Buddy Strings

Buddy Strings - Problem

Given two strings s and goal, return true if you can swap exactly two letters in s so that 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".

Input & Output

Example 1 — Valid Swap

$ Input: s = "ab", goal = "ba"

› Output: true

💡 Note: We can swap s[0] and s[1] to get "ba". The character 'a' at position 0 becomes position 1, and 'b' at position 1 becomes position 0.

Example 2 — Invalid Different Characters

$ Input: s = "ab", goal = "ab"

› Output: false

💡 Note: The strings are already identical, but we need exactly one swap. Since there are no duplicate characters, no valid swap exists.

Example 3 — Valid with Duplicates

$ Input: s = "aa", goal = "aa"

› Output: true

💡 Note: The strings are identical and contain duplicate characters. We can swap the two 'a's (positions 0 and 1) to get the same string.

Constraints

1 ≤ s.length, goal.length ≤ 2 × 10⁴
s and goal consist of lowercase letters

Visualization

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Asked in

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The key insight is that buddy strings are possible in exactly two cases: (1) strings differ at exactly 2 positions and those characters can be swapped, or (2) strings are identical with at least one duplicate character. Best approach uses frequency counting with difference analysis. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(n)

For each pair of indices (i, j) where i < j, swap characters at these positions in s and check if the result equals goal. This explores all possible single swaps.

Character Frequency Analysis

⏱️ Time: O(n) Space: O(1)

First check if both strings have the same character frequencies. Then find positions where characters differ. Valid if exactly 0 differences (with duplicates) or exactly 2 differences (swappable).

Brute Force — Algorithm Steps

Step 1: Check if strings have same length (early return false if not)
Step 2: Try every pair of indices (i, j) where i < j
Step 3: For each pair, swap s[i] and s[j] and check if result equals goal

Visualization

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Step-by-Step Walkthrough

Initial Strings

Compare s and goal

Try Swaps

Test each possible swap pair

Check Match

See if any swap creates goal

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool solution(char* s, char* goal) {
    int len_s = strlen(s);
    int len_goal = strlen(goal);
    
    if (len_s != len_goal) {
        return false;
    }
    
    // Try every possible swap
    for (int i = 0; i < len_s; i++) {
        for (int j = i + 1; j < len_s; j++) {
            // Create swapped string
            char* swapped = malloc((len_s + 1) * sizeof(char));
            strcpy(swapped, s);
            
            // Swap characters
            char temp = swapped[i];
            swapped[i] = swapped[j];
            swapped[j] = temp;
            
            if (strcmp(swapped, goal) == 0) {
                free(swapped);
                return true;
            }
            
            free(swapped);
        }
    }
    
    return false;
}

int main() {
    char s[1001], goal[1001];
    fgets(s, sizeof(s), stdin);
    fgets(goal, sizeof(goal), stdin);
    
    // Remove newlines
    s[strcspn(s, "\n")] = 0;
    goal[strcspn(goal, "\n")] = 0;
    
    bool result = solution(s, goal);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Nested loops O(n²) for pairs, plus O(n) string comparison for each pair

⚠ Quadratic Growth

Space Complexity

O(n)

Creating temporary strings for each swap attempt

⚡ Linearithmic Space

59.1K Views

Medium Frequency

~15 min Avg. Time

1.8K Likes

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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