Doubt in a question

#1

Hi sir, I was trying a question on leetcode. Canyou pls look at my code and tell what am i doing wrong?

Link to Question:

Link to my Solution:

Please reply asap.

#2

The logic you have used is partially right , do what is sort string a & b, then keep a count of mis matched characters, like if a[i]!=b[i] you keep a count. If count is exactly equal to 2 then return true, else for any other case, return false.
Also if length is not same for both string return false.

#3

Sir, this approach will not satisfy Example 2 as Input: A = “ab”, B = “ab” Output: false

#4

For this your count will be 0 right? So in it , you can check if count is 0 or 2, then you can return true else false.

#5

Sir i have already tried this

#6

we need to maintaion all test case which involves Test Case 1: A=“aa” B=“aa” for which output is true and Test case 2: A=“ab” and B=“ab” for which output is false

#7

as in fos first test case, on swapping ‘a’ and ‘a’ we get B. But in test case 2, on swapping ‘a’ and ‘b’ we get A=“ba” where as B=“ab”, which are not same

#8

Okay i got it, no need of sorting, rest everything told by me was right.
If A.length() != B.length() : no possible swap

If A == B , we need swap two same characters. Check is duplicated char in A .

In other cases, we find index for A[i] != B[i] . There should be only 2 diffs and it’s our one swap.

C++:

    bool buddyStrings(string A, string B) {
        if (A.length() != B.length()) return false;
        if (A == B && set<char>(A.begin(), A.end()).size() < A.size()) return true;
        vector<int> dif;
        for (int i = 0; i < A.length(); ++i) if (A[i] != B[i]) dif.push_back(i);
        return dif.size() == 2 && A[dif[0]] == B[dif[1]] && A[dif[1]] == B[dif[0]];
    }

It’s getting accepted

#9

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