Today I have done Bulls and Cows on leetcode and leetcode's November LeetCoding Challenge with cpp.

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.

Example 3:

Input: secret = "1", guess = "0"
Output: "0A0B"

Example 4:

Input: secret = "1", guess = "1"
Output: "1A0B"

Constraints:

• 1 <= secret.length, guess.length <= 1000
• secret.length == guess.length
• secret and guess consist of digits only.

Solution

nothing to say

class Solution {
public:
    string getHint(string secret, string guess) {
        int bulls = 0, cows = 0;
        map<char, int> sMap, gMap;
        for(int i = 0; i < secret.size(); ++i) {
            if(secret[i] == guess[i]) bulls += 1;
            else {
                sMap[secret[i]] += 1;
                gMap[guess[i]] += 1;
            }
        }
        for(auto &p : sMap) {
            if(gMap.count(p.first)) {
                cows += min(gMap[p.first], p.second);
            }
        }
        return to_string(bulls) + "A" + to_string(cows) + "B";
    }
};

November LeetCoding Challenge 5

Description

Minimum Cost to Move Chips to The Same Position

We have n chips, where the position of the ith chip is position[i].

We need to move all the chips to the same position. In one step, we can change the position of the ith chip from position[i] to:

• position[i] + 2 or position[i] - 2 with cost = 0.
• position[i] + 1 or position[i] - 1 with cost = 1.

Return the minimum cost needed to move all the chips to the same position.

Example 1:

Input: position = [1,2,3]
Output: 1
Explanation: First step: Move the chip at position 3 to position 1 with cost = 0.
Second step: Move the chip at position 2 to position 1 with cost = 1.
Total cost is 1.

Example 2:

Input: position = [2,2,2,3,3]
Output: 2
Explanation: We can move the two chips at poistion 3 to position 2. Each move has cost = 1. The total cost = 2.

Example 3:

Input: position = [1,1000000000]
Output: 1

Constraints:

• 1 <= position.length <= 100
• 1 <= position[i] <= 10^9

Solution

parity

class Solution {
public:
    int minCostToMoveChips(vector<int>& position) {
        int cnt[2] = {0, 0};
        for(auto i : position) {
            cnt[i&1] += 1;
        }
        return min(cnt[0], cnt[1]);
    }
};