16.15. Master Mind
Description
The Game of Master Mind is played as follows:
The computer has four slots, and each slot will contain a ball that is red (R). yellow (Y). green (G) or blue (B). For example, the computer might have RGGB (Slot #1 is red, Slots #2 and #3 are green, Slot #4 is blue).
You, the user, are trying to guess the solution. You might, for example, guess YRGB.
When you guess the correct color for the correct slot, you get a "hit:' If you guess a color that exists but is in the wrong slot, you get a "pseudo-hit:' Note that a slot that is a hit can never count as a pseudo-hit.
For example, if the actual solution is RGBY and you guess GGRR, you have one hit and one pseudo-hit. Write a method that, given a guess and a solution, returns the number of hits and pseudo-hits.
Given a sequence of colors solution
, and a guess
, write a method that return the number of hits and pseudo-hit answer
, where answer[0]
is the number of hits and answer[1]
is the number of pseudo-hit.
Example:
Input: solution="RGBY",guess="GGRR"
Output: [1,1]
Explanation: hit once, pseudo-hit once.
Note:
len(solution) = len(guess) = 4
- There are only
"R"
,"G"
,"B"
,"Y"
insolution
andguess
.
Solutions
Solution 1: Hash Table
We simultaneously traverse both strings, count the number of corresponding characters that are the same, and accumulate them in $x$. Then we record the characters and their frequencies in both strings in hash tables $cnt1$ and $cnt2$, respectively.
Next, we traverse both hash tables, count the number of common characters, and accumulate them in $y$. The answer is then $[x, y - x]$.
The time complexity is $O(C)$, and the space complexity is $O(C)$. Here, $C=4$ for this problem.
Python3
class Solution:
def masterMind(self, solution: str, guess: str) -> List[int]:
x = sum(a == b for a, b in zip(solution, guess))
y = sum((Counter(solution) & Counter(guess)).values())
return [x, y - x]
Java
class Solution {
public int[] masterMind(String solution, String guess) {
int x = 0, y = 0;
Map<Character, Integer> cnt1 = new HashMap<>();
Map<Character, Integer> cnt2 = new HashMap<>();
for (int i = 0; i < 4; ++i) {
char a = solution.charAt(i), b = guess.charAt(i);
x += a == b ? 1 : 0;
cnt1.merge(a, 1, Integer::sum);
cnt2.merge(b, 1, Integer::sum);
}
for (char c : "RYGB".toCharArray()) {
y += Math.min(cnt1.getOrDefault(c, 0), cnt2.getOrDefault(c, 0));
}
return new int[] {x, y - x};
}
}
C++
class Solution {
public:
vector<int> masterMind(string solution, string guess) {
int x = 0, y = 0;
unordered_map<char, int> cnt1;
unordered_map<char, int> cnt2;
for (int i = 0; i < 4; ++i) {
x += solution[i] == guess[i];
cnt1[solution[i]]++;
cnt2[guess[i]]++;
}
for (char c : "RYGB") y += min(cnt1[c], cnt2[c]);
return vector<int>{x, y - x};
}
};
Go
func masterMind(solution string, guess string) []int {
var x, y int
cnt1 := map[byte]int{}
cnt2 := map[byte]int{}
for i := range solution {
a, b := solution[i], guess[i]
if a == b {
x++
}
cnt1[a]++
cnt2[b]++
}
for _, c := range []byte("RYGB") {
y += min(cnt1[c], cnt2[c])
}
return []int{x, y - x}
}
JavaScript
/**
* @param {string} solution
* @param {string} guess
* @return {number[]}
*/
var masterMind = function (solution, guess) {
let counts1 = { R: 0, G: 0, B: 0, Y: 0 };
let counts2 = { R: 0, G: 0, B: 0, Y: 0 };
let res1 = 0;
for (let i = 0; i < solution.length; i++) {
let s1 = solution[i],
s2 = guess[i];
if (s1 === s2) {
res1++;
} else {
counts1[s1] += 1;
counts2[s2] += 1;
}
}
let res2 = Object.keys(counts1).reduce((a, c) => a + Math.min(counts1[c], counts2[c]), 0);
return [res1, res2];
};
Swift
class Solution {
func masterMind(_ solution: String, _ guess: String) -> [Int] {
var x = 0
var y = 0
var cnt1: [Character: Int] = [:]
var cnt2: [Character: Int] = [:]
for i in solution.indices {
let a = solution[i]
let b = guess[i]
if a == b {
x += 1
}
cnt1[a, default: 0] += 1
cnt2[b, default: 0] += 1
}
let colors = "RYGB"
for c in colors {
let minCount = min(cnt1[c, default: 0], cnt2[c, default: 0])
y += minCount
}
return [x, y - x]
}
}