299. Bulls and Cows LeetCode Solution

Last updated: February 17, 2025

In this guide, you will get 299. Bulls and Cows LeetCode Solution with the best time and space complexity. The solution to Bulls and Cows problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Bulls and Cows solution in C++
Bulls and Cows solution in Java
Bulls and Cows solution in Python
Additional Resources

299. Bulls and CowsLeetCode Solution image

Problem Statement of 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.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(10)

299. Bulls and Cows LeetCode Solution in 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]);

    return to_string(A) + "A" + to_string(B) + "B";
  }
};
/* code provided by PROGIEZ */

299. Bulls and Cows LeetCode Solution in Java

class Solution {
  public String getHint(String secret, String guess) {
    int A = 0;
    int B = 0;
    int[] count1 = new int[10];
    int[] count2 = new int[10];

    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";
  }
}
// code provided by PROGIEZ

299. Bulls and Cows LeetCode Solution in 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)
 # code by PROGIEZ

Additional Resources

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