Stock Buy and Sell - Multiple Transaction Allowed
Last Updated :
23 Jul, 2025
Given an array prices[] of size n denoting the cost of stock on each day, the task is to find the maximum total profit if we can buy and sell the stocks any number of times.
Note: We can only sell a stock which we have bought earlier and we cannot hold multiple stocks on any day.
Examples:
Input: prices[] = {100, 180, 260, 310, 40, 535, 695}
Output: 865
Explanation: Buy the stock on day 0 and sell it on day 3 => 310 - 100 = 210
Buy the stock on day 4 and sell it on day 6 => 695 - 40 = 655
Maximum Profit = 210 + 655 = 865
Input: prices[] = {4, 2, 2, 2, 4}
Output: 2
Explanation: Buy the stock on day 3 and sell it on day 4 => 4 - 2 = 2
Maximum Profit = 2
[Naive Approach] Using Recursion - Exponential Time
We consider every valid pair (A pair of indexes i and j such that price[i] < price[j] and j > i), we get the profit of the pair as (price[j] - price[i]) and add recursively compute profits for i-1 and j+1. Finally we return the maximum of all profits obtained by all valid pairs.
C++
#include <iostream>
#include <vector>
using namespace std;
int maxProfitRec(vector<int> &price, int start, int end) {
int res = 0;
// Consider every valid pair, find the profit with it
// and recursively find profits on left and right of it
for (int i = start; i < end; i++) {
for (int j = i + 1; j <= end; j++) {
if (price[j] > price[i]) {
int curr = (price[j] - price[i]) +
maxProfitRec(price, start, i - 1) +
maxProfitRec(price, j + 1, end);
res = max(res, curr);
}
}
}
return res;
}
// Wrapper function
int maximumProfit(vector<int> &prices) {
return maxProfitRec(prices, 0, prices.size()-1);
}
int main() {
vector<int> prices = {100, 180, 260, 310, 40, 535, 695};
cout << maximumProfit(prices);
return 0;
}
#include <stdio.h>
// Function to find the maximum profit
int maxProfitRec(int price[], int start, int end) {
int res = 0;
// Consider every valid pair, find the profit with it
// and recursively find profits on left and right of it
for (int i = start; i < end; i++) {
for (int j = i + 1; j <= end; j++) {
if (price[j] > price[i]) {
int curr = (price[j] - price[i]) +
maxProfitRec(price, start, i - 1) +
maxProfitRec(price, j + 1, end);
if (curr > res) {
res = curr;
}
}
}
}
return res;
}
// Wrapper function
int maximumProfit(int prices[], int size) {
return maxProfitRec(prices, 0, size - 1);
}
int main() {
int prices[] = {100, 180, 260, 310, 40, 535, 695};
int size = sizeof(prices) / sizeof(prices[0]);
printf("%d\n", maximumProfit(prices, size));
return 0;
}
import java.util.Arrays;
class GfG {
// Function to find the maximum profit
static int maxProfitRec(int[] price, int start, int end) {
int res = 0;
// Consider every valid pair, find the profit with it
// and recursively find profits on left and right of it
for (int i = start; i < end; i++) {
for (int j = i + 1; j <= end; j++) {
if (price[j] > price[i]) {
int curr = (price[j] - price[i]) +
maxProfitRec(price, start, i - 1) +
maxProfitRec(price, j + 1, end);
res = Math.max(res, curr);
}
}
}
return res;
}
// Wrapper function
static int maximumProfit(int[] prices) {
return maxProfitRec(prices, 0, prices.length - 1);
}
public static void main(String[] args) {
int[] prices = {100, 180, 260, 310, 40, 535, 695};
System.out.println(maximumProfit(prices));
}
}
# Function to find the maximum profit
def maxProfitRec(price, start, end):
res = 0
# Consider every valid pair, find the profit with it
# and recursively find profits on left and right of it
for i in range(start, end):
for j in range(i + 1, end + 1):
if price[j] > price[i]:
curr = (price[j] - price[i]) + \
maxProfitRec(price, start, i - 1) + \
maxProfitRec(price, j + 1, end)
res = max(res, curr)
return res
# Wrapper function
def maximumProfit(prices):
return maxProfitRec(prices, 0, len(prices) - 1)
if __name__ == "__main__":
prices = [100, 180, 260, 310, 40, 535, 695]
print(maximumProfit(prices))
using System;
class GfG {
// Function to find the maximum profit
static int maxProfitRec(int[] price, int start, int end) {
int res = 0;
// Consider every valid pair, find the profit with it
// and recursively find profits on left and right of it
for (int i = start; i < end; i++) {
for (int j = i + 1; j <= end; j++) {
if (price[j] > price[i]) {
int curr = (price[j] - price[i]) +
maxProfitRec(price, start, i - 1) +
maxProfitRec(price, j + 1, end);
if (curr > res) {
res = curr;
}
}
}
}
return res;
}
// Wrapper function
static int maximumProfit(int[] prices) {
return maxProfitRec(prices, 0, prices.Length - 1);
}
static void Main() {
int[] prices = {100, 180, 260, 310, 40, 535, 695};
Console.WriteLine(maximumProfit(prices));
}
}
// Function to find the maximum profit
function maxProfitRec(price, start, end) {
let res = 0;
// Consider every valid pair, find the profit with it
// and recursively find profits on left and right of it
for (let i = start; i < end; i++) {
for (let j = i + 1; j <= end; j++) {
if (price[j] > price[i]) {
let curr = (price[j] - price[i]) +
maxProfitRec(price, start, i - 1) +
maxProfitRec(price, j + 1, end);
res = Math.max(res, curr);
}
}
}
return res;
}
// Wrapper function
function maximumProfit(prices) {
return maxProfitRec(prices, 0, prices.length - 1);
}
const prices = [100, 180, 260, 310, 40, 535, 695];
console.log(maximumProfit(prices));
[Better Approach] Local Minima and Maxima - O(n) Time and O(1) Space
When the prices are going down, we do not do anything and let the prices go down. When the prices reach a local minimum value (a value after which the prices go up), we buy the stock. When the prices are going up, we let the prices go up and once the prices reach a local maxima, we sell the stock.
The idea is to traverse the array from left to right and do following.
- Find local minima and then a local maxima.
- Compute the difference between two and add to the result.
Working:
Below is the implementation of the algorithm
C++
#include <iostream>
#include <vector>
using namespace std;
int maximumProfit(vector<int>& prices) {
int n = prices.size();
int lMin = prices[0]; // Local Minima
int lMax = prices[0]; // Local Maxima
int res = 0;
int i = 0;
while (i < n - 1) {
// Find local minima
while (i < n - 1 && prices[i] >= prices[i + 1]) { i++; }
lMin = prices[i];
// Local Maxima
while (i < n - 1 && prices[i] <= prices[i + 1]) { i++; }
lMax = prices[i];
// Add current profit
res = res + (lMax - lMin);
}
return res;
}
int main() {
vector<int> prices = {100, 180, 260, 310, 40, 535, 695};
cout << maximumProfit(prices);
return 0;
}
#include <stdio.h>
// Function to calculate the maximum profit
int maximumProfit(int prices[], int n) {
int lMin = prices[0]; // Local Minima
int lMax = prices[0]; // Local Maxima
int res = 0;
int i = 0;
while (i < n - 1) {
// Find local minima
while (i < n - 1 && prices[i] >= prices[i + 1]) { i++; }
lMin = prices[i];
// Local Maxima
while (i < n - 1 && prices[i] <= prices[i + 1]) { i++; }
lMax = prices[i];
// Add current profit
res += (lMax - lMin);
}
return res;
}
// Driver Code
int main() {
int prices[] = {100, 180, 260, 310, 40, 535, 695};
int n = sizeof(prices) / sizeof(prices[0]);
printf("%d\n", maximumProfit(prices, n));
return 0;
}
class GfG {
// Function to calculate the maximum profit
static int maximumProfit(int[] prices) {
int n = prices.length;
int lMin = prices[0]; // Local Minima
int lMax = prices[0]; // Local Maxima
int res = 0;
int i = 0;
while (i < n - 1) {
// Find local minima
while (i < n - 1 && prices[i] >= prices[i + 1]) { i++; }
lMin = prices[i];
// Local Maxima
while (i < n - 1 && prices[i] <= prices[i + 1]) { i++; }
lMax = prices[i];
// Add current profit
res += (lMax - lMin);
}
return res;
}
public static void main(String[] args) {
int[] prices = {100, 180, 260, 310, 40, 535, 695};
System.out.println(maximumProfit(prices));
}
}
# Function to calculate the maximum profit
def maximumProfit(prices):
n = len(prices)
lMin = prices[0] # Local Minima
lMax = prices[0] # Local Maxima
res = 0
i = 0
while i < n - 1:
# Find local minima
while i < n - 1 and prices[i] >= prices[i + 1]:
i += 1
lMin = prices[i]
# Local Maxima
while i < n - 1 and prices[i] <= prices[i + 1]:
i += 1
lMax = prices[i]
# Add current profit
res += (lMax - lMin)
return res
# Driver Code
if __name__ == "__main__":
prices = [100, 180, 260, 310, 40, 535, 695]
print(maximumProfit(prices))
using System;
class GfG {
// Function to calculate the maximum profit
static int maximumProfit(int[] prices) {
int n = prices.Length;
int lMin = prices[0]; // Local Minima
int lMax = prices[0]; // Local Maxima
int res = 0;
int i = 0;
while (i < n - 1) {
// Find local minima
while (i < n - 1 && prices[i] >= prices[i + 1]) { i++; }
lMin = prices[i];
// Local Maxima
while (i < n - 1 && prices[i] <= prices[i + 1]) { i++; }
lMax = prices[i];
// Add current profit
res += (lMax - lMin);
}
return res;
}
// Driver Code
static void Main() {
int[] prices = {100, 180, 260, 310, 40, 535, 695};
Console.WriteLine(maximumProfit(prices));
}
}
// Function to calculate the maximum profit
function maximumProfit(prices) {
let n = prices.length;
let lMin = prices[0]; // Local Minima
let lMax = prices[0]; // Local Maxima
let res = 0;
let i = 0;
while (i < n - 1) {
// Find local minima
while (i < n - 1 && prices[i] >= prices[i + 1]) { i++; }
lMin = prices[i];
// Local Maxima
while (i < n - 1 && prices[i] <= prices[i + 1]) { i++; }
lMax = prices[i];
// Add current profit
res += (lMax - lMin);
}
return res;
}
const prices = [100, 180, 260, 310, 40, 535, 695];
console.log(maximumProfit(prices));
Time Complexity: O(n)
Auxiliary Space: O(1)
[Expected Approach] Accumulate Profit - O(n) Time and O(1) Space
This is mainly an optimization over the previous solution. Instead of selling at local maxima, we keep selling while the prices are going up. This way we accumulate the same profit and avoid some condition checks required for computing local minima and maxima.
Traverse price[] from i = 1 to price.size() - 1
- res = 0
- if price[i] > price[i - 1]
- res = res + price[i] - price[i - 1]
Below is the implementation of the algorithm:
C++
// C++ Program to find the max profit when multiple
// transactions are allowed
#include <iostream>
#include <vector>
using namespace std;
int maximumProfit(const vector<int>& prices) {
int res = 0;
// Keep on adding the difference between
// adjacent when the prices a
for (int i = 1; i < prices.size(); i++) {
if (prices[i] > prices[i - 1])
res += prices[i] - prices[i - 1];
}
return res;
}
int main() {
vector<int> prices = { 100, 180, 260, 310, 40, 535, 695 };
cout << maximumProfit(prices) << endl;
return 0;
}
// C Program to find the max profit when multiple
// transactions are allowed
#include <stdio.h>
int maximumProfit(const int* prices, int n) {
int res = 0;
// Keep on adding the difference between
// adjacent when the prices a
for (int i = 1; i < n; i++) {
if (prices[i] > prices[i - 1])
res += prices[i] - prices[i - 1];
}
return res;
}
int main() {
int prices[] = { 100, 180, 260, 310, 40, 535, 695 };
int size = sizeof(prices) / sizeof(prices[0]);
printf("%d\n", maximumProfit(prices, size));
return 0;
}
// Java Program to find the max profit when multiple
// transactions are allowed
import java.util.Arrays;
class GfG {
static int maximumProfit(int[] prices) {
int res = 0;
// Keep on adding the difference between
// adjacent when the prices a
for (int i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1])
res += prices[i] - prices[i - 1];
}
return res;
}
public static void main(String[] args) {
int[] prices = { 100, 180, 260, 310, 40, 535, 695 };
System.out.println(maximumProfit(prices));
}
}
# Python Program to find the max profit when multiple
# transactions are allowed
def maximumProfit(prices):
res = 0
# Keep on adding the difference between
# adjacent when the prices a
for i in range(1, len(prices)):
if prices[i] > prices[i - 1]:
res += prices[i] - prices[i - 1]
return res
if __name__ == "__main__":
prices = [100, 180, 260, 310, 40, 535, 695]
print(maximumProfit(prices))
// C# Program to find the max profit when multiple
// transactions are allowed
using System;
using System.Collections.Generic;
class GfG {
static int maximumProfit(List<int> prices) {
int res = 0;
// Keep on adding the difference between
// adjacent when the prices a
for (int i = 1; i < prices.Count; i++) {
if (prices[i] > prices[i - 1])
res += prices[i] - prices[i - 1];
}
return res;
}
static void Main(string[] args) {
List<int> prices =
new List<int> { 100, 180, 260, 310, 40, 535, 695 };
Console.WriteLine(maximumProfit(prices));
}
}
// JavaScript Program to find the max profit when multiple
// transactions are allowed
function maximumProfit(prices) {
let res = 0;
// Keep on adding the difference between
// adjacent when the prices a
for (let i = 1; i < prices.length; i++) {
if (prices[i] > prices[i - 1])
res += prices[i] - prices[i - 1];
}
return res;
}
const prices = [100, 180, 260, 310, 40, 535, 695];
console.log(maximumProfit(prices));
Time Complexity: O(n)
Auxiliary Space: O(1)
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