# Program to find minimum number of days to wait to make profit in python

Suppose we have a list of prices representing the daily stock market prices of a company in chronological sequence. We have to find a same length list where the value at index i will be the minimum number of days we would have to wait until we make a profit. If there is no such way to make a profit the value should be 0.

So, if the input is like prices = [4, 3, 5, 9, 7, 6], then the output will be [2, 1, 1, 0, 0, 0]

To solve this, we will follow these steps:

• ans := a list of size same as prices and fill with 0
• q := a new list
• for each index i and price p in prices, do
• while q is not empty and p > second value of last item of q, do
• j := first item of last element of q
• ans[j] := i - j
• delete last element from q
• insert (i, p) at the end of q
• return ans

Let us see the following implementation to get better understanding:

## Example

Live Demo

class Solution:
def solve(self, prices):
ans = [0 for _ in prices]
q = []
for i, p in enumerate(prices):
while q and p > q[-1][1]:
j = q[-1][0]
ans[j] = i - j
q.pop()
q.append((i, p))
return ans

ob = Solution()
prices = [4, 3, 5, 9, 7, 6]
print(ob.solve(prices))

## Input

[4, 3, 5, 9, 7, 6]

## Output

[2, 1, 1, 0, 0, 0]