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recursion-backtracking

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更新日2026年5月17日 09:22

Generate all valid arrangements using backtracking with pruning. Use when solving permutations, combinations, subsets, N-queens, sudoku, or constraint satisfaction problems.

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When to use

Use backtracking for constraint satisfaction and combinatorial generation: permutations, combinations, subsets, N-queens, sudoku, valid arrangements.

Rules

Pattern: make a choice, recurse, undo the choice (backtrack)
Prune early — skip branches that already violate constraints to avoid exploring dead ends
For subsets: at each element, choose to include or exclude it (2^n total)
For permutations: choose each unused element at each position (n! total)
ALWAYS pass state by reference and undo mutations rather than copying
NEVER copy state — it's wasteful and slow
If the problem says "generate all" or "find all valid," backtracking is usually the right approach

Complexity

Subsets: 2^n. Permutations: n!.

Example

Subsets of [1,2,3]: at each element, branch include/exclude. def solve(i): if i==n: yield list(path); return; path.append(nums[i]); solve(i+1); path.pop(); solve(i+1).

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name	recursion-backtracking
description	Generate all valid arrangements using backtracking with pruning. Use when solving permutations, combinations, subsets, N-queens, sudoku, or constraint satisfaction problems.
topic	Backtracking
token_cost	100
related	["dynamic-programming","dfs-vs-bfs"]
keywords	["permutation","combination","subset","backtrack","constraint","generate","valid","recursive","pruning","n-queens","sudoku","exhaustive","all","solutions","choose","pick","arrangement","password","sequence"]

When to use

Use backtracking for constraint satisfaction and combinatorial generation: permutations, combinations, subsets, N-queens, sudoku, valid arrangements.

Rules

Pattern: make a choice, recurse, undo the choice (backtrack)
Prune early — skip branches that already violate constraints to avoid exploring dead ends
For subsets: at each element, choose to include or exclude it (2^n total)
For permutations: choose each unused element at each position (n! total)
ALWAYS pass state by reference and undo mutations rather than copying
NEVER copy state — it's wasteful and slow
If the problem says "generate all" or "find all valid," backtracking is usually the right approach

Complexity

Subsets: 2^n. Permutations: n!.

Example

Subsets of [1,2,3]: at each element, branch include/exclude. def solve(i): if i==n: yield list(path); return; path.append(nums[i]); solve(i+1); path.pop(); solve(i+1).