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weekly-contest-185

A

Statement

Metadata

给你一个混合了数字和字母的字符串 s,其中的字母均为小写英文字母。

请你将该字符串重新格式化,使得任意两个相邻字符的类型都不同。也就是说,字母后面应该跟着数字,而数字后面应该跟着字母。

请你返回 重新格式化后 的字符串;如果无法按要求重新格式化,则返回一个 空字符串

 

示例 1:

输入:s = "a0b1c2"
输出:"0a1b2c"
解释:"0a1b2c" 中任意两个相邻字符的类型都不同。 "a0b1c2", "0a1b2c", "0c2a1b" 也是满足题目要求的答案。

示例 2:

输入:s = "leetcode"
输出:""
解释:"leetcode" 中只有字母,所以无法满足重新格式化的条件。

示例 3:

输入:s = "1229857369"
输出:""
解释:"1229857369" 中只有数字,所以无法满足重新格式化的条件。

示例 4:

输入:s = "covid2019"
输出:"c2o0v1i9d"

示例 5:

输入:s = "ab123"
输出:"1a2b3"

 

提示:

  • 1 <= s.length <= 500
  • s 仅由小写英文字母和/或数字组成。

Metadata

You are given an alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).

You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.

Return the reformatted string or return an empty string if it is impossible to reformat the string.

 

Example 1:

Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

 

Constraints:

  • 1 <= s.length <= 500
  • s consists of only lowercase English letters and/or digits.

B

Statement

Metadata
  • Link: 点菜展示表
  • Difficulty: Medium
  • Tag: 数组 哈希表 字符串 有序集合 排序

给你一个数组 orders,表示客户在餐厅中完成的订单,确切地说, orders[i]=[customerNamei,tableNumberi,foodItemi] ,其中 customerNamei 是客户的姓名,tableNumberi 是客户所在餐桌的桌号,而 foodItemi 是客户点的餐品名称。

请你返回该餐厅的 点菜展示表在这张表中,表中第一行为标题,其第一列为餐桌桌号 “Table” ,后面每一列都是按字母顺序排列的餐品名称。接下来每一行中的项则表示每张餐桌订购的相应餐品数量,第一列应当填对应的桌号,后面依次填写下单的餐品数量。

注意:客户姓名不是点菜展示表的一部分。此外,表中的数据行应该按餐桌桌号升序排列。

 

示例 1:

输入:orders = [["David","3","Ceviche"],["Corina","10","Beef Burrito"],["David","3","Fried Chicken"],["Carla","5","Water"],["Carla","5","Ceviche"],["Rous","3","Ceviche"]]
输出:[["Table","Beef Burrito","Ceviche","Fried Chicken","Water"],["3","0","2","1","0"],["5","0","1","0","1"],["10","1","0","0","0"]] 
解释:
点菜展示表如下所示:
Table,Beef Burrito,Ceviche,Fried Chicken,Water
3    ,0           ,2      ,1            ,0
5    ,0           ,1      ,0            ,1
10   ,1           ,0      ,0            ,0
对于餐桌 3:David 点了 "Ceviche" 和 "Fried Chicken",而 Rous 点了 "Ceviche"
而餐桌 5:Carla 点了 "Water" 和 "Ceviche"
餐桌 10:Corina 点了 "Beef Burrito" 

示例 2:

输入:orders = [["James","12","Fried Chicken"],["Ratesh","12","Fried Chicken"],["Amadeus","12","Fried Chicken"],["Adam","1","Canadian Waffles"],["Brianna","1","Canadian Waffles"]]
输出:[["Table","Canadian Waffles","Fried Chicken"],["1","2","0"],["12","0","3"]] 
解释:
对于餐桌 1:Adam 和 Brianna 都点了 "Canadian Waffles"
而餐桌 12:James, Ratesh 和 Amadeus 都点了 "Fried Chicken"

示例 3:

输入:orders = [["Laura","2","Bean Burrito"],["Jhon","2","Beef Burrito"],["Melissa","2","Soda"]]
输出:[["Table","Bean Burrito","Beef Burrito","Soda"],["2","1","1","1"]]

 

提示:

  • 1 <= orders.length <= 5 * 10^4
  • orders[i].length == 3
  • 1 <= customerNamei.length, foodItemi.length <= 20
  • customerNameifoodItemi 由大小写英文字母及空格字符 ' ' 组成。
  • tableNumberi1500 范围内的整数。

Metadata

Given the array orders, which represents the orders that customers have done in a restaurant. More specifically orders[i]=[customerNamei,tableNumberi,foodItemi] where customerNamei is the name of the customer, tableNumberi is the table customer sit at, and foodItemi is the item customer orders.

Return the restaurant's “display table. The “display table” is a table whose row entries denote how many of each food item each table ordered. The first column is the table number and the remaining columns correspond to each food item in alphabetical order. The first row should be a header whose first column is “Table”, followed by the names of the food items. Note that the customer names are not part of the table. Additionally, the rows should be sorted in numerically increasing order.

 

Example 1:


Input: orders = [["David","3","Ceviche"],["Corina","10","Beef Burrito"],["David","3","Fried Chicken"],["Carla","5","Water"],["Carla","5","Ceviche"],["Rous","3","Ceviche"]]
Output: [["Table","Beef Burrito","Ceviche","Fried Chicken","Water"],["3","0","2","1","0"],["5","0","1","0","1"],["10","1","0","0","0"]] 
Explanation:
The displaying table looks like:
Table,Beef Burrito,Ceviche,Fried Chicken,Water
3    ,0           ,2      ,1            ,0
5    ,0           ,1      ,0            ,1
10   ,1           ,0      ,0            ,0
For the table 3: David orders "Ceviche" and "Fried Chicken", and Rous orders "Ceviche".
For the table 5: Carla orders "Water" and "Ceviche".
For the table 10: Corina orders "Beef Burrito". 

Example 2:


Input: orders = [["James","12","Fried Chicken"],["Ratesh","12","Fried Chicken"],["Amadeus","12","Fried Chicken"],["Adam","1","Canadian Waffles"],["Brianna","1","Canadian Waffles"]]
Output: [["Table","Canadian Waffles","Fried Chicken"],["1","2","0"],["12","0","3"]] 
Explanation: 
For the table 1: Adam and Brianna order "Canadian Waffles".
For the table 12: James, Ratesh and Amadeus order "Fried Chicken".

Example 3:


Input: orders = [["Laura","2","Bean Burrito"],["Jhon","2","Beef Burrito"],["Melissa","2","Soda"]]
Output: [["Table","Bean Burrito","Beef Burrito","Soda"],["2","1","1","1"]]

 

Constraints:

  • 1 <= orders.length <= 5 * 10^4
  • orders[i].length == 3
  • 1 <= customerNamei.length, foodItemi.length <= 20
  • customerNamei and foodItemi consist of lowercase and uppercase English letters and the space character.
  • tableNumberi is a valid integer between 1 and 500.

Solution

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define SZ(x) int((x).size())
#define endl "\n"
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head
constexpr int N = 1e5 + 10;
int n, mp[220];

string s;

class Solution {
public:
    int minNumberOfFrogs(string croakOfFrogs) {
        s = croakOfFrogs;
        int cnt[10] = {0};
        int res = 0;
        mp['c'] = 0;
        mp['r'] = 1;
        mp['o'] = 2;
        mp['a'] = 3;
        mp['k'] = 4;
        for (auto &c : s) {
            if (c == 'c')
                ++cnt[0];
            else {
                if (cnt[mp[c] - 1] == 0)
                    return -1;
                --cnt[mp[c] - 1];
                ++cnt[mp[c]];
            }
            int sum = cnt[0] + cnt[1] + cnt[2] + cnt[3];
            chmax(res, sum);
        }
        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

Metadata
  • Link: 数青蛙
  • Difficulty: Medium
  • Tag: 字符串 计数

给你一个字符串 croakOfFrogs,它表示不同青蛙发出的蛙鸣声(字符串 "croak" )的组合。由于同一时间可以有多只青蛙呱呱作响,所以 croakOfFrogs 中会混合多个 “croak” 请你返回模拟字符串中所有蛙鸣所需不同青蛙的最少数目。

注意:要想发出蛙鸣 "croak",青蛙必须 依序 输出 ‘c’, ’r’, ’o’, ’a’, ’k’ 这 5 个字母。如果没有输出全部五个字母,那么它就不会发出声音。

如果字符串 croakOfFrogs 不是由若干有效的 "croak" 字符混合而成,请返回 -1

 

示例 1:

输入:croakOfFrogs = "croakcroak"
输出:1 
解释:一只青蛙 “呱呱” 两次

示例 2:

输入:croakOfFrogs = "crcoakroak"
输出:2 
解释:最少需要两只青蛙,“呱呱” 声用黑体标注
第一只青蛙 "crcoakroak"
第二只青蛙 "crcoakroak"

示例 3:

输入:croakOfFrogs = "croakcrook"
输出:-1
解释:给出的字符串不是 "croak" 的有效组合。

示例 4:

输入:croakOfFrogs = "croakcroa"
输出:-1

 

提示:

  • 1 <= croakOfFrogs.length <= 10^5
  • 字符串中的字符只有 'c', 'r', 'o', 'a' 或者 'k'

Metadata

You are given the string croakOfFrogs, which represents a combination of the string "croak" from different frogs, that is, multiple frogs can croak at the same time, so multiple "croak" are mixed.

Return the minimum number of different frogs to finish all the croaks in the given string.

A valid "croak" means a frog is printing five letters 'c', 'r', 'o', 'a', and 'k' sequentially. The frogs have to print all five letters to finish a croak. If the given string is not a combination of a valid "croak" return -1.

 

Example 1:

Input: croakOfFrogs = "croakcroak"
Output: 1 
Explanation: One frog yelling "croak" twice.

Example 2:

Input: croakOfFrogs = "crcoakroak"
Output: 2 
Explanation: The minimum number of frogs is two. 
The first frog could yell "crcoakroak".
The second frog could yell later "crcoakroak".

Example 3:

Input: croakOfFrogs = "croakcrook"
Output: -1
Explanation: The given string is an invalid combination of "croak" from different frogs.

 

Constraints:

  • 1 <= croakOfFrogs.length <= 105
  • croakOfFrogs is either 'c', 'r', 'o', 'a', or 'k'.

D

Statement

Metadata

给你三个整数 nmk 。下图描述的算法用于找出正整数数组中最大的元素。

请你生成一个具有下述属性的数组 arr

  • arr 中有 n 个整数。
  • 1 <= arr[i] <= m 其中 (0 <= i < n)
  • 将上面提到的算法应用于 arrsearch_cost 的值等于 k

返回上述条件下生成数组 arr方法数 ,由于答案可能会很大,所以 必须10^9 + 7 取余。

 

示例 1:

输入:n = 2, m = 3, k = 1
输出:6
解释:可能的数组分别为 [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3]

示例 2:

输入:n = 5, m = 2, k = 3
输出:0
解释:没有数组可以满足上述条件

示例 3:

输入:n = 9, m = 1, k = 1
输出:1
解释:可能的数组只有 [1, 1, 1, 1, 1, 1, 1, 1, 1]

示例 4:

输入:n = 50, m = 100, k = 25
输出:34549172
解释:不要忘了对 1000000007 取余

示例 5:

输入:n = 37, m = 17, k = 7
输出:418930126

 

提示:

  • 1 <= n <= 50
  • 1 <= m <= 100
  • 0 <= k <= n

Metadata

You are given three integers n, m and k. Consider the following algorithm to find the maximum element of an array of positive integers:

You should build the array arr which has the following properties:

  • arr has exactly n integers.
  • 1 <= arr[i] <= m where (0 <= i < n).
  • After applying the mentioned algorithm to arr, the value search_cost is equal to k.

Return the number of ways to build the array arr under the mentioned conditions. As the answer may grow large, the answer must be computed modulo 109 + 7.

 

Example 1:

Input: n = 2, m = 3, k = 1
Output: 6
Explanation: The possible arrays are [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3]

Example 2:

Input: n = 5, m = 2, k = 3
Output: 0
Explanation: There are no possible arrays that satisify the mentioned conditions.

Example 3:

Input: n = 9, m = 1, k = 1
Output: 1
Explanation: The only possible array is [1, 1, 1, 1, 1, 1, 1, 1, 1]

 

Constraints:

  • 1 <= n <= 50
  • 1 <= m <= 100
  • 0 <= k <= n


最后更新: January 15, 2023
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