299. Bulls and Cows
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
andguess
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 $|\Sigma| = 10$.
Python3
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"
Java
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);
}
}
C++
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";
}
};
Go
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)
}
TypeScript
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`;
}
PHP
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";
}
}