Problem Description

The problem presents us with a string s and a list of pairs, where each pair contains two indices of the string s. We are allowed to swap characters at any of the given pair indices as many times as we wish. The objective is to determine the lexicographically smallest string that can be formed after making any number of swaps among the pairs.

Flowchart Walkthrough

Let's use the algorithm flowchart to determine the appropriate algorithm for solving Leetcode problem 1202, Smallest String With Swaps. You can refer to the flowchart by clicking here. Here’s the step-by-step analysis:

Is it a graph?

• Yes: The problem provides pairs that can be swapped, which can be interpreted as edges in a graph, with each index in the string representing a node.

Is it a tree?

• No: The graph is not necessarily a hierarchy or a single connected component without cycles. Swaps can form multiple connected components.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The graph involves undirected connections since swapping is bidirectional and can include cycles.

Is the problem related to shortest paths?

• No: The problem is to rearrange characters to form the lexicographically smallest string, not to find shortest paths.

Does the problem involve connectivity?

• Yes: We need to find connected components in the graph where each connected component can be independently rearranged.

Does the problem have small constraints?

• Yes: Since the question allows us to handle the components individually and often DFS is tractable with problems involving manipulation of connected components under manageable constraint sizes, this is a viable route.

Conclusion: The flowchart suggests using DFS/backtracking for this problem, as it involves exploring connected components and rearranging elements within them to achieve the desired condition, which fits well with the characteristics of DFS. Depth-First Search is suitable for exploring each connected component thoroughly, which is crucial for determining which indices in the string can be considered together for the purpose of forming the smallest lexicographical sequence.

Intuition

To arrive at the solution, we need to think about how the operations will affect the string s. Since any number of swaps can be made, if there is a way to swap between any two indices through a series of swaps, we can effectively consider those characters as part of the same group. This leads us to the concept of "connected components".

A connected component in this context is a set of indices where any index can reach any other index within the same set through the pairs given. Once we've identified these connected components, within each component, we can sort the characters lexicographically and place them back into their original positions. This is possible because swaps within a component can rearrange characters in any order.

To implement this, we use a union-find data structure (also known as the disjoint-set data structure). With union-find, we can efficiently join two indices together and identify which component an index belongs to.

The solution approach is as follows:

1. Initialize a union-find structure to keep track of the connected components. This structure will map an index to its root or representative index of its connected component.

2. Iterate through the pairs and perform the union operation for each pair, effectively joining two indices into the same connected component.

3. Organize the characters into groups by their root or representative index.

4. Sort each group of characters in reverse lexicographical order. This makes it easier to pop the smallest character when rebuilding the string.

5. Finally, build the result string by picking the smallest available character from each index's connected component. Since we sorted groups in reverse order, we pop characters from the end of each list, ensuring they are selected in increasing lexicographical order.

The provided function find is a helper function for the union-find structure that finds the root of an index, and during the process, applies path compression to keep the structure efficient by flattening the hierarchy (making each member of a component directly point to the root).

Learn more about Depth-First Search, Breadth-First Search, Union Find and Sorting patterns.

Solution Approach

The solution primarily utilizes the Union-Find algorithm and an efficient array-based approach to organize and sort the characters based on their connected components.

Here's the walkthrough of the solution approach:

1. Initialization of Union-Find Structure: We begin by creating an array p which will act as the parent array for the Union-Find algorithm. Each element's initial value is its own index, implying that it is its own root.

   p = list(range(n)) # n is the length of string s

2. Union Operations: Iterate through the given pairs and perform the union operations. Instead of the traditional Union-Find with rank and path compression, the solution uses a simpler version that just assigns a new root to a component without considering the rank.

   for a, b in pairs:
       p[find(a)] = find(b)

   The find function locates the root of an index and applies path compression by setting the p[x] entry directly to the root of p[x].

3. Grouping Characters By Component: Once all pairs are processed, we need to group the characters according to their representative or root index. A dictionary d with default list values is used to append characters at the same group index.

   d = defaultdict(list)
   for i, c in enumerate(s):
       d[find(i)].append(c)

4. Sorting Groups: Each list within dictionary d is sorted in reverse order. Sorting in reverse order enables us to pop elements from the end of the list later, which is faster than popping from the beginning.

   for i in d.keys():
       d[i].sort(reverse=True)

5. Rebuilding the String: The final string is formed by iterating over the original string's indices, finding the root of each index, and popping the smallest character from the corresponding group.

   return "".join(d[find(i)].pop() for i in range(n))

The solution effectively groups indices of the string into connected components, sorts characters within each component, and then reconstructs the string by choosing the lexicographically smallest character available from each component for each position in the string. It's a smart application of the Union-Find algorithm and sorting to get the lexicographically smallest permutation satisfying the constraints.

Example Walkthrough

Let's consider a simple example to illustrate the solution approach with a string s = "dcab" and pairs pairs = [(0, 3), (1, 2)].

1. Initialization of Union-Find Structure: We first initialize the parent array p for Union-Find. The string length n is 4, so p = [0, 1, 2, 3].

2. Union Operations: Next, we perform union operations with the given pairs.

   • For pair (0, 3), we call find(0) and find(3) which return 0 and 3, and then we set p[0] to root of 3, which is 3. Now p = [3, 1, 2, 3].
   • For pair (1, 2), we call find(1) and find(2) which return 1 and 2, and then we set p[1] to root of 2, which is 2. Now p = [3, 2, 2, 3].

3. Grouping Characters By Component: We group the characters by their roots. Resulting in d = {2: ['a', 'b'], 3: ['d', 'c']}.

4. Sorting Groups: Sort each group in reverse order:

   • For group at root 2, sort ['a', 'b'] to get ['b', 'a'].
   • For group at root 3, sort ['d', 'c'] to get ['d', 'c']. Now we have d = {2: ['b', 'a'], 3: ['d', 'c']}.

5. Rebuilding the String: We now rebuild the string by iterating over each index remembering to pop from the end of the list in order to maintain the lexicographical order.

   • For index 0, its root is 3, so we pop from d[3] to get 'c' (d[3] = ['d']).
   • For index 1, its root is 2, so we pop from d[2] to get 'a' (d[2] = ['b']).
   • For index 2, its root is also 2, so we pop again from d[2] to get 'b' (d[2] = []).
   • For index 3, its root is 3, so we pop from d[3] to get 'd' (d[3] = []).

Finally, the rebuilt string is "cabd", which is the lexicographically smallest string we can obtain through the allowed swaps.

The entire process utilizes the union-find algorithm to efficiently manage connected components and applies sorting to ensure the smallest characters are placed first. The resulting string is created by choosing the smallest character allowed by the swap operations at each index.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code can be broken down into the following parts:

1. Union-Find Initialization: Initializing the parent list p with n elements where n is the length of string s takes O(n) time.

2. Union-Find Path Compression: For each pair in pairs, we potentially perform a union-by-rank and path compression. In the worst-case scenario, all elements can be connected, which ends up being nearly O(alpha(n)) per operation where alpha(n) is the Inverse Ackermann function. It is very slowly growing and is less than 5 for all practical values of n. Since there can be m pairs, this step takes at most O(m * alpha(n)) time.

3. Grouping letters by connected components: Enumerating s and finding the root of each index takes O(n * alpha(n)) time. Appending characters to the corresponding lists in the dictionary d takes O(1) time per character, resulting in O(n) for this step.

4. Sorting the letters: Each list in the dictionary d is sorted in reverse order. If k is the maximum size of the lists to be sorted, this step takes O(k * log(k)) time for each connected component. Since in the worst case there can be n/k such components, the overall sorting complexity can be O((n/k) * k * log(k)) = O(n * log(k)).

5. Reconstructing the result string: Popping an element from each list in d and forming a new string takes O(n) time.

Summing these up, the worst-case time complexity of the algorithm is O(n + m * alpha(n) + n * alpha(n) + n * log(k) + n). Since alpha(n) is a very slow-growing function and can be considered constant for all practical purposes, and log(k) <= log(n), we can simplify this to O(n * log(n) + m).

Space Complexity

The space complexity can be broken down into the following:

1. Parent array p: Uses O(n) space.

2. Dictionary d: Contains at most n lists. In the worst case, each list has one character, so the total space taken by d is O(n).

3. Auxiliary space for sorting: In the in-place sorting algorithm used by Python's .sort(), the space complexity is O(1). However, if we consider the space for the call stack due to recursive sorting algorithms in some implementations, this could give an additional space of O(log(k)) per connected component which sums up to O(n) in total.

Hence, the total space complexity is O(n).

Learn more about how to find time and space complexity quickly using problem constraints.