Bulls and Cows

medium string hash map counting

Problem

Given the secret and the guess (same length, digits only), return "xAyB" where x = bulls (digit and position match) and y = cows (digit in the secret but at a different position).

Inputsecret = "1807", guess = "7810"

Output"1A3B"

'8' is a bull at index 1. 1, 0, 7 in the guess match remaining secret digits at different positions → 3 cows.

secret

guess

def get_hint(secret, guess):
    bulls = 0
    cs, cg = [0]*10, [0]*10
    for a, b in zip(secret, guess):
        if a == b: bulls += 1
        else:
            cs[int(a)] += 1
            cg[int(b)] += 1
    cows = sum(min(cs[i], cg[i]) for i in range(10))
    return f"{bulls}A{cows}B"

function getHint(secret, guess) {
  let bulls = 0;
  const cs = new Array(10).fill(0), cg = new Array(10).fill(0);
  for (let i = 0; i < secret.length; i++) {
    if (secret[i] === guess[i]) bulls++;
    else {
      cs[+secret[i]]++;
      cg[+guess[i]]++;
    }
  }
  let cows = 0;
  for (let d = 0; d < 10; d++) cows += Math.min(cs[d], cg[d]);
  return bulls + "A" + cows + "B";
}

class Solution {
    public String getHint(String secret, String guess) {
        int bulls = 0;
        int[] cs = new int[10], cg = new int[10];
        for (int i = 0; i < secret.length(); i++) {
            char a = secret.charAt(i), b = guess.charAt(i);
            if (a == b) bulls++;
            else { cs[a - '0']++; cg[b - '0']++; }
        }
        int cows = 0;
        for (int d = 0; d < 10; d++) cows += Math.min(cs[d], cg[d]);
        return bulls + "A" + cows + "B";
    }
}

string getHint(string secret, string guess) {
    int bulls = 0;
    int cs[10] = {0}, cg[10] = {0};
    for (int i = 0; i < (int)secret.size(); i++) {
        if (secret[i] == guess[i]) bulls++;
        else { cs[secret[i]-'0']++; cg[guess[i]-'0']++; }
    }
    int cows = 0;
    for (int d = 0; d < 10; d++) cows += min(cs[d], cg[d]);
    return to_string(bulls) + "A" + to_string(cows) + "B";
}

Explanation

A bull is a digit that matches both value and position; a cow is a digit that exists somewhere in the secret but is in the wrong spot. We can find both in a single pass plus a quick tally.

We loop over the positions together. Whenever secret[i] equals guess[i] we increment bulls right away. When they differ, the digits are not bulls but might still be cows, so we record them in two small count arrays of size 10: cs for leftover secret digits and cg for leftover guess digits.

After the loop, a digit can form a cow only as many times as it appears in both leftover piles. So for each digit d from 0 to 9 we add min(cs[d], cg[d]) to cows.

The min is the key: it stops us from over-counting when a digit appears a different number of times on each side.

Example: secret="1807", guess="7810". The 8 at index 1 is a bull. The leftover digits 1, 0, 7 each appear in both piles, so cows = 3, giving "1A3B".

Time: O(n) Space: O(1)