**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[26][26];
int ans = 0;
boolean[] inA = new boolean[26];
boolean[] inB = new boolean[26];
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):
badpairs = set()
ans = 0
for (a, b) in zip(s, t):
if a == b:
continue
badpairs.add((a, b))
ans += 1
for x, y in badpairs:
if (y, x) in badpairs:
return ans - 2
for x, _ in badpairs:
for ch in ascii_lowercase:
if (ch, x) in badpairs:
return ans - 1
return ans
```

## View More Similar Problems

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

View Solution →## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →