What is the time complexity of K-Similar Strings?

The optimal solution has a time complexity of O(n^2 * 2^n).

What is the space complexity of K-Similar Strings?

The space complexity is O(2^n).

What data structures are used to solve K-Similar Strings?

This problem uses Hash Table, String, Breadth-First Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving K-Similar Strings?

Insight 1: The problem can be modeled as a graph search where each state is a possible string that can be obtained from `s1` through swaps. The goal is to find the shortest path from `s1` to `s2`. Insight 2: Breadth-First Search (BFS) is suitable for finding the shortest path in an unweighted graph, making it an ideal algorithm for this problem.

Which companies ask LeetCode 854. K-Similar Strings in interviews?

This problem is frequently asked at DoorDash.

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K-Similar Strings - LeetCode 854 Solution

K-Similar Strings - Complete Solution Guide

K-Similar Strings is LeetCode problem 854, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Strings s1 and s2 are k -similar (for some non-negative integer k ) if we can swap the positions of two letters in s1 exactly k times so that the resulting string equals s2 . Given two anagrams s1 and s2 , return the smallest k for which s1 and s2 are k -similar . Example 1: Input: s1 = "ab", s2 = "ba" Output: 1 Explanation: The two string are 1-similar because we can use one swap to change s1 to s2: "ab" --> "ba". Example 2: Input: s1 = "abc", s2 = "bca" Output: 2 Explanation: The two strings a

Detailed Explanation

The problem asks to find the minimum number of swaps required to transform string `s1` into string `s2`, where `s1` and `s2` are anagrams of each other. A swap involves exchanging the positions of two characters in `s1`. The goal is to find the 'k' value, where `s1` and `s2` are considered k-similar.

Solution Approach

The solution uses Breadth-First Search (BFS) to explore the possible swap combinations and find the minimum number of swaps needed to transform `s1` into `s2`. The algorithm maintains a queue of states to explore, where each state is a pair of the current string and the number of swaps performed so far. A visited set is used to avoid revisiting states.

Step-by-Step Algorithm

Step 1: Preprocess the strings: Create filtered versions of `s1` and `s2` containing only the characters that are at different positions between the two original strings. This reduces the search space.
Step 2: Initialize BFS: Create a queue and add the initial state (filtered `s1` and swap count 0). Also, create a set to keep track of visited states to prevent cycles.
Step 3: Iterate using BFS: While the queue is not empty, dequeue a state (current string and swap count).
Step 4: Check for the goal state: If the current string is equal to the filtered `s2`, return the swap count (k).
Step 5: Find the first mismatched character: Iterate through the current string to find the index `i` of the first character that is different from the character at the same index in filtered `s2`.
Step 6: Explore possible swaps: Iterate through the remaining characters of the current string (starting from `i+1`). If a character at index `j` is equal to the character at index `i` in the target string `s2`, perform a swap between indices `i` and `j`.
Step 7: Add the new state to the queue: If the new string generated from the swap has not been visited before, add it to the queue with an incremented swap count. Mark the new string as visited.
Step 8: Backtrack (undo the swap): After exploring the swap, revert it back to the original state to explore other possible swaps. This ensures all possible swaps are considered.
Step 9: If the queue becomes empty and the target string is not found, it means that it is not reachable. Return -1 in this case although since the input strings are anagrams this case cannot happen.

Key Insights

Insight 1: The problem can be modeled as a graph search where each state is a possible string that can be obtained from `s1` through swaps. The goal is to find the shortest path from `s1` to `s2`.
Insight 2: Breadth-First Search (BFS) is suitable for finding the shortest path in an unweighted graph, making it an ideal algorithm for this problem.
Insight 3: To optimize, only consider swapping characters that are not in their correct positions. Pre-filtering these positions simplifies the search space and reduces the number of unnecessary operations.

Complexity Analysis

Time Complexity: O(n^2 * 2^n)

Space Complexity: O(2^n)

Topics

This problem involves: Hash Table, String, Breadth-First Search.

Companies

Asked at: DoorDash.

K-Similar Strings | LeetCode 854

Hard

Hash Table String Breadth-First Search

Time: O(n^2 * 2^n)Space: O(2^n)

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Queue;
import java.util.Set;

class Solution {
    public int kSimilarity(String s1, String s2) {
        if (s1.equals(s2)) {
            return 0;
        }

        List<Character> s1List = new ArrayList<>();
        List<Character> s2List = new ArrayList<>();
        for (int i = 0; i < s1.length(); i++) {
            if (s1.charAt(i) != s2.charAt(i)) {
                s1List.add(s1.charAt(i));
                s2List.add(s2.charAt(i));
            }
        }

        s1 = "";
        s2 = "";
        for (char c : s1List) {
            s1 += c;
        }
        for (char c : s2List) {
            s2 += c;
        }

        Queue<Pair> q = new ArrayDeque<>();
        q.offer(new Pair(s1, 0));
        Set<String> visited = new HashSet<>();
        visited.add(s1);

        int n = s1.length();

        while (!q.isEmpty()) {
            Pair current = q.poll();
            String currentS = current.s;
            int k = current.k;

            if (currentS.equals(s2)) {
                return k;
            }

            int i = 0;
            while (currentS.charAt(i) == s2.charAt(i)) {
                i++;
            }

            List<Character> currentSList = new ArrayList<>();
            for (char c : currentS.toCharArray()) {
                currentSList.add(c);
            }

            for (int j = i + 1; j < n; j++) {
                if (currentSList.get(j) == s2.charAt(i)) {
                    char temp = currentSList.get(i);
                    currentSList.set(i, currentSList.get(j));
                    currentSList.set(j, temp);
                    
                    StringBuilder sb = new StringBuilder();
                    for(char c : currentSList){
                        sb.append(c);
                    }
                    String nextS = sb.toString();

                    if (!visited.contains(nextS)) {
                        visited.add(nextS);
                        q.offer(new Pair(nextS, k + 1));
                    }

                    temp = currentSList.get(i);
                    currentSList.set(i, currentSList.get(j));
                    currentSList.set(j, temp);
                }
            }
        }

        return -1;
    }

    private static class Pair {
        String s;
        int k;

        public Pair(String s, int k) {
            this.s = s;
            this.k = k;
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 854: K-Similar Strings - Detailed explanation and solution approaches.

Topics

Hash Table String Breadth-First Search

Companies

DoorDash

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:854

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Preprocess the strings: Create filtered versions of `s1` and `s2` containing only the characters that are at different positions between the two original strings. This reduces the search space.
2
Step 2: Initialize BFS: Create a queue and add the initial state (filtered `s1` and swap count 0). Also, create a set to keep track of visited states to prevent cycles.
3
Step 3: Iterate using BFS: While the queue is not empty, dequeue a state (current string and swap count).
4
Step 4: Check for the goal state: If the current string is equal to the filtered `s2`, return the swap count (k).
5
Step 5: Find the first mismatched character: Iterate through the current string to find the index `i` of the first character that is different from the character at the same index in filtered `s2`.
6
Step 6: Explore possible swaps: Iterate through the remaining characters of the current string (starting from `i+1`). If a character at index `j` is equal to the character at index `i` in the target string `s2`, perform a swap between indices `i` and `j`.
7
Step 7: Add the new state to the queue: If the new string generated from the swap has not been visited before, add it to the queue with an incremented swap count. Mark the new string as visited.
8
Step 8: Backtrack (undo the swap): After exploring the swap, revert it back to the original state to explore other possible swaps. This ensures all possible swaps are considered.
9
Step 9: If the queue becomes empty and the target string is not found, it means that it is not reachable. Return -1 in this case although since the input strings are anagrams this case cannot happen.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n^2 * 2^n)

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity arises from the combination of exploring all possible states reachable by swaps (O(2^n)), and the operations performed within each state (O(n^2) for finding mismatched characters and performing swaps). The space complexity is determined by the maximum number of states stored in the queue and visited set, which can be exponential in the length of the input strings.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

A* Search

A* search could be used instead of BFS. A heuristic function (e.g., the number of mismatched characters divided by 2, since each swap can correct at most 2 characters) can guide the search towards the goal state. However, the heuristic must be admissible and consistent to guarantee optimality.

Time: Potentially better than BFS in some cases, but still exponential in the worst case. Depends on the quality of the heuristic.Space: Potentially similar to BFS, depending on how well the heuristic prunes the search space.

Trade-offs: A* search might perform better than BFS if a good heuristic is available to guide the search. However, designing a suitable heuristic can be challenging, and a poorly designed heuristic can lead to worse performance than BFS. BFS is guaranteed to find the optimal solution, while A* search's optimality depends on the heuristic function.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph search where each state is a possible string that can be obtained from `s1` through swaps. The goal is to find the shortest path from `s1` to `s2`.
2Insight 2: Breadth-First Search (BFS) is suitable for finding the shortest path in an unweighted graph, making it an ideal algorithm for this problem.
3Insight 3: To optimize, only consider swapping characters that are not in their correct positions. Pre-filtering these positions simplifies the search space and reduces the number of unnecessary operations.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty strings or identical strings. The code explicitly checks if `s1` and `s2` are identical and returns 0 in that case, handling the trivial case where no swaps are needed.
Edge case 2: All characters are in the correct position after filtering. The preprocessing step ensures that the strings `s1` and `s2` only contain characters that are out of place. If no such characters exist, the strings are already equal after filtering, and the algorithm correctly identifies this and returns 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling duplicate states properly. Without a visited set, the BFS can get stuck in a cycle, leading to infinite loops and incorrect results.

✗Common mistake 2: Not performing backtracking after exploring a swap. Failing to undo the swap after exploring a state can lead to missing other possible swap combinations.

✗Common mistake 3: Not filtering out the common characters between the string. This will cause unnecessary operations and lead to TLE.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing data transfer sequences. In scenarios where data needs to be transferred from one state to another with minimal operations, this type of algorithm can be used to find the shortest sequence of data swaps.

🏦Real-world scenario 2: Task scheduling. The problem of minimizing the number of operations (swaps) to reorder tasks to meet a specific schedule can be solved using similar concepts.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "K-Similar Strings"?

For this hard-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?