backtracking.all_combinations¶
In this problem, we want to determine all possible combinations of k numbers out of 1 … n. We use backtracking to solve this problem.
Time complexity: O(C(n,k)) which is O(n choose k) = O((n!/(k! * (n - k)!))),
Attributes¶
Functions¶
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Generates all possible combinations of k numbers out of 1 ... n using itertools. |
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Helper function to recursively build all combinations. |
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Generates all possible combinations of k numbers out of 1 ... n using backtracking. |
Module Contents¶
- backtracking.all_combinations.combination_lists(n: int, k: int) list[list[int]] ¶
Generates all possible combinations of k numbers out of 1 … n using itertools.
>>> combination_lists(n=4, k=2) [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]
- backtracking.all_combinations.create_all_state(increment: int, total_number: int, level: int, current_list: list[int], total_list: list[list[int]]) None ¶
Helper function to recursively build all combinations.
>>> create_all_state(1, 4, 2, [], result := []) >>> result [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]] >>> create_all_state(1, 3, 3, [], result := []) >>> result [[1, 2, 3]] >>> create_all_state(2, 2, 1, [1], result := []) >>> result [[1, 2]] >>> create_all_state(1, 0, 0, [], result := []) >>> result [[]] >>> create_all_state(1, 4, 0, [1, 2], result := []) >>> result [[1, 2]] >>> create_all_state(5, 4, 2, [1, 2], result := []) >>> result []
- backtracking.all_combinations.generate_all_combinations(n: int, k: int) list[list[int]] ¶
Generates all possible combinations of k numbers out of 1 … n using backtracking.
>>> generate_all_combinations(n=4, k=2) [[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]] >>> generate_all_combinations(n=0, k=0) [[]] >>> generate_all_combinations(n=10, k=-1) Traceback (most recent call last): ... ValueError: k must not be negative >>> generate_all_combinations(n=-1, k=10) Traceback (most recent call last): ... ValueError: n must not be negative >>> generate_all_combinations(n=5, k=4) [[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 4, 5], [1, 3, 4, 5], [2, 3, 4, 5]] >>> generate_all_combinations(n=3, k=3) [[1, 2, 3]] >>> generate_all_combinations(n=3, k=1) [[1], [2], [3]] >>> generate_all_combinations(n=1, k=0) [[]] >>> generate_all_combinations(n=1, k=1) [[1]] >>> from itertools import combinations >>> all(generate_all_combinations(n, k) == combination_lists(n, k) ... for n in range(1, 6) for k in range(1, 6)) True
- backtracking.all_combinations.tests¶