Bulls and Cows

Medium

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ArraysStringsTwo Pointers

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.

Try coding it in LeetCode

Bulls and Cows

Medium

Asked by:

16 views

Topics:

ArraysStringsTwo Pointers

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the secret and guess strings have different lengths?
What characters will the secret and guess strings contain (e.g., only digits 0-9)?
If there are multiple valid 'bulls and cows' counts, should I return any valid answer or a specific one?
Are the secret and guess strings guaranteed to be non-empty?
What should I return if either the secret or guess string is null or empty?

Brute Force Solution

Approach

The brute force strategy for Bulls and Cows is like guessing a secret number by trying every possible combination. We carefully compare the guess with the secret number to find matching digits in the correct positions (Bulls) and matching digits in the wrong positions (Cows).

Here's how the algorithm would work step-by-step:

Start with the first possible guess for the secret number.
Compare each digit of the guess with each digit of the secret number.
Count how many digits in the guess are exactly the same as the secret number's digits and in the same place; this is the number of Bulls.
Now, look for digits that appear in both the guess and the secret number, but are in different positions. Be careful not to double-count any digits that were already counted as Bulls; this is the number of Cows.
If the number of Bulls and Cows matches what we were given as a hint, then this guess might be the secret number.
Keep trying every possible guess and comparing it to the secret number, counting Bulls and Cows each time.
Continue until you have tried all possible guesses.

Code Implementation

def bulls_and_cows_brute_force(secret_number, guess):    number_of_bulls = 0
    number_of_cows = 0
    secret_number_list = list(secret_number)
    guess_list = list(guess)

    # First, find the bulls
    for index in range(len(secret_number)):
        if secret_number_list[index] == guess_list[index]:

            number_of_bulls += 1
            guess_list[index] = None # Avoid re-counting
            secret_number_list[index] = None # Avoid re-counting

    # Now find the cows. Iterate through the guess
    for guess_index in range(len(guess_list)):
        if guess_list[guess_index] is not None:

            # Look for matches in the secret number
            for secret_index in range(len(secret_number_list)):
                if secret_number_list[secret_index] is not None and guess_list[guess_index] == secret_number_list[secret_index]:

                    number_of_cows += 1
                    secret_number_list[secret_index] = None # Mark as used
                    break # avoid double counting

    return str(number_of_bulls) + "B" + str(number_of_cows) + "C"

Big(O) Analysis

Time Complexity

O(n*m) – The outer loop iterates 'n' times, where 'n' is the length of the guess string. The inner loop iterates 'm' times, where 'm' is the length of the secret string, comparing each digit of the guess with each digit of the secret. Therefore, for each of the 'n' digits in the guess, we perform 'm' comparisons. Hence, the time complexity is O(n*m).

Space Complexity

O(1) – The brute force approach primarily uses a fixed number of variables to store the counts of bulls and cows during each comparison between the guess and the secret number. Since we are only counting and comparing, and not storing any data structures dependent on the length of the secret or guess, the space required does not scale with the input size. Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The goal is to efficiently determine the number of 'bulls' (correct digits in the correct positions) and 'cows' (correct digits in the wrong positions) between a guess and a secret number. Instead of comparing each digit multiple times, the optimal approach involves counting the occurrences of each digit and then cleverly deducting to avoid double-counting.

Here's how the algorithm would work step-by-step:

First, we need to go through both the guess and the secret number at the same time, looking for 'bulls'. A 'bull' is when a digit in the guess matches the digit in the secret number at the same position. Count the number of bulls we find.
Next, we need to count how many times each digit appears in both the guess and the secret number, ignoring the digits we already counted as bulls.
Now, we want to find the 'cows'. A 'cow' is a digit that exists in both the guess and the secret number, but is in the wrong place. For each digit, find the minimum number of times it appears in the guess and the secret number. Sum these minimum counts across all the digits. This gives us the total number of digits present in both numbers.
Finally, subtract the number of bulls we found earlier from the total count of digits present. The remaining number is the number of 'cows'.
Return the number of bulls and cows as our final answer.

Code Implementation

def solve_bulls_and_cows(secret, guess):
    bull_count = 0
    digit_counts = {}

    # First pass: count bulls and populate digit counts.
    for i in range(len(secret)): 
        if secret[i] == guess[i]:
            bull_count += 1
        else:
            if secret[i] in digit_counts:
                digit_counts[secret[i]] = digit_counts[secret[i]] + 1
            else:
                digit_counts[secret[i]] = 1

    cow_count = 0
    remaining_guess_digits = {}

    for i in range(len(guess)): 
        if secret[i] != guess[i]:
            if guess[i] in remaining_guess_digits:
                remaining_guess_digits[guess[i]] = remaining_guess_digits[guess[i]] + 1
            else:
                remaining_guess_digits[guess[i]] = 1

    # Calculate cows based on digit counts in secret and guess.
    for digit, count in remaining_guess_digits.items(): 
        if digit in digit_counts:
            cow_count += min(count, digit_counts[digit])

    return f"{bull_count}A{cow_count}B"

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the guess and secret strings once to identify bulls, which takes O(n) time where n is the length of the strings. Then, it counts digit frequencies, which can be considered O(1) since there are only 10 possible digits. Finally, calculating cows involves iterating through the digit counts, which is also O(1). Therefore, the dominant operation is the initial pass for identifying bulls, resulting in an overall time complexity of O(n).

Space Complexity

O(1) – The algorithm uses constant auxiliary space. While the explanation mentions counting digit occurrences, a fixed-size array or hash map (of size 10, representing digits 0-9) is sufficient regardless of the input string length N. Therefore, the extra space required does not scale with the input size, making it O(1).

Edge Cases

Case	How to Handle
Secret or guess strings are null or empty	Return an empty string or throw an IllegalArgumentException if either input is null or empty, since no bulls or cows can be determined.
Secret and guess strings have different lengths	Return an error or throw an IllegalArgumentException, since a valid comparison requires equal lengths.
Secret and guess contain non-digit characters	Throw an IllegalArgumentException or filter out non-digit characters before processing to ensure correct calculations.
Secret and guess strings are very long (e.g., 1000+ characters)	The solution should iterate in O(N) time where N is the length of the strings; using a hashmap is acceptable.
Secret and guess are identical	Return a string indicating the length of the string as the number of bulls and zero cows.
Secret and guess share no common digits	Return a string with 0 bulls and 0 cows as output.
Secret has duplicate digits not present in guess, or vice-versa	The hashmap will accurately count cows even when one string has more duplicates of a digit than the other.
Integer overflow in the bull or cow count if excessively long input	Since the length of the strings is constrained to digits only, an integer overflow is highly unlikely with standard integer sizes, but consider using a larger data type if string length limitations are removed.