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Python/data_structures/stacks/stock_span_problem.py

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"""
The stock span problem is a financial problem where we have a series of n daily
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price quotes for a stock and we need to calculate span of stock's price for all n days.
The span Si of the stock's price on a given day i is defined as the maximum
number of consecutive days just before the given day, for which the price of the stock
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on the current day is less than or equal to its price on the given day.
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"""
def calculate_span(price: list[int]) -> list[int]:
"""
Calculate the span values for a given list of stock prices.
Args:
price: List of stock prices.
Returns:
List of span values.
>>> calculate_span([10, 4, 5, 90, 120, 80])
[1, 1, 2, 4, 5, 1]
>>> calculate_span([100, 50, 60, 70, 80, 90])
[1, 1, 2, 3, 4, 5]
>>> calculate_span([5, 4, 3, 2, 1])
[1, 1, 1, 1, 1]
>>> calculate_span([1, 2, 3, 4, 5])
[1, 2, 3, 4, 5]
>>> calculate_span([10, 20, 30, 40, 50])
[1, 2, 3, 4, 5]
>>> calculate_span([100, 80, 60, 70, 60, 75, 85])
[1, 1, 1, 2, 1, 4, 6]
"""
n = len(price)
s = [0] * n
# Create a stack and push index of fist element to it
st = []
st.append(0)
# Span value of first element is always 1
s[0] = 1
# Calculate span values for rest of the elements
for i in range(1, n):
# Pop elements from stack while stack is not
# empty and top of stack is smaller than price[i]
while len(st) > 0 and price[st[-1]] <= price[i]:
st.pop()
# If stack becomes empty, then price[i] is greater
# than all elements on left of it, i.e. price[0],
# price[1], ..price[i-1]. Else the price[i] is
# greater than elements after top of stack
s[i] = i + 1 if len(st) <= 0 else (i - st[-1])
# Push this element to stack
st.append(i)
return s
# A utility function to print elements of array
def print_array(arr, n):
for i in range(n):
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print(arr[i], end=" ")
# Driver program to test above function
price = [10, 4, 5, 90, 120, 80]
# Calculate the span values
S = calculate_span(price)
# Print the calculated span values
print_array(S, len(price))