Question

299. Bulls and Cows

中文文档

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 $O(|\Sigma|)$, where $|\Sigma|$ is the size of the character set. In this problem, the character set is digits, so $|\Sigma| = 10$.

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"

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);
  }
}

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";
  }
};

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)
}

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`;
}

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";
  }
}