## Description

https://leetcode.com/problems/most-common-word/

You are playing the **Bulls and Cows** game with your friend.

You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:

- The number of “bulls”, which are digits in the guess that are in the correct position.
- The number of “cows”, which are digits in the guess that are in your secret number but are located in the wrong position. Specifically, the non-bull digits in the guess that could be rearranged such that they become bulls.

Given the secret number `secret`

and your friend’s guess `guess`

, return *the hint for your friend’s guess*.

The hint should be formatted as `"xAyB"`

, where `x`

is the number of bulls and `y`

is the number of cows. Note that both `secret`

and `guess`

may contain duplicate digits.

**Example 1:**

Input:secret = "1807", guess = "7810"Output:"1A3B"Explanation:Bulls are connected with a '|' and cows are underlined: "1807" | "7810"

**Example 2:**

Input:secret = "1123", guess = "0111"Output:"1A1B"Explanation:Bulls are connected with a '|' and cows are underlined: "1123" "1123" | or | "0111" "0111" Note that only one of the two unmatched 1s is counted as a cow since the non-bull digits can only be rearranged to allow one 1 to be a bull.

**Example 3:**

Input:secret = "1", guess = "0"Output:"0A0B"

**Example 4:**

Input:secret = "1", guess = "1"Output:"1A0B"

**Constraints:**

`1 <= secret.length, guess.length <= 1000`

`secret.length == guess.length`

`secret`

and`guess`

consist of digits only.

## Explanation

two pass

## Python Solution

```
class Solution:
def getHint(self, secret: str, guess: str) -> str:
bulls = 0
cows = 0
secret_dict = {}
for c1 in secret:
secret_dict[c1] = secret_dict.get(c1, 0) + 1
for c1, c2 in zip(secret, guess):
if c1 == c2:
bulls += 1
secret_dict[c1] = secret_dict.get(c1) - 1
for c1, c2 in zip(secret, guess):
if c1 != c2 and c2 in secret_dict and secret_dict[c2] > 0:
cows += 1
secret_dict[c2] = secret_dict.get(c2) - 1
return "{}A{}B".format(bulls, cows)
```

- Time Complexity: ~N
- Space Complexity: ~1