What exactly defines a swap in the Buddy Strings problem?

A swap is defined as selecting two distinct indices i and j in s and exchanging s[i] with s[j].

Can Buddy Strings return true if the strings are identical?

Yes, but only if there is at least one character that occurs more than once, allowing a valid swap.

Why use a hash table instead of just comparing mismatched indices?

A hash table allows detection of duplicate characters efficiently, which is required when strings are identical.

What should be done if s and goal have different lengths?

Return false immediately because no single swap can reconcile differing string lengths.

Does the pattern 'Hash Table plus String' apply to all variations of Buddy Strings?

Yes, it is essential for counting character frequencies and evaluating swap conditions efficiently.

#859

Easy

auto_awesome哈希·表·结合·string

LeetCode 题解工作台

亲密字符串

给你两个字符串 s 和 goal ，只要我们可以通过交换 s 中的两个字母得到与 goal 相等的结果，就返回 true ；否则返回 false 。交换字母的定义是：取两个下标 i 和 j （下标从 0 开始）且满足 i != j ，接着交换 s[i] 和 s[j] 处的字符。例如，在 "abc…

哈希表字符串

题目描述

给你两个字符串 s 和 goal ，只要我们可以通过交换 s 中的两个字母得到与 goal 相等的结果，就返回 true ；否则返回 false 。

交换字母的定义是：取两个下标 i 和 j （下标从 0 开始）且满足 i != j ，接着交换 s[i] 和 s[j] 处的字符。

例如，在 "abcd" 中交换下标 0 和下标 2 的元素可以生成 "cbad" 。

示例 1：

输入：s = "ab", goal = "ba"
输出：true
解释：你可以交换 s[0] = 'a' 和 s[1] = 'b' 生成 "ba"，此时 s 和 goal 相等。

示例 2：

输入：s = "ab", goal = "ab"
输出：false
解释：你只能交换 s[0] = 'a' 和 s[1] = 'b' 生成 "ba"，此时 s 和 goal 不相等。

示例 3：

输入：s = "aa", goal = "aa"
输出：true
解释：你可以交换 s[0] = 'a' 和 s[1] = 'a' 生成 "aa"，此时 s 和 goal 相等。

提示：

1 <= s.length, goal.length <= 2 * 10⁴
s 和 goal 由小写英文字母组成

lightbulb

解题思路

方法一：字符统计

首先，先理解亲密字符串的意思：

若两个字符串的长度或字符出现的频数不等，一定不是亲密字符串；
若两个字符串对应位置不相等的字符数量为 2，或者数量为 0 并且字符串存在两个相同字符，则是亲密字符串。

因此，我们先判断两个字符串长度，若不等，直接返回 false。

接着，统计两个字符串的字符频数，记为 cnt1 和 cnt2，若 cnt1 不等于 cnt2，直接返回 false。

然后枚举两个字符串，统计对应位置不相等的字符数量，若为 2，则返回 true；若为 0，且字符串存在两个相同字符，则返回 true。

时间复杂度 $O(n)$ ，空间复杂度 $O(C)$ 。其中 $n$ 是字符串 s 或 goal 的长度；而 $C$ 为字符集大小。

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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()))

speed

复杂度分析

指标	值
时间	complexity is O(n) because each string is traversed once and character counts are tracked. Space complexity is O(1) or O(26) since only lowercase letters are counted using a fixed-size hash table.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Asks if swapping two letters is sufficient when s and goal are identical with duplicates.
question_mark
Probes whether the candidate handles strings of different lengths correctly.
question_mark
Wants reasoning behind using a hash table for character frequency versus simple mismatch tracking.

warning

常见陷阱

外企场景

error
Failing to check for duplicate characters when s equals goal, missing valid swap cases.
error
Assuming any two mismatches can be swapped without verifying order matches goal.
error
Overcomplicating with nested loops instead of linear scan plus hash table.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Allow swapping any number of letters and check if s can become goal using minimal swaps.
arrow_right_alt
Determine the number of distinct swaps needed to convert s to goal instead of just verifying one.
arrow_right_alt
Extend to uppercase and mixed-case strings, adjusting hash table or frequency array accordingly.

help

常见问题

外企场景

继续练习

#854 相似度为 K 的字符串 #839 相似字符串组 #884 两句话中的不常见单词