LeetCode #854 — HARD

K-Similar Strings

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 are 2-similar because we can use two swaps to change s1 to s2: "abc" --> "bac" --> "bca".

Constraints:

1 <= s1.length <= 20
s2.length == s1.length
s1 and s2 contain only lowercase letters from the set {'a', 'b', 'c', 'd', 'e', 'f'}.
s2 is an anagram of s1.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map

Example 1

"ab"
"ba"

Example 2

"abc"
"bca"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #854: K-Similar Strings
class Solution {
    public int kSimilarity(String s1, String s2) {
        Deque<String> q = new ArrayDeque<>();
        Set<String> vis = new HashSet<>();
        q.offer(s1);
        vis.add(s1);
        int ans = 0;
        while (true) {
            for (int i = q.size(); i > 0; --i) {
                String s = q.pollFirst();
                if (s.equals(s2)) {
                    return ans;
                }
                for (String nxt : next(s, s2)) {
                    if (!vis.contains(nxt)) {
                        vis.add(nxt);
                        q.offer(nxt);
                    }
                }
            }
            ++ans;
        }
    }

    private List<String> next(String s, String s2) {
        int i = 0, n = s.length();
        char[] cs = s.toCharArray();
        for (; cs[i] == s2.charAt(i); ++i) {
        }

        List<String> res = new ArrayList<>();
        for (int j = i + 1; j < n; ++j) {
            if (cs[j] == s2.charAt(i) && cs[j] != s2.charAt(j)) {
                swap(cs, i, j);
                res.add(new String(cs));
                swap(cs, i, j);
            }
        }
        return res;
    }

    private void swap(char[] cs, int i, int j) {
        char t = cs[i];
        cs[i] = cs[j];
        cs[j] = t;
    }
}

// Accepted solution for LeetCode #854: K-Similar Strings
func kSimilarity(s1 string, s2 string) int {
	next := func(s string) []string {
		i := 0
		res := []string{}
		for ; s[i] == s2[i]; i++ {
		}
		for j := i + 1; j < len(s1); j++ {
			if s[j] == s2[i] && s[j] != s2[j] {
				res = append(res, s[:i]+string(s[j])+s[i+1:j]+string(s[i])+s[j+1:])
			}
		}
		return res
	}

	q := []string{s1}
	vis := map[string]bool{s1: true}
	ans := 0
	for {
		for i := len(q); i > 0; i-- {
			s := q[0]
			q = q[1:]
			if s == s2 {
				return ans
			}
			for _, nxt := range next(s) {
				if !vis[nxt] {
					vis[nxt] = true
					q = append(q, nxt)
				}
			}
		}
		ans++
	}
}

# Accepted solution for LeetCode #854: K-Similar Strings
class Solution:
    def kSimilarity(self, s1: str, s2: str) -> int:
        def next(s):
            i = 0
            while s[i] == s2[i]:
                i += 1
            res = []
            for j in range(i + 1, n):
                if s[j] == s2[i] and s[j] != s2[j]:
                    res.append(s2[: i + 1] + s[i + 1 : j] + s[i] + s[j + 1 :])
            return res

        q = deque([s1])
        vis = {s1}
        ans, n = 0, len(s1)
        while 1:
            for _ in range(len(q)):
                s = q.popleft()
                if s == s2:
                    return ans
                for nxt in next(s):
                    if nxt not in vis:
                        vis.add(nxt)
                        q.append(nxt)
            ans += 1

// Accepted solution for LeetCode #854: K-Similar Strings
struct Solution;

use std::collections::HashSet;
use std::collections::VecDeque;

impl Solution {
    fn k_similarity(a: String, b: String) -> i32 {
        let n = a.len();
        let a: Vec<char> = a.chars().collect();
        let b: Vec<char> = b.chars().collect();
        let mut queue = VecDeque::new();
        let mut visited = HashSet::new();
        visited.insert(a.clone());
        queue.push_back(a);
        let mut res = 0;
        while !queue.is_empty() {
            'outer: for _ in 0..queue.len() {
                let mut front = queue.pop_front().unwrap();
                let mut i = 0;
                while i < n && front[i] == b[i] {
                    i += 1;
                }
                if i == n {
                    return res;
                } else {
                    for j in i + 1..n {
                        if front[j] == b[i] && front[i] == b[j] {
                            front.swap(i, j);
                            if visited.insert(front.clone()) {
                                queue.push_back(front.clone());
                            }
                            front.swap(i, j);
                            continue 'outer;
                        }
                    }
                    for j in i + 1..n {
                        if front[j] == b[i] {
                            front.swap(i, j);
                            if visited.insert(front.clone()) {
                                queue.push_back(front.clone());
                            }
                            front.swap(i, j);
                        }
                    }
                }
            }
            res += 1;
        }
        0
    }
}

#[test]
fn test() {
    let a = "ab".to_string();
    let b = "ba".to_string();
    let res = 1;
    assert_eq!(Solution::k_similarity(a, b), res);
    let a = "abc".to_string();
    let b = "bca".to_string();
    let res = 2;
    assert_eq!(Solution::k_similarity(a, b), res);
    let a = "abac".to_string();
    let b = "baca".to_string();
    let res = 2;
    assert_eq!(Solution::k_similarity(a, b), res);
    let a = "aabc".to_string();
    let b = "abca".to_string();
    let res = 2;
    assert_eq!(Solution::k_similarity(a, b), res);
}

// Accepted solution for LeetCode #854: K-Similar Strings
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #854: K-Similar Strings
// class Solution {
//     public int kSimilarity(String s1, String s2) {
//         Deque<String> q = new ArrayDeque<>();
//         Set<String> vis = new HashSet<>();
//         q.offer(s1);
//         vis.add(s1);
//         int ans = 0;
//         while (true) {
//             for (int i = q.size(); i > 0; --i) {
//                 String s = q.pollFirst();
//                 if (s.equals(s2)) {
//                     return ans;
//                 }
//                 for (String nxt : next(s, s2)) {
//                     if (!vis.contains(nxt)) {
//                         vis.add(nxt);
//                         q.offer(nxt);
//                     }
//                 }
//             }
//             ++ans;
//         }
//     }
// 
//     private List<String> next(String s, String s2) {
//         int i = 0, n = s.length();
//         char[] cs = s.toCharArray();
//         for (; cs[i] == s2.charAt(i); ++i) {
//         }
// 
//         List<String> res = new ArrayList<>();
//         for (int j = i + 1; j < n; ++j) {
//             if (cs[j] == s2.charAt(i) && cs[j] != s2.charAt(j)) {
//                 swap(cs, i, j);
//                 res.add(new String(cs));
//                 swap(cs, i, j);
//             }
//         }
//         return res;
//     }
// 
//     private void swap(char[] cs, int i, int j) {
//         char t = cs[i];
//         cs[i] = cs[j];
//         cs[j] = t;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.