LeetCode 0299. Bulls and Cows Solution in Java, Python, C++, JavaScript, Go & Rust | Explanation + Code

Description

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.

Solutions

Solution 1: Counting

We create two counters, cnt1 and cnt2, to count the occurrence of each digit in the secret number and the friend's guess respectively. At the same time, we create a variable x to count the number of bulls.

Then we iterate through the secret number and the friend's guess. If the current digit is the same, we increment x by one. Otherwise, we increment the count of the current digit in the secret number and the friend's guess respectively.

Finally, we iterate through each digit in cnt1, take the minimum count of the current digit in cnt1 and cnt2, and add this minimum value to the variable y.

In the end, we return the values of x and y.

The time complexity is O(n), where n is the length of the secret number and the friend's guess. The space complexity is O(|Σ|), where |Σ| is the size of the character set. In this problem, the character set is digits, so |Σ| = 10.

PythonJavaC++GoTypeScriptPHP

class Solution: def getHint(self, secret: str, guess: str) -> str: cnt1, cnt2 = Counter(), Counter() x = 0 for a, b in zip(secret, guess): if a == b: x += 1 else: cnt1[a] += 1 cnt2[b] += 1 y = sum(min(cnt1[c], cnt2[c]) for c in cnt1) return f"{x}A{y}B"(code-box)

class Solution { public String getHint(String secret, String guess) { int x = 0, y = 0; int[] cnt1 = new int[10]; int[] cnt2 = new int[10]; for (int i = 0; i < secret.length(); ++i) { int a = secret.charAt(i) - '0', b = guess.charAt(i) - '0'; if (a == b) { ++x; } else { ++cnt1[a]; ++cnt2[b]; } } for (int i = 0; i < 10; ++i) { y += Math.min(cnt1[i], cnt2[i]); } return String.format("%dA%dB", x, y); } }(code-box)

class Solution { public: string getHint(string secret, string guess) { int x = 0, y = 0; int cnt1[10]{}; int cnt2[10]{}; for (int i = 0; i < secret.size(); ++i) { int a = secret[i] - '0', b = guess[i] - '0'; if (a == b) { ++x; } else { ++cnt1[a]; ++cnt2[b]; } } for (int i = 0; i < 10; ++i) { y += min(cnt1[i], cnt2[i]); } return to_string(x) + "A" + to_string(y) + "B"; } };(code-box)

func getHint(secret string, guess string) string { x, y := 0, 0 cnt1 := [10]int{} cnt2 := [10]int{} for i, c := range secret { a, b := int(c-'0'), int(guess[i]-'0') if a == b { x++ } else { cnt1[a]++ cnt2[b]++ } } for i, c := range cnt1 { y += min(c, cnt2[i]) } return fmt.Sprintf("%dA%dB", x, y) }(code-box)

function getHint(secret: string, guess: string): string { const cnt1: number[] = Array(10).fill(0); const cnt2: number[] = Array(10).fill(0); let x: number = 0; for (let i = 0; i < secret.length; ++i) { if (secret[i] === guess[i]) { ++x; } else { ++cnt1[+secret[i]]; ++cnt2[+guess[i]]; } } let y: number = 0; for (let i = 0; i < 10; ++i) { y += Math.min(cnt1[i], cnt2[i]); } return `${x}A${y}B`; }(code-box)

class Solution { /** * @param String $secret * @param String $guess * @return String */ function getHint($secret, $guess) { $cnt1 = array_fill(0, 10, 0); $cnt2 = array_fill(0, 10, 0); $x = 0; for ($i = 0; $i < strlen($secret); ++$i) { if ($secret[$i] === $guess[$i]) { ++$x; } else { ++$cnt1[(int) $secret[$i]]; ++$cnt2[(int) $guess[$i]]; } } $y = 0; for ($i = 0; $i < 10; ++$i) { $y += min($cnt1[$i], $cnt2[$i]); } return "{$x}A{$y}B"; } }(code-box)