LeetCode #859 — EASY

Buddy Strings

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

The Problem

Problem Statement

Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, otherwise, return false.

Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j].

For example, swapping at indices 0 and 2 in "abcd" results in "cbad".

Example 1:

Input: s = "ab", goal = "ba"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'b' to get "ba", which is equal to goal.

Example 2:

Input: s = "ab", goal = "ab"
Output: false
Explanation: The only letters you can swap are s[0] = 'a' and s[1] = 'b', which results in "ba" != goal.

Example 3:

Input: s = "aa", goal = "aa"
Output: true
Explanation: You can swap s[0] = 'a' and s[1] = 'a' to get "aa", which is equal to goal.

Constraints:

1 <= s.length, goal.length <= 2 * 10⁴
s and goal consist of lowercase letters.

Step 01

Brute Force Baseline

Problem summary: Given two strings s and goal, return true if you can swap two letters in s so the result is equal to goal, otherwise, return false. Swapping letters is defined as taking two indices i and j (0-indexed) such that i != j and swapping the characters at s[i] and s[j]. For example, swapping at indices 0 and 2 in "abcd" results in "cbad".

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

"ab"
"ab"

Example 3

"aa"
"aa"

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

Upper-end input sizes

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 #859: Buddy Strings
class Solution {
    public boolean buddyStrings(String s, String goal) {
        int m = s.length(), n = goal.length();
        if (m != n) {
            return false;
        }
        int diff = 0;
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < n; ++i) {
            int a = s.charAt(i), b = goal.charAt(i);
            ++cnt1[a - 'a'];
            ++cnt2[b - 'a'];
            if (a != b) {
                ++diff;
            }
        }
        boolean f = false;
        for (int i = 0; i < 26; ++i) {
            if (cnt1[i] != cnt2[i]) {
                return false;
            }
            if (cnt1[i] > 1) {
                f = true;
            }
        }
        return diff == 2 || (diff == 0 && f);
    }
}

// Accepted solution for LeetCode #859: Buddy Strings
func buddyStrings(s string, goal string) bool {
	m, n := len(s), len(goal)
	if m != n {
		return false
	}
	diff := 0
	cnt1 := make([]int, 26)
	cnt2 := make([]int, 26)
	for i := 0; i < n; i++ {
		cnt1[s[i]-'a']++
		cnt2[goal[i]-'a']++
		if s[i] != goal[i] {
			diff++
		}
	}
	f := false
	for i := 0; i < 26; i++ {
		if cnt1[i] != cnt2[i] {
			return false
		}
		if cnt1[i] > 1 {
			f = true
		}
	}
	return diff == 2 || (diff == 0 && f)
}

# Accepted solution for LeetCode #859: Buddy Strings
class Solution:
    def buddyStrings(self, s: str, goal: str) -> bool:
        m, n = len(s), len(goal)
        if m != n:
            return False
        cnt1, cnt2 = Counter(s), Counter(goal)
        if cnt1 != cnt2:
            return False
        diff = sum(s[i] != goal[i] for i in range(n))
        return diff == 2 or (diff == 0 and any(v > 1 for v in cnt1.values()))

// Accepted solution for LeetCode #859: Buddy Strings
struct Solution;

use std::collections::HashSet;

impl Solution {
    fn buddy_strings(a: String, b: String) -> bool {
        let a: Vec<char> = a.chars().collect();
        let b: Vec<char> = b.chars().collect();
        let n = a.len();
        let m = b.len();
        if n != m {
            return false;
        }
        if a == b {
            let mut hs: HashSet<char> = HashSet::new();
            let mut sum = 0;
            for &c in &a {
                if !hs.insert(c) {
                    sum += 1;
                }
            }
            sum != 0
        } else {
            let mut pair: Vec<usize> = vec![];
            for i in 0..n {
                if a[i] != b[i] {
                    pair.push(i);
                }
            }
            if pair.len() == 2 {
                let i = pair[0];
                let j = pair[1];
                a[i] == b[j] && a[j] == b[i]
            } else {
                false
            }
        }
    }
}

#[test]
fn test() {
    let a = "ab".to_string();
    let b = "ba".to_string();
    assert_eq!(Solution::buddy_strings(a, b), true);
    let a = "ab".to_string();
    let b = "ab".to_string();
    assert_eq!(Solution::buddy_strings(a, b), false);
    let a = "aa".to_string();
    let b = "aa".to_string();
    assert_eq!(Solution::buddy_strings(a, b), true);
    let a = "aaaaaaabc".to_string();
    let b = "aaaaaaacb".to_string();
    assert_eq!(Solution::buddy_strings(a, b), true);
    let a = "".to_string();
    let b = "aa".to_string();
    assert_eq!(Solution::buddy_strings(a, b), false);
}

// Accepted solution for LeetCode #859: Buddy Strings
function buddyStrings(s: string, goal: string): boolean {
    const m = s.length;
    const n = goal.length;
    if (m != n) {
        return false;
    }
    const cnt1 = new Array(26).fill(0);
    const cnt2 = new Array(26).fill(0);
    let diff = 0;
    for (let i = 0; i < n; ++i) {
        cnt1[s.charCodeAt(i) - 'a'.charCodeAt(0)]++;
        cnt2[goal.charCodeAt(i) - 'a'.charCodeAt(0)]++;
        if (s[i] != goal[i]) {
            ++diff;
        }
    }
    for (let i = 0; i < 26; ++i) {
        if (cnt1[i] != cnt2[i]) {
            return false;
        }
    }
    return diff == 2 || (diff == 0 && cnt1.some(v => v > 1));
}

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.