# Swap Characters Once to Minimize Differences - Google Top Interview Questions

### Problem Statement :

```You are given two lowercase alphabet strings s and t of the same length.

Given that you can make at most one swap between two characters in s, return the minimum number of indices where s[i] ≠ t[i].

Constraints

1 ≤ n ≤ 100,000 where n is the length of s and t

Example 1

Input

s = "abbz"

t = "zcca"

Output

2

Explanation

We can swap "a" and "z" to turn s into "zbba". Then there's 2 characters that differ between t.

Example 2

Input
s = "zfba"

t = "zbca"

Output

1

Explanation

We can swap "f" and "b" to turn s into "zbfa". Then there's only 1 character that differ between t.

Example 3

Input

s = "abc"

t = "abc"

Output

0

Explanation

We don't need to swap since all the characters in s already match with t.```

### Solution :

```                        ```Solution in C++ :

const int Z = 26;

int mismatched[Z];

int solve(string s, string t) {
memset(mismatched, 0, sizeof(mismatched));
int mismatched_pairs = 0;
for (int i = 0; i < (int)s.size(); i++) {
mismatched_pairs += s[i] != t[i];
if (s[i] != t[i]) mismatched[s[i] - 'a'] |= 1 << (t[i] - 'a');
}

bool can_swap_one = false;
for (int i = 0; i < Z; i++) {
if (!mismatched[i]) continue;
int i_bit = 1 << i;
for (int j = 0; j < Z; j++) {
if ((mismatched[i] & (1 << j)) && (mismatched[j] & i_bit)) return mismatched_pairs - 2;
can_swap_one |= mismatched[j] & i_bit;
}
}

return mismatched_pairs - can_swap_one;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(String a, String b) {
int N = a.length();
boolean[][] occ = new boolean;
int ans = 0;
boolean[] inA = new boolean;
boolean[] inB = new boolean;
for (int i = 0; i < N; i++) {
int v1 = a.charAt(i) - 'a';
int v2 = b.charAt(i) - 'a';
if (v1 != v2) {
ans++;
occ[v1][v2] = true;
inA[v1] = true;
inB[v2] = true;
}
}
for (int i = 0; i < 26; i++) {
for (int j = i + 1; j < 26; j++) {
if (occ[i][j] && occ[j][i]) {
// there exists indices where letters i and j can be swapped.
return ans - 2;
}
}
}

for (int i = 0; i < 26; i++) {
if (inA[i] && inB[i]) {
// a swap exists to get a pair of letter i.
return ans - 1;
}
}

return ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, s, t):
ans = 0
for (a, b) in zip(s, t):
if a == b:
continue
ans += 1
return ans - 2
for ch in ascii_lowercase:
return ans - 1
return ans```
```

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