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Permutation problem

The permutation problem is a typical application of the backtracking algorithm. It is defined as finding all possible arrangements of elements from a given set (such as an array or string).

The table below lists several example data, including the input arrays and their corresponding permutations.

Table   Permutation examples

Input array Permutations
\([1]\) \([1]\)
\([1, 2]\) \([1, 2], [2, 1]\)
\([1, 2, 3]\) \([1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]\)

Cases without equal elements

Question

Enter an integer array without duplicate elements and return all possible permutations.

From the perspective of the backtracking algorithm, we can imagine the process of generating permutations as a series of choices. Suppose the input array is \([1, 2, 3]\), if we first choose \(1\), then \(3\), and finally \(2\), we obtain the permutation \([1, 3, 2]\). Backtracking means undoing a choice and then continuing to try other choices.

From the code perspective, the candidate set choices contains all elements of the input array, and the state state contains elements that have been selected so far. Please note that each element can only be chosen once, thus all elements in state must be unique.

As shown in the figure below, we can unfold the search process into a recursive tree, where each node represents the current state state. Starting from the root node, after three rounds of choices, we reach the leaf nodes, each corresponding to a permutation.

Permutation recursive tree

Pruning of repeated choices

To ensure that each element is selected only once, we consider introducing a boolean array selected, where selected[i] indicates whether choices[i] has been selected. We base our pruning operations on this array:

  • After making the choice choice[i], we set selected[i] to \(\text{True}\), indicating it has been chosen.
  • When iterating through the choice list choices, skip all nodes that have already been selected, i.e., prune.

As shown in the figure below, suppose we choose 1 in the first round, 3 in the second round, and 2 in the third round, we need to prune the branch of element 1 in the second round and elements 1 and 3 in the third round.

Permutation pruning example

Observing the figure above, this pruning operation reduces the search space size from \(O(n^n)\) to \(O(n!)\).

Code implementation

After understanding the above information, we can "fill in the blanks" in the framework code. To shorten the overall code, we do not implement individual functions within the framework code separately, but expand them in the backtrack() function:

[file]{permutations_i}-[class]{}-[func]{permutations_i}

Considering cases with equal elements

Question

Enter an integer array, which may contain duplicate elements, and return all unique permutations.

Suppose the input array is \([1, 1, 2]\). To differentiate the two duplicate elements \(1\), we mark the second \(1\) as \(\hat{1}\).

As shown in the figure below, half of the permutations generated by the above method are duplicates.

Duplicate permutations

So, how do we eliminate duplicate permutations? Most directly, consider using a hash set to deduplicate permutation results. However, this is not elegant, as branches generating duplicate permutations are unnecessary and should be identified and pruned in advance, which can further improve algorithm efficiency.

Pruning of equal elements

Observing the figure below, in the first round, choosing \(1\) or \(\hat{1}\) results in identical permutations under both choices, thus we should prune \(\hat{1}\).

Similarly, after choosing \(2\) in the first round, choosing \(1\) and \(\hat{1}\) in the second round also produces duplicate branches, so we should also prune \(\hat{1}\) in the second round.

Essentially, our goal is to ensure that multiple equal elements are only selected once in each round of choices.

Duplicate permutations pruning

Code implementation

Based on the code from the previous problem, we consider initiating a hash set duplicated in each round of choices, used to record elements that have been tried in that round, and prune duplicate elements:

[file]{permutations_ii}-[class]{}-[func]{permutations_ii}

Assuming all elements are distinct from each other, there are \(n!\) (factorial) permutations of \(n\) elements; when recording results, it is necessary to copy a list of length \(n\), using \(O(n)\) time. Thus, the time complexity is \(O(n!n)\).

The maximum recursion depth is \(n\), using \(O(n)\) frame space. Selected uses \(O(n)\) space. At any one time, there can be up to \(n\) duplicated, using \(O(n^2)\) space. Therefore, the space complexity is \(O(n^2)\).

Comparison of the two pruning methods

Please note, although both selected and duplicated are used for pruning, their targets are different.

  • Repeated choice pruning: There is only one selected throughout the search process. It records which elements are currently in the state, aiming to prevent an element from appearing repeatedly in state.
  • Equal element pruning: Each round of choices (each call to the backtrack function) contains a duplicated. It records which elements have been chosen in the current traversal (for loop), aiming to ensure equal elements are selected only once.

The figure below shows the scope of the two pruning conditions. Note, each node in the tree represents a choice, and the nodes from the root to the leaf form a permutation.

Scope of the two pruning conditions