# Maximize profit in buying and selling stocks with Rest condition

The price of a stock on each day is given in an array arr[] for N days, the task is to find the maximum profit that can be made by buying and selling the stocks in those days with conditions that the stock must be sold before buying again and stock cannot be bought on the next day of selling a stock. (i.e, rest for at least one day).
Examples:

Input: arr[] = {2, 4, 5, 0, 2}
Output:
Explanation: Buy at cost 2 and sell at cost 4, profit = 2. Buy at cost 0 and sell at cost 2, profit = 2. Total profit = 4.
Input: arr[] = {2, 0, 5, 1, 8}
Output:

Approach :
Consider three states: REST, HOLD and SOLD.

REST means not doing anything.
HOLD means holding stock (bought a stock and have not sold).
SOLD means state after selling the stock.

To reach the REST state, do nothing.
To reach the HOLD state, buy the stock.
To reach the SOLD state, sell the stock for which first there is a need to buy.
By the above actions, a transition diagram is made.

By using this transition diagram the profit can be found out.
On ith day:

• rest[i] denotes maximum profit made by resting on day i. Since ith day is rest day, the stock might have been sold on (i-1)th day or might not have been sold. The value of rest[i] = max( rest[i-1], sold[i-1]).
• hold[i] denotes maximum profit made by buying on ith day or by buying on some day before ith and resting on ith day. So, the value of hold[i] = max( hold[i-1], rest[i-1]+price[i]).
• sold[i] denotes maximum profit by selling ith day. Stock must have been bought on some day before selling it on the ith day. Hence sold[i] = hold[i-1] + price[i]

Hence, the final answer will be the

maximum of sold[n-1] and rest[n-1].

Below is the implementation of the above approach:

## C++

 // C++ program for the above problem #include using namespace std;    int maxProfit(int prices[], int n) {     // If there is only one day     // for buying and selling     // no profit can be made     if (n <= 1)         return 0;        // Array to store Maxprofit by     // resting on given day     int rest[n] = { 0 };        // Array to store Maxprofit by     // buying or resting on the     // given day     int hold[n] = { 0 };        // Array to store Maxprofit by     // selling on given day     int sold[n] = { 0 };        // Initially there will 0 profit     rest[0] = 0;        // Buying on 1st day results     // in negative profit     hold[0] = -prices[0];        // zero profit since selling     // before buying isn't possible     sold[0] = 0;        for (int i = 1; i < n; i++) {            // max of profit on (i-1)th         // day by resting and profit         // on (i-1)th day by selling.         rest[i] = max(rest[i - 1],                       sold[i - 1]);            // max of profit by resting         // on ith day and         // buying on ith day.         hold[i] = max(hold[i - 1],                       rest[i - 1]                           - prices[i]);            // max of profit by selling         // on ith day         sold[i] = hold[i - 1] + prices[i];     }        // maxprofit     return max(rest[n - 1],                sold[n - 1]); }    // Driver Code int main() {     int price[] = { 2, 4,                     5, 0, 2 };     int n = sizeof(price)             / sizeof(price[0]);     cout << maxProfit(price, n)          << endl;     return 0; }

## Java

 // Java program for the above problem class GFG{    static int maxProfit(int prices[], int n) {            // If there is only one day     // for buying and selling     // no profit can be made     if (n <= 1)         return 0;        // Array to store Maxprofit by     // resting on given day     int rest[] = new int[n];        // Array to store Maxprofit by     // buying or resting on the     // given day     int hold[] = new int[9];        // Array to store Maxprofit by     // selling on given day     int sold[] = new int[9];        // Initially there will 0 profit     rest[0] = 0;        // Buying on 1st day results     // in negative profit     hold[0] = -prices[0];        // Zero profit since selling     // before buying isn't possible     sold[0] = 0;        for(int i = 1; i < n; i++)     {                  // max of profit on (i-1)th        // day by resting and profit        // on (i-1)th day by selling.        rest[i] = Math.max(rest[i - 1],                           sold[i - 1]);                  // max of profit by resting        // on ith day and        // buying on ith day.        hold[i] = Math.max(hold[i - 1],                            rest[i - 1] -                          prices[i]);                  // max of profit by selling        // on ith day        sold[i] = hold[i - 1] + prices[i];     }            // maxprofit     return Math.max(rest[n - 1],                     sold[n - 1]); }    // Driver Code public static void main(String[] args) {     int price[] = { 2, 4, 5, 0, 2 };     int n = price.length;            System.out.print(maxProfit(price, n) + "\n"); } }    // This code is contributed by amal kumar choubey

## C#

 // C# program for the above problem  using System;    class GFG{        static int maxProfit(int[] prices, int n)  {             // If there is only one day      // for buying and selling      // no profit can be made      if (n <= 1)          return 0;         // Array to store Maxprofit by      // resting on given day      int[] rest = new int[n];         // Array to store Maxprofit by      // buying or resting on the      // given day      int[] hold = new int[9];         // Array to store Maxprofit by      // selling on given day      int[] sold = new int[9];         // Initially there will 0 profit      rest[0] = 0;         // Buying on 1st day results      // in negative profit      hold[0] = -prices[0];         // Zero profit since selling      // before buying isn't possible      sold[0] = 0;         for(int i = 1; i < n; i++)      {                     // max of profit on (i-1)th          // day by resting and profit          // on (i-1)th day by selling.          rest[i] = Math.Max(rest[i - 1],                             sold[i - 1]);                         // max of profit by resting          // on ith day and          // buying on ith day.          hold[i] = Math.Max(hold[i - 1],                             rest[i - 1] -                           prices[i]);                         // max of profit by selling          // on ith day          sold[i] = hold[i - 1] + prices[i];      }             // maxprofit      return Math.Max(rest[n - 1],                      sold[n - 1]);  }     // Driver code static void Main()  {     int[] price = { 2, 4, 5, 0, 2 };      int n = price.Length;         Console.WriteLine(maxProfit(price, n)); } }    // This code is contributed by divyeshrabadiya07

Output:

4

Time Complexity: O(N)
Auxiliary Space: O(N)

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