# Bulls and Cows LeetCode Solution | Easy Approach Share:

Bulls and Cows 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.
```

Constraints:

• `1 <= secret.length, guess.length <= 1000`
• `secret.length == guess.length`
• `secret` and `guess` consist of digits only.

Time: O(n)
Space: O(10)

### C++

``````class Solution {
public:
string getHint(string secret, string guess) {
int A = 0;
int B = 0;
vector<int> count1(10);
vector<int> count2(10);

for (int i = 0; i < secret.length(); ++i)
if (secret[i] == guess[i])
++A;
else {
++count1[secret[i] - '0'];
++count2[guess[i] - '0'];
}

for (int i = 0; i < 10; ++i)
B += min(count1[i], count2[i]);

}
};
``````

### Java

`````` class Solution {
public String getHint(String secret, String guess) {
int A = 0;
int B = 0;
int[] count1 = new int;
int[] count2 = new int;

for (int i = 0; i < secret.length(); ++i)
if (secret.charAt(i) == guess.charAt(i))
++A;
else {
++count1[secret.charAt(i) - '0'];
++count2[guess.charAt(i) - '0'];
}

for (int i = 0; i < 10; ++i)
B += Math.min(count1[i], count2[i]);

return String.valueOf(A) + "A" + String.valueOf(B) + "B";
}
}

``````

### Python

``````class Solution:
def getHint(self, secret: str, guess: str) -> str:
bulls = sum(map(operator.eq, secret, guess))
bovine = sum(min(secret.count(x), guess.count(x)) for x in set(guess))
return '%dA%dB' % (bulls, bovine - bulls)

``````

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