Bulls & Cows | Leet Code

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Bulls And Cows | Leet Code

You are playing the following Bulls and Cows game with your friend: You write down a number and ask your friend to guess what the number is. Each time your friend makes a guess, you provide a hint that indicates how many digits in said guess match your secret number exactly in both digit and position (called “bulls”) and how many digits match the secret number but locate in the wrong position (called “cows”). Your friend will use successive guesses and hints to eventually derive the secret number.

Write a function to return a hint according to the secret number and friend’s guess, use A to indicate the bulls and B to indicate the cows.

Please note that both secret number and friend’s guess may contain duplicate digits.

Example 1:

Input: secret = “3452”, guess = “5431”

Output: “1A2B”

Note: You may assume that the secret number and your friend’s guess only contain digits, and their lengths are always equal.


We create two arrays as secretCows and guessCows and an integer variable bulls. First we check for number of bulls in the guess string and as we found the bulls we increment the value of bulls by one. If we do not found bulls then we add that character into both the arrays, After doing this we create a integer variable cow = 0. Next to check for number of cows ,we agani traverse through the arrays and if both the arrays has some same specific value we change the value of cows.



class Solution {
    public String getHint(String secret, String guess) {
        int[] secretCows = new int[10];
        int[] guessCows = new int[10];
        int bulls = 0;
        for (int i = 0; i < secret.length(); i++) {
            if (guess.charAt(i) == secret.charAt(i)) {
            } else {
        int cows = 0;
        for (int i = 0; i < 10; i++) {
            cows += Math.min(secretCows[i], guessCows[i]);
        return bulls + "A" + cows + "B";

Complexity Analysis:

  • Time Complexity: O(n)

  • Space Complexity: O(1).

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