854. K-Similar Strings
Problem Description
In this problem, we are given two strings s1
and s2
that are anagrams of each other. This means s1
and s2
consist of the same letters in a possibly different order. The concept of k
-similarity between the strings is introduced where k
represents the minimum number of swaps between any two positions in string s1
needed to transform it into string s2
.
The task is to find the smallest non-negative integer k
such that after exactly k
swaps of pairs of letters in s1
, we obtain s2
. Note that during each swap, the two letters to be swapped in s1
need not be adjacent. The challenge lies in determining the most efficient sequence of swaps that results in s1
becoming s2
, using the fewest number of swaps possible.
Flowchart Walkthrough
First, let's determine the appropriate algorithm using the Flowchart. Here's a step-by-step examination:
Is it a graph?
- Yes: The transformation process between strings can be conceptualized as a graph where each string is a node and there's an edge between two nodes if you can change one string into the other by swapping characters.
Is it a tree?
- No: Although transformations follow a specific rule set, this doesn't inherently create a hierarchical tree as multiple paths might exist between any two nodes (multiple swaps can be made to reach from one string to another).
Is the problem related to directed acyclic graphs (DAGs)?
- No: Even though transformations are directed (one valid swap leads to another string), the swaps can revert strings to previous states, possible cycles avoid defining it strictly as acyclic.
Is the problem related to shortest paths?
- Yes: The key objective is to find the minimum number of swaps to transform one string into another, akin to finding the shortest path in a graph.
Is the graph weighted?
- No: Each swap counts equally as a single move, evoking an unweighted graph scenario where each edge (swap) has the same 'weight' or 'cost'.
Conclusion: Following the flowchart leads to the use of BFS as the best approach in this context of unweighted shortest-path problems in a graph. Thus, Breadth-First Search (BFS) is the suggested algorithm for efficiently solving Leetcode 854. K-Similar Strings.
Intuition
To solve this problem, we use a Breadth-First Search (BFS) approach. We think of each string that can be formed by swapping letters in s1
as a node in a graph, with an edge between two nodes if one can be transformed into the other by a single swap. The problem then becomes a shortest path problem, where we want to find the fewest number of edges (swaps) required to reach s2
starting from s1
.
With BFS, we traverse the graph level by level. Each level of the traversal represents an incremental increase in the number of swaps (increment in k
). At each step, we generate all possible strings that are one swap away from the current string while ensuring two conditions:
- We only swap letters that would bring us closer to the string
s2
(the letter being swapped ins1
must match the letter in the same position ins2
and not be in its final position ins1
). - We haven't already visited the newly created string in our traversal; this avoids unnecessary work and ensures we are always heading in the direction of increasing
k
optimally.
We utilize a queue to maintain the strings at the current level of BFS and a set vis
to keep track of visited strings to prevent revisiting them. As soon as we find s2
during our traversal, we have found the smallest k
, which represents the length of the shortest path in our graph of strings, and we return this value.
The rationale for why BFS works well here is because it guarantees that the first time we reach s2
, it's through the shortest path possible, hence the minimum number of swaps k
.
Learn more about Breadth-First Search patterns.
Solution Approach
The solution makes use of the Breadth-First Search (BFS) algorithm, which is a standard approach for finding the shortest path in an unweighted graph. Here, the nodes of the graph are different permutations of the string s1
, and the edges represent a valid swap that can transform one permutation into another.
In the provided Python code, this approach is implemented as follows:
-
BFS is carried out using a queue named
q
— adeque
data structure that allows for the efficient addition and removal of elements from both ends. Initially, the queue contains only the starting strings1
. -
A set named
vis
stores all the visited strings to avoid repeated processing and to keep track of the unique permutations that have already been considered. -
The function
next
generates all possible strings that can be reached from the current strings
with a single swap that gets us closer to the target strings2
. To do this,next
finds the first indexi
where the letters ins
ands2
differ. Then it looks for indicesj
beyondi
where swapping the letter at positioni
with the letter at positionj
is beneficial, namely where:s[j]
is equal tos2[i]
, meaning that this letter is desired at positioni
to matchs2
.s[j]
is not already in its final position (s[j] != s2[j]
), ensuring that the swap will contribute to making the strings more similar.
-
With the BFS traversal, the algorithm goes through all strings at the current level of similarity before increasing the value of
ans
, which represents the number of swapsk
. Each outer while loop represents another level of depth in the search, hence another swap. -
The inner loop iterates through all strings in the current level of the
q
, performing swaps via thenext
function to get the next level of permutations and checking if any of these permutations matchs2
. -
When the
s2
string is finally found,ans
is returned, which by the nature of BFS, will be the minimum number of swaps required. -
The loop terminates as soon as
s2
is encountered because BFS guarantees that the first occurrence ofs2
will have been reached via the shortest path (minimum number of swaps).
Through this approach, even though the problem space could be large, the algorithm ensures an efficient traversal by only considering useful swaps and by avoiding the revisitation of previously encountered strings.
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Start EvaluatorExample Walkthrough
Let's consider a small example to illustrate the solution approach. Suppose we have s1 = "123"
and s2 = "321"
. We need to find the minimum number of swaps in s1
to transform it into s2
.
Beginning with the initial string s1 = "123"
, we enqueue this string into the queue q
and mark "123" as visited in the set vis
.
First Iteration
q = ["123"]
,vis = {"123"}
, andans = 0
(no swaps done yet).- We dequeue "123" and generate next possible strings.
- We notice that
s1[0] != s2[0]
, so we look for characters ins1
that matchs2[0]
. We find "3" ats1[2]
matchess2[0]
. - However, swapping "1" and "3" in "123" yields "321", which is exactly
s2
. Since we've founds2
,ans = 1
(one swap). - Our search ends here because we have transformed
s1
intos2
.
End of the process with ans = 1
. The minimum number of swaps required is 1.
In this case, we found the answer quickly due to the simplicity of the example. In a more complex scenario, BFS would continue exploring further levels with more swaps until s2
is reached. This approach ensures the first time s2
is encountered during the BFS, it would be the result of the minimum number of swaps needed, hence we would return that as the smallest non-negative integer k
.
Solution Implementation
1from collections import deque
2
3class Solution:
4 def kSimilarity(self, s1: str, s2: str) -> int:
5 # Helper function to generate the next state of strings that are one swap closer to s2
6 def generate_next_states(current_string):
7 i = 0
8 # Find the first position where characters differ
9 while current_string[i] == s2[i]:
10 i += 1
11
12 next_states = []
13 # Loop through the string to find swap positions
14 for j in range(i + 1, string_length):
15 # Swap only if it brings current_string closer to s2
16 if current_string[j] == s2[i] and current_string[j] != s2[j]:
17 swapped_string = (current_string[:i] + current_string[j] +
18 current_string[i+1:j] + current_string[i] +
19 current_string[j+1:])
20 next_states.append(swapped_string)
21 return next_states
22
23 # Initialize the queue with starting string s1
24 queue = deque([s1])
25 # Set to store visited strings
26 visited = set([s1])
27 # Number of swaps made
28 swaps = 0
29 string_length = len(s1)
30
31 # Breadth-first search loop
32 while True:
33 queue_length = len(queue)
34 for _ in range(queue_length):
35 current_string = queue.popleft()
36 # If the current string equals s2, we've found the minimum swaps
37 if current_string == s2:
38 return swaps
39 # Generate and iterate over next possible swap states
40 for next_state in generate_next_states(current_string):
41 # Enqueue the next state if not visited
42 if next_state not in visited:
43 visited.add(next_state)
44 queue.append(next_state)
45 # Increment the swap counter after each breadth level is processed
46 swaps += 1
47
1class Solution {
2
3 // Method to find K similarity between strings s1 and s2.
4 // It calculates the minimum number of adjacent swaps to make string s1 equal to s2
5 public int kSimilarity(String s1, String s2) {
6 Deque<String> queue = new ArrayDeque<>(); // Use a queue to store strings for BFS
7 Set<String> visited = new HashSet<>(); // Use a set to keep track of visited strings
8 queue.offer(s1);
9 visited.add(s1);
10 int stepCount = 0; // Initialize a step counter
11
12 // Perform BFS until s1 can be transformed into s2
13 while (true) {
14 int queueSize = queue.size();
15 // Process all strings at the current step
16 for (int i = queueSize; i > 0; --i) {
17 String currentString = queue.pollFirst();
18 // If the current string matches s2, return the number of steps taken
19 if (currentString.equals(s2)) {
20 return stepCount;
21 }
22 // Generate all possible next moves and add them to the queue if not visited
23 for (String nextString : generateNextStrings(currentString, s2)) {
24 if (!visited.contains(nextString)) {
25 visited.add(nextString);
26 queue.offer(nextString);
27 }
28 }
29 }
30 stepCount++; // Increment step count after each breadth of BFS
31 }
32 }
33
34 // Helper method to generate the next moves by swapping characters in s to move towards s2
35 private List<String> generateNextStrings(String s, String s2) {
36 int i = 0;
37 int length = s.length();
38 char[] chars = s.toCharArray();
39 // Find the first non-matching character position
40 while (chars[i] == s2.charAt(i)) {
41 i++;
42 }
43
44 List<String> nextMoves = new ArrayList<>();
45 // Generate all valid swaps to move characters closer to their final positions
46 for (int j = i + 1; j < length; ++j) {
47 // Swap if the character in 's' can be swapped to match the corresponding one in 's2'
48 // and it is not in its final position already
49 if (chars[j] == s2.charAt(i) && chars[j] != s2.charAt(j)) {
50 swap(chars, i, j); // Perform the swap
51 nextMoves.add(new String(chars)); // Add the new string after the swap
52 swap(chars, i, j); // Swap back for the next iteration
53 }
54 }
55 return nextMoves;
56 }
57
58 // Helper method to swap two characters in an array
59 private void swap(char[] chars, int i, int j) {
60 char temp = chars[i];
61 chars[i] = chars[j];
62 chars[j] = temp;
63 }
64}
65
1#include <string>
2#include <vector>
3#include <queue>
4#include <unordered_set>
5
6using namespace std;
7
8class Solution {
9public:
10 // Function that returns the minimum number of adjacent swaps required to transform
11 // string s1 into string s2 where the strings are k-similar.
12 int kSimilarity(string s1, string s2) {
13 queue<string> workQueue; // Queue to hold strings to be processed
14 workQueue.push(s1); // Starting with s1
15
16 unordered_set<string> visited; // Set to keep track of visited strings to avoid repetition
17 visited.insert(s1);
18
19 int steps = 0; // Counter to track the number of swaps made
20
21 // Main processing loop
22 while (true) {
23 int levelSize = workQueue.size(); // Process every string in the current level
24 for (int i = 0; i < levelSize; ++i) {
25 auto currentString = workQueue.front();
26 workQueue.pop();
27
28 // If the current string is equal to s2, we have found the solution
29 if (currentString == s2) {
30 return steps;
31 }
32
33 // Generate all possible strings that are one swap closer to s2
34 vector<string> neighbors = findNeighbors(currentString, s2);
35 for (auto& neighbor : neighbors) {
36 // If we have not visited this string before, add it to the queue and mark it as visited
37 if (visited.count(neighbor) == 0) {
38 visited.insert(neighbor);
39 workQueue.push(neighbor);
40 }
41 }
42 }
43 steps++; // Increment the number of swaps (levels in BFS) after processing the current level
44 }
45 }
46
47 // Helper function to generate all possible strings that are one swap closer to s2
48 vector<string> findNeighbors(string& currentString, string& target) {
49 int i = 0;
50 int n = currentString.size();
51
52 // Find the first position where the characters differ between the two strings
53 for (; i < n && currentString[i] == target[i]; ++i) {}
54
55 vector<string> result; // Vector to hold all the valid next strings
56
57 // Try swapping the character at index 'i' with each character 'j' that is different from target[j]
58 // and is the same as target[i], moving us one step closer to s2 with each swap
59 for (int j = i + 1; j < n; ++j) {
60 if (currentString[j] == target[i] && currentString[j] != target[j]) {
61 swap(currentString[i], currentString[j]); // Execute the swap
62 result.push_back(currentString); // Add the new string to the result list
63 swap(currentString[i], currentString[j]); // Swap back for next iteration
64 }
65 }
66 return result;
67 }
68};
69
1type StringQueue = string[]; // Alias for an array used as a queue
2type StringSet = Set<string>; // Alias for a set of strings
3
4let workQueue: StringQueue = []; // Queue to hold strings to be processed
5let visited: StringSet = new Set<string>(); // Set to keep track of visited strings to avoid repetition
6
7let steps: number = 0; // Counter to track the number of swaps made
8
9// Function that adds an element to the queue
10function enqueue(queue: StringQueue, element: string) {
11 queue.push(element);
12}
13
14// Function that removes and returns the first element from the queue
15function dequeue(queue: StringQueue): string | undefined {
16 return queue.shift();
17}
18
19// Function that returns the length of the queue
20function queueSize(queue: StringQueue): number {
21 return queue.length;
22}
23
24// Function that returns the minimum number of adjacent swaps required to transform
25// string s1 into string s2 where the strings are k-similar
26function kSimilarity(s1: string, s2: string): number {
27 enqueue(workQueue, s1); // Starting with s1
28 visited.add(s1);
29
30 while (true) {
31 let levelSize = queueSize(workQueue); // Process every string in the current level
32 for (let i = 0; i < levelSize; i++) {
33 let currentString = dequeue(workQueue);
34
35 if (currentString === s2) {
36 // If the current string is equal to s2, we have found the solution
37 return steps;
38 }
39
40 let neighbors = findNeighbors(currentString, s2);
41 for (let neighbor of neighbors) {
42 if (!visited.has(neighbor)) {
43 // If we have not visited this string before, add it to the queue and mark it as visited
44 visited.add(neighbor);
45 enqueue(workQueue, neighbor);
46 }
47 }
48 }
49 steps++; // Increment the number of swaps (levels in BFS) after processing the current level
50 }
51}
52
53// Helper function to generate all possible strings that are one swap closer to s2
54function findNeighbors(currentString: string, target: string): string[] {
55 let pos = 0;
56 const n = currentString.length;
57
58 // Find the first position where the characters differ between the two strings
59 while (pos < n && currentString.charAt(pos) === target.charAt(pos)) {
60 pos++;
61 }
62
63 let results: string[] = []; // Array to hold all the valid next strings
64
65 // Try swapping the character at index 'pos' with each character 'j'
66 // that is different from target[j] and is the same as target[pos],
67 // moving us one step closer to s2 with each swap
68 for (let j = pos + 1; j < n; j++) {
69 if (
70 currentString.charAt(j) === target.charAt(pos) &&
71 currentString.charAt(j) !== target.charAt(j)
72 ) {
73 let swappedString =
74 currentString.substring(0, pos) +
75 currentString.charAt(j) +
76 currentString.substring(pos + 1, j) +
77 currentString.charAt(pos) +
78 currentString.substring(j + 1);
79 results.push(swappedString); // Add the new string to the results array
80 }
81 }
82 return results;
83}
84
Time and Space Complexity
The given Python code is solving the k-Similarity problem, which aims to determine the minimum number of adjacent swaps needed to transform one string into another, given that the strings are of the same length and consist of the same characters.
Time Complexity:
The time complexity of this algorithm primarily depends on the number of levels we traverse in the BFS and the number of branches we generate at each level.
- Breadth-First Search (BFS) Traversal: The code uses a BFS traversal to explore all possibilities via swapping operations. The BFS traversal could, in the worst case, explore all permutations of the string
s1
, which would beO(n!)
wheren
is the length of the strings1
. - Branching Factor in BFS: At each level of BFS, the
next()
function can potentially generate multiple branches. This branching factor can be at mostn
, wheren
is the length of the strings since we could theoretically swap the first misaligned character with any of the othern - 1
remaining characters that match the required character ins2
. - Combining the above two, a loose upper bound for the time complexity of the BFS traversal with branching is
O(n * n!)
, but this is a very pessimistic upper bound as it assumes we explore all permutations ofs1
, and in each step, we have the maximum amount of valid swaps.
However, the BFS guarantees that we find the shortest path, so we will likely find a solution before exploring all permutations. We can say that the average time complexity is better than this upper bound but it's hard to quantify without more information about the properties of the specific strings s1
and s2
.
Space Complexity:
The space complexity is influenced by two factors:
- Queue Storage (q): The maximum size of the queue could theoretically approach the number of all possible permutations in the worst case, which is
O(n!)
. - Visited Set (vis): The visited set can hold a unique entry for each visited permutation, which, in the worst case, can also be
O(n!)
.
Therefore, the worst-case space complexity of the algorithm is O(n!)
. This will likely be significantly less in practice due to the pruning of the BFS search space, meaning we do not typically visit all permutations.
Learn more about how to find time and space complexity quickly using problem constraints.
How does quick sort divide the problem into subproblems?
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