Changes the code To return the list in dynamic_programming/subset_generation.py (#10191)

* Changing the code to return tuple

* Changing the code to return tuple

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* Update dynamic_programming/subset_generation.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Adding doctests in subset_generation.py

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* Update subset_generation.py

* Update subset_generation.py

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* Update subset_generation.py

* Update subset_generation.py

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* Update dynamic_programming/subset_generation.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update stock_span_problem.py

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* Update subset_generation.py

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* Update subset_generation.py

* Update subset_generation.py

* Update subset_generation.py

* Update subset_generation.py

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Co-authored-by: Christian Clauss <cclauss@me.com>
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@ -1,44 +1,60 @@
# Print all subset combinations of n element in given set of r element.
def combination_util(arr, n, r, index, data, i):
def subset_combinations(elements: list[int], n: int) -> list:
"""
Current combination is ready to be printed, print it
arr[] ---> Input Array
data[] ---> Temporary array to store current combination
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in data[]
r ---> Size of a combination to be printed
Compute n-element combinations from a given list using dynamic programming.
Args:
elements: The list of elements from which combinations will be generated.
n: The number of elements in each combination.
Returns:
A list of tuples, each representing a combination of n elements.
>>> subset_combinations(elements=[10, 20, 30, 40], n=2)
[(10, 20), (10, 30), (10, 40), (20, 30), (20, 40), (30, 40)]
>>> subset_combinations(elements=[1, 2, 3], n=1)
[(1,), (2,), (3,)]
>>> subset_combinations(elements=[1, 2, 3], n=3)
[(1, 2, 3)]
>>> subset_combinations(elements=[42], n=1)
[(42,)]
>>> subset_combinations(elements=[6, 7, 8, 9], n=4)
[(6, 7, 8, 9)]
>>> subset_combinations(elements=[10, 20, 30, 40, 50], n=0)
[()]
>>> subset_combinations(elements=[1, 2, 3, 4], n=2)
[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
>>> subset_combinations(elements=[1, 'apple', 3.14], n=2)
[(1, 'apple'), (1, 3.14), ('apple', 3.14)]
>>> subset_combinations(elements=['single'], n=0)
[()]
>>> subset_combinations(elements=[], n=9)
[]
>>> from itertools import combinations
>>> all(subset_combinations(items, n) == list(combinations(items, n))
... for items, n in (
... ([10, 20, 30, 40], 2), ([1, 2, 3], 1), ([1, 2, 3], 3), ([42], 1),
... ([6, 7, 8, 9], 4), ([10, 20, 30, 40, 50], 1), ([1, 2, 3, 4], 2),
... ([1, 'apple', 3.14], 2), (['single'], 0), ([], 9)))
True
"""
if index == r:
for j in range(r):
print(data[j], end=" ")
print(" ")
return
# When no more elements are there to put in data[]
if i >= n:
return
# current is included, put next at next location
data[index] = arr[i]
combination_util(arr, n, r, index + 1, data, i + 1)
# current is excluded, replace it with
# next (Note that i+1 is passed, but
# index is not changed)
combination_util(arr, n, r, index, data, i + 1)
# The main function that prints all combinations
# of size r in arr[] of size n. This function
# mainly uses combinationUtil()
r = len(elements)
if n > r:
return []
dp: list[list[tuple]] = [[] for _ in range(r + 1)]
def print_combination(arr, n, r):
# A temporary array to store all combination one by one
data = [0] * r
# Print all combination using temporary array 'data[]'
combination_util(arr, n, r, 0, data, 0)
dp[0].append(())
for i in range(1, r + 1):
for j in range(i, 0, -1):
for prev_combination in dp[j - 1]:
dp[j].append(tuple(prev_combination) + (elements[i - 1],))
try:
return sorted(dp[n])
except TypeError:
return dp[n]
if __name__ == "__main__":
# Driver code to check the function above
arr = [10, 20, 30, 40, 50]
print_combination(arr, len(arr), 3)
# This code is contributed by Ambuj sahu
from doctest import testmod
testmod()
print(f"{subset_combinations(elements=[10, 20, 30, 40], n=2) = }")