``+``1``):` `            ``# If buying the stock at ith day and``            ``# selling it at jth day is profitable``            ``if` `(price[j] > price[i]):``                ` `                ``# Update the current profit``                ``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``if` `__name__ ``=``=` `'__main__'``:``    ``price ``=` `[``100``, ``180``, ``260``, ``310``, ``40``, ``535``, ``695``];``    ``n ``=` `len``(price);` `    ``print``(maxProfit(price, ``0``, n ``-` `1``));` `# This code is contributed by Rajput-Ji`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `// Function to return the maximum profit``// that can be made after buying and``// selling the given stocks``static` `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 = Math.Max(profit, curr_profit);``            ``}``        ``}``    ``}``    ``return` `profit;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]price = { 100, 180, 260, 310,``                    ``40, 535, 695 };``    ``int` `n = price.Length;` `    ``Console.Write(maxProfit(price, 0, n - 1));``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
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.

1. Find the local minima and store it as starting index. If not exists, return.
2. Find the local maxima. and store it as an ending index. If we reach the end, set the end as the ending index.
3. Update the solution (Increment count of buy-sell pairs)
4. Repeat the above steps if the end is not reached.

## C++

 `// C++ Program to find best buying and selling days``#include ``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);` `    ``// Function call``    ``stockBuySell(price, n);` `    ``return` `0;``}` `// This is code is contributed by rathbhupendra`

## C

 `// Program to find best buying and selling days``#include ` `// solution structure``struct` `Interval {``    ``int` `buy;``    ``int` `sell;``};` `// 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``;` `    ``int` `count = 0; ``// count of solution pairs` `    ``// solution vector``    ``Interval sol[n / 2 + 1];` `    ``// 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``        ``sol[count].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``        ``sol[count].sell = i - 1;` `        ``// Increment count of buy/sell pairs``        ``count++;``    ``}` `    ``// print solution``    ``if` `(count == 0)``        ``printf``(``"There is no day when buying the stock will make profitn"``);``    ``else` `{``        ``for` `(``int` `i = 0; i < count; i++)``            ``printf``(``"Buy on day: %dt Sell on day: %dn"``, sol[i].buy, sol[i].sell);``    ``}` `    ``return``;``}` `// Driver program to test above functions``int` `main()``{``    ``// stock prices on consecutive days``    ``int` `price[] = { 100, 180, 260, 310, 40, 535, 695 };``    ``int` `n = ``sizeof``(price) / ``sizeof``(price);` `    ``// function call``    ``stockBuySell(price, n);` `    ``return` `0;``}`

## Java

 `// Program to find best buying and selling days``import` `java.util.ArrayList;` `// Solution structure``class` `Interval {``    ``int` `buy, sell;``}` `class` `StockBuySell {``    ``// 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``;` `        ``int` `count = ``0``;` `        ``// solution array``        ``ArrayList sol = ``new` `ArrayList();` `        ``// 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``;` `            ``Interval e = ``new` `Interval();``            ``e.buy = i++;``            ``// Store the index of minima` `            ``// 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``            ``e.sell = i - ``1``;``            ``sol.add(e);` `            ``// Increment number of buy/sell``            ``count++;``        ``}` `        ``// print solution``        ``if` `(count == ``0``)``            ``System.out.println(``"There is no day when buying the stock "``                               ``+ ``"will make profit"``);``        ``else``            ``for` `(``int` `j = ``0``; j < count; j++)``                ``System.out.println(``"Buy on day: "` `+ sol.get(j).buy``                                   ``+ ``"        "``                                   ``+ ``"Sell on day : "` `+ sol.get(j).sell);` `        ``return``;``    ``}` `    ``public` `static` `void` `main(String args[])``    ``{``        ``StockBuySell stock = ``new` `StockBuySell();` `        ``// stock prices on consecutive days``        ``int` `price[] = { ``100``, ``180``, ``260``, ``310``, ``40``, ``535``, ``695` `};``        ``int` `n = price.length;` `        ``// function call``        ``stock.stockBuySell(price, n);``    ``}``}` `// This code has been contributed by Mayank Jaiswal`

## Python3

 `# Python3 Program to find``# best buying and selling days` `# This function finds the buy sell``# schedule for maximum profit``def` `stockBuySell(price, n):``    ` `    ``# Prices must be given for at least two days``    ``if` `(n ``=``=` `1``):``        ``return``    ` `    ``# Traverse through given price array``    ``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``)) ``and``                ``(price[i ``+` `1``] <``=` `price[i])):``            ``i ``+``=` `1``        ` `        ``# If we reached the end, break``        ``# as no further solution possible``        ``if` `(i ``=``=` `n ``-` `1``):``            ``break``        ` `        ``# Store the index of minima``        ``buy ``=` `i``        ``i ``+``=` `1``        ` `        ``# Find Local Maxima``        ``# Note that the limit is (n-1) as we are``        ``# comparing to previous element``        ``while` `((i < n) ``and` `(price[i] >``=` `price[i ``-` `1``])):``            ``i ``+``=` `1``            ` `        ``# Store the index of maxima``        ``sell ``=` `i ``-` `1``        ` `        ``print``(``"Buy on day: "``,buy,``"\t"``,``                ``"Sell on day: "``,sell)``        ` `# Driver code` `# Stock prices on consecutive days``price ``=` `[``100``, ``180``, ``260``, ``310``, ``40``, ``535``, ``695``]``n ``=` `len``(price)` `# Function call``stockBuySell(price, n)` `# This is code contributed by SHUBHAMSINGH10`

## C#

 `// C# program to find best buying and selling days``using` `System;``using` `System.Collections.Generic;` `// Solution structure``class` `Interval``{``    ``public` `int` `buy, sell;``}` `public` `class` `StockBuySell``{``    ``// 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``;` `        ``int` `count = 0;` `        ``// solution array``        ``List sol = ``new` `List();` `        ``// 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``;` `            ``Interval e = ``new` `Interval();``            ``e.buy = i++;``            ``// Store the index of minima` `            ``// 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``            ``e.sell = i - 1;``            ``sol.Add(e);` `            ``// Increment number of buy/sell``            ``count++;``        ``}` `        ``// print solution``        ``if` `(count == 0)``            ``Console.WriteLine(``"There is no day when buying the stock "``                            ``+ ``"will make profit"``);``        ``else``            ``for` `(``int` `j = 0; j < count; j++)``                ``Console.WriteLine(``"Buy on day: "` `+ sol[j].buy``                                ``+ ``"     "``                                ``+ ``"Sell on day : "` `+ sol[j].sell);` `        ``return``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String []args)``    ``{``        ``StockBuySell stock = ``new` `StockBuySell();` `        ``// stock prices on consecutive days``        ``int` `[]price = { 100, 180, 260, 310, 40, 535, 695 };``        ``int` `n = price.Length;` `        ``// function call``        ``stock.stockBuySell(price, n);``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output
```Buy on day: 0     Sell on day: 3
Buy 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++

 `#include ``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);``    ``cout << maxProfit(prices, N) << endl;``    ``return` `0;``}``// This code is contributed by Kingshuk Deb`
Output
`865`

Time Complexity: O(n)
Auxiliary Space: O(1)

Another Approach:

The idea is to find the minimum element in the array and subtract with all elements in the array and find the maximum value.

First, we store the first element of the array in a variable.
Now compare the first element with all the elements of the array and find the minimum element.
Then store the minimum element of the array in that variable.
Then subtract the minimum element with the entire array and find the maximum value and then at last return the max value.

## C++

 `#include ``using` `namespace` `std;` `int` `maxProfit(``int` `prices[], ``int` `N)``{``    ``int` `n = N;``    ``int` `cost = 0;``    ``int` `maxCost = 0;` `    ``if` `(n == 0)``    ``{``        ``return` `0;``    ``}``    ` `    ``// Store the first element of array``    ``// in a variable``    ``int` `min_price = prices;` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Now compare first element with all``        ``// the element of array and find the``        ``// minimum element``        ``min_price = min(min_price, prices[i]);` `        ``// Since min_price is smallest element of the``        ``// array so subtract with every element of the``        ``// array and return the maxCost``        ``cost = prices[i] - min_price;` `        ``maxCost = max(maxCost, cost);``    ``}``    ``return` `maxCost;``}` `// Driver Code   ``int` `main()``{``  ` `    ``// Stock prices on consecutive days``    ``int` `prices[] = { 7, 1, 5, 3, 6, 4 };``    ``int` `N = ``sizeof``(prices) / ``sizeof``(prices);``    ` `    ``cout << maxProfit(prices, N);` `    ``return` `0;``}` `// This code is contributed by divyeshrabadiya07`

## Java

 `import` `java.io.*;``import` `java.util.*;` `class` `BuyStock {` `    ``public` `static` `int` `maxProfit(``int``[] prices)``    ``{``        ``int` `n = prices.length;``        ``int` `cost = ``0``;``        ``int` `maxCost = ``0``;` `        ``if` `(n == ``0``) {``            ``return` `0``;``        ``}` `        ``// store the first element of array in a variable` `        ``int` `min_price = prices[``0``];` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// now compare first element with all the``            ``// element of array and find the minimum element` `            ``min_price = Math.min(min_price, prices[i]);` `            ``// since min_price is smallest element of the``            ``// array so subtract with every element of the``            ``// array and return the maxCost` `            ``cost = prices[i] - min_price;` `            ``maxCost = Math.max(maxCost, cost);``        ``}``        ``return` `maxCost;``    ``}``  ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// stock prices on consecutive days``        ``int` `prices[] = { ``7``, ``1``, ``5``, ``3``, ``6``, ``4` `};``        ``System.out.print(maxProfit(prices));``    ``}``}`

## Python3

 `def` `maxProfit(prices):``    ` `    ``n ``=` `len``(prices)``    ``cost ``=` `0``    ``maxcost ``=` `0` `    ``if` `(n ``=``=` `0``):``        ``return` `0` `    ``# Store the first element of``    ``# array in a variable``    ``min_price ``=` `prices[``0``]` `    ``for` `i ``in` `range``(n):``        ` `        ``# Now compare first element with all``        ``# the element of array and find the``        ``# minimum element``        ``min_price ``=` `min``(min_price, prices[i])` `        ``# Since min_price is smallest element``        ``# of the array so subtract with every``        ``# element of the array and return the``        ``# maxCost``        ``cost ``=` `prices[i] ``-` `min_price` `        ``maxcost ``=` `max``(maxcost, cost)` `    ``return` `maxcost` `# Driver Code``prices ``=` `[ ``7``, ``1``, ``5``, ``3``, ``6``, ``4` `]` `# Stock prices on consecutive days``print``(maxProfit(prices))` `# This code is contributed by avanitrachhadiya2155`

## C#

 `using` `System;` `class` `BuyStock {` `    ``public` `static` `int` `MaxProfit(``int``[] prices)``    ``{``        ``int` `n = prices.Length;``        ``int` `cost = 0;``        ``int` `MaxCost = 0;` `        ``if` `(n == 0) {``            ``return` `0;``        ``}` `        ``// store the first element of array in a variable``        ``int` `Min_price = prices;` `        ``for` `(``int` `i = 0; i < n; i++) {` `            ``// now compare first element with all the``            ``// element of array and find the Minimum element``            ``Min_price = Math.Min(Min_price, prices[i]);` `            ``// since Min_price is smallest element of the``            ``// array so subtract with every element of the``            ``// array and return the MaxCost``            ``cost = prices[i] - Min_price;` `            ``MaxCost = Math.Max(MaxCost, cost);``        ``}``        ``return` `MaxCost;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``// stock prices on consecutive days``        ``int` `[]prices = { 7, 1, 5, 3, 6, 4 };``        ``Console.Write(MaxProfit(prices));``    ``}``}` `// This code is contributed by shivanisinghss2110`

## Javascript

 ``
Output
`5`

Time Complexity: O(n)
Space complexity: O(1)