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biweekly-contest-83

A

Statement

Metadata

给你一个整数数组 ranks 和一个字符数组 suit 。你有 5 张扑克牌，第 i 张牌大小为 ranks[i] ，花色为 suits[i] 。

下述是从好到坏你可能持有的 手牌类型 ：

  1. "Flush"：同花，五张相同花色的扑克牌。
  2. "Three of a Kind"：三条，有 3 张大小相同的扑克牌。
  3. "Pair"：对子，两张大小一样的扑克牌。
  4. "High Card"：高牌，五张大小互不相同的扑克牌。

请你返回一个字符串，表示给定的 5 张牌中，你能组成的 最好手牌类型 。

注意：返回的字符串 大小写 需与题目描述相同。

 

示例 1：

输入：ranks = [13,2,3,1,9], suits = ["a","a","a","a","a"]
输出："Flush"
解释：5 张扑克牌的花色相同，所以返回 "Flush" 。

示例 2：

输入：ranks = [4,4,2,4,4], suits = ["d","a","a","b","c"]
输出："Three of a Kind"
解释：第一、二和四张牌组成三张相同大小的扑克牌，所以得到 "Three of a Kind" 。
注意我们也可以得到 "Pair" ，但是 "Three of a Kind" 是更好的手牌类型。
有其他的 3 张牌也可以组成 "Three of a Kind" 手牌类型。

示例 3：

输入：ranks = [10,10,2,12,9], suits = ["a","b","c","a","d"]
输出："Pair"
解释：第一和第二张牌大小相同，所以得到 "Pair" 。
我们无法得到 "Flush" 或者 "Three of a Kind" 。

 

提示：

  • ranks.length == suits.length == 5
  • 1 <= ranks[i] <= 13
  • 'a' <= suits[i] <= 'd'
  • 任意两张扑克牌不会同时有相同的大小和花色。

Metadata

You are given an integer array ranks and a character array suits. You have 5 cards where the ith card has a rank of ranks[i] and a suit of suits[i].

The following are the types of poker hands you can make from best to worst:

  1. "Flush": Five cards of the same suit.
  2. "Three of a Kind": Three cards of the same rank.
  3. "Pair": Two cards of the same rank.
  4. "High Card": Any single card.

Return a string representing the best type of poker hand you can make with the given cards.

Note that the return values are case-sensitive.

 

Example 1:

Input: ranks = [13,2,3,1,9], suits = ["a","a","a","a","a"]
Output: "Flush"
Explanation: The hand with all the cards consists of 5 cards with the same suit, so we have a "Flush".

Example 2:

Input: ranks = [4,4,2,4,4], suits = ["d","a","a","b","c"]
Output: "Three of a Kind"
Explanation: The hand with the first, second, and fourth card consists of 3 cards with the same rank, so we have a "Three of a Kind".
Note that we could also make a "Pair" hand but "Three of a Kind" is a better hand.
Also note that other cards could be used to make the "Three of a Kind" hand.

Example 3:

Input: ranks = [10,10,2,12,9], suits = ["a","b","c","a","d"]
Output: "Pair"
Explanation: The hand with the first and second card consists of 2 cards with the same rank, so we have a "Pair".
Note that we cannot make a "Flush" or a "Three of a Kind".

 

Constraints:

  • ranks.length == suits.length == 5
  • 1 <= ranks[i] <= 13
  • 'a' <= suits[i] <= 'd'
  • No two cards have the same rank and suit.

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class Solution {
public:
    string bestHand(vector<int> &ranks, vector<char> &suits) {
        {
            bool Flush = true;

            for (int i = 1; i < 5; i++) {
                if (suits[i] != suits[0]) {
                    Flush = false;
                    break;
                }
            }

            if (Flush) {
                return "Flush";
            }
        }

        sort(all(ranks));

        for (int i = 2; i < 5; i++) {
            if (ranks[i] == ranks[i - 1] && ranks[i] == ranks[i - 2]) {
                return "Three of a Kind";
            }
        }

        for (int i = 1; i < 5; i++) {
            if (ranks[i] == ranks[i - 1]) {
                return "Pair";
            }
        }

        return "High Card";
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

Metadata

给你一个整数数组 nums ，返回全部为 0 的 子数组 数目。

子数组 是一个数组中一段连续非空元素组成的序列。

 

示例 1：

输入：nums = [1,3,0,0,2,0,0,4]
输出：6
解释：
子数组 [0] 出现了 4 次。
子数组 [0,0] 出现了 2 次。
不存在长度大于 2 的全 0 子数组，所以我们返回 6 。

示例 2：

输入：nums = [0,0,0,2,0,0]
输出：9
解释：
子数组 [0] 出现了 5 次。
子数组 [0,0] 出现了 3 次。
子数组 [0,0,0] 出现了 1 次。
不存在长度大于 3 的全 0 子数组，所以我们返回 9 。

示例 3：

输入：nums = [2,10,2019]
输出：0
解释：没有全 0 子数组，所以我们返回 0 。

 

提示：

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109

Metadata

Given an integer array nums, return the number of subarrays filled with 0.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [1,3,0,0,2,0,0,4]
Output: 6
Explanation: 
There are 4 occurrences of [0] as a subarray.
There are 2 occurrences of [0,0] as a subarray.
There is no occurrence of a subarray with a size more than 2 filled with 0. Therefore, we return 6.

Example 2:

Input: nums = [0,0,0,2,0,0]
Output: 9
Explanation:
There are 5 occurrences of [0] as a subarray.
There are 3 occurrences of [0,0] as a subarray.
There is 1 occurrence of [0,0,0] as a subarray.
There is no occurrence of a subarray with a size more than 3 filled with 0. Therefore, we return 9.

Example 3:

Input: nums = [2,10,2019]
Output: 0
Explanation: There is no subarray filled with 0. Therefore, we return 0.

 

Constraints:

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <numeric>
#include <vector>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class Solution {
public:
    long long zeroFilledSubarray(vector<int> &nums) {
        int n = int(nums.size());
        auto f = vector<ll>(nums.size() + 5, 0);
        for (int i = 1; i <= n; i++) {
            int x = nums[i - 1];
            if (x != 0) {
                f[i] = 0;
            } else {
                f[i] = f[i - 1] + 1;
            }
        }

        return accumulate(f.begin(), f.end(), 0ll);
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

Metadata

设计一个数字容器系统，可以实现以下功能：

  • 在系统中给定下标处 插入 或者 替换 一个数字。
  • 返回 系统中给定数字的最小下标。

请你实现一个 NumberContainers 类：

  • NumberContainers() 初始化数字容器系统。
  • void change(int index, int number) 在下标 index 处填入 number 。如果该下标 index 处已经有数字了，那么用 number 替换该数字。
  • int find(int number) 返回给定数字 number 在系统中的最小下标。如果系统中没有 number ，那么返回 -1 。

 

示例：

输入：
["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]
[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]
输出：
[null, -1, null, null, null, null, 1, null, 2]
解释：
NumberContainers nc = new NumberContainers();
nc.find(10); // 没有数字 10 ，所以返回 -1 。
nc.change(2, 10); // 容器中下标为 2 处填入数字 10 。
nc.change(1, 10); // 容器中下标为 1 处填入数字 10 。
nc.change(3, 10); // 容器中下标为 3 处填入数字 10 。
nc.change(5, 10); // 容器中下标为 5 处填入数字 10 。
nc.find(10); // 数字 10 所在的下标为 1 ，2 ，3 和 5 。因为最小下标为 1 ，所以返回 1 。
nc.change(1, 20); // 容器中下标为 1 处填入数字 20 。注意，下标 1 处之前为 10 ，现在被替换为 20 。
nc.find(10); // 数字 10 所在下标为 2 ，3 和 5 。最小下标为 2 ，所以返回 2 。

 

提示：

  • 1 <= index, number <= 109
  • 调用 change 和 find 的 总次数 不超过 105 次。

Metadata

Design a number container system that can do the following:

  • Insert or Replace a number at the given index in the system.
  • Return the smallest index for the given number in the system.

Implement the NumberContainers class:

  • NumberContainers() Initializes the number container system.
  • void change(int index, int number) Fills the container at index with the number. If there is already a number at that index, replace it.
  • int find(int number) Returns the smallest index for the given number, or -1 if there is no index that is filled by number in the system.

 

Example 1:

Input
["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]
[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]
Output
[null, -1, null, null, null, null, 1, null, 2]
Explanation
NumberContainers nc = new NumberContainers();
nc.find(10); // There is no index that is filled with number 10. Therefore, we return -1.
nc.change(2, 10); // Your container at index 2 will be filled with number 10.
nc.change(1, 10); // Your container at index 1 will be filled with number 10.
nc.change(3, 10); // Your container at index 3 will be filled with number 10.
nc.change(5, 10); // Your container at index 5 will be filled with number 10.
nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.
nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. 
nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.

 

Constraints:

  • 1 <= index, number <= 109
  • At most 105 calls will be made in total to change and find.

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <vector>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class NumberContainers {
public:
    NumberContainers() {
        num_map.clear();
        ix_map.clear();
    }

    void change(int index, int number) {
        if (num_map.count(index) != 0) {
            ix_map[num_map[index]].erase(index);
        }

        num_map[index] = number;
        ix_map[number].insert(index);
    }

    int find(int number) {
        if (ix_map.count(number) == 0 || ix_map[number].empty()) {
            return -1;
        }

        auto it = ix_map[number].begin();
        return *it;
    }

    map<int, int> num_map;
    map<int, set<int>> ix_map;
};

/**
 * Your NumberContainers object will be instantiated and called as such:
 * NumberContainers* obj = new NumberContainers();
 * obj->change(index,number);
 * int param_2 = obj->find(number);
 */

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

Metadata

给你一个长度为 n 的整数数组 rolls 和一个整数 k 。你扔一个 k 面的骰子 n 次，骰子的每个面分别是 1 到 k ，其中第 i 次扔得到的数字是 rolls[i] 。

请你返回 无法 从 rolls 中得到的 最短 骰子子序列的长度。

扔一个 k 面的骰子 len 次得到的是一个长度为 len 的 骰子子序列 。

注意 ，子序列只需要保持在原数组中的顺序，不需要连续。

 

示例 1：

输入：rolls = [4,2,1,2,3,3,2,4,1], k = 4
输出：3
解释：所有长度为 1 的骰子子序列 [1] ，[2] ，[3] ，[4] 都可以从原数组中得到。
所有长度为 2 的骰子子序列 [1, 1] ，[1, 2] ，… ，[4, 4] 都可以从原数组中得到。
子序列 [1, 4, 2] 无法从原数组中得到，所以我们返回 3 。
还有别的子序列也无法从原数组中得到。

示例 2：

输入：rolls = [1,1,2,2], k = 2
输出：2
解释：所有长度为 1 的子序列 [1] ，[2] 都可以从原数组中得到。
子序列 [2, 1] 无法从原数组中得到，所以我们返回 2 。
还有别的子序列也无法从原数组中得到，但 [2, 1] 是最短的子序列。

示例 3：

输入：rolls = [1,1,3,2,2,2,3,3], k = 4
输出：1
解释：子序列 [4] 无法从原数组中得到，所以我们返回 1 。
还有别的子序列也无法从原数组中得到，但 [4] 是最短的子序列。

 

提示：

  • n == rolls.length
  • 1 <= n <= 105
  • 1 <= rolls[i] <= k <= 105

Metadata

You are given an integer array rolls of length n and an integer k. You roll a k sided dice numbered from 1 to k, n times, where the result of the ith roll is rolls[i].

Return the length of the shortest sequence of rolls that cannot be taken from rolls.

A sequence of rolls of length len is the result of rolling a k sided dice len times.

Note that the sequence taken does not have to be consecutive as long as it is in order.

 

Example 1:

Input: rolls = [4,2,1,2,3,3,2,4,1], k = 4
Output: 3
Explanation: Every sequence of rolls of length 1, [1], [2], [3], [4], can be taken from rolls.
Every sequence of rolls of length 2, [1, 1], [1, 2], …, [4, 4], can be taken from rolls.
The sequence [1, 4, 2] cannot be taken from rolls, so we return 3.
Note that there are other sequences that cannot be taken from rolls.

Example 2:

Input: rolls = [1,1,2,2], k = 2
Output: 2
Explanation: Every sequence of rolls of length 1, [1], [2], can be taken from rolls.
The sequence [2, 1] cannot be taken from rolls, so we return 2.
Note that there are other sequences that cannot be taken from rolls but [2, 1] is the shortest.

Example 3:

Input: rolls = [1,1,3,2,2,2,3,3], k = 4
Output: 1
Explanation: The sequence [4] cannot be taken from rolls, so we return 1.
Note that there are other sequences that cannot be taken from rolls but [4] is the shortest.

 

Constraints:

  • n == rolls.length
  • 1 <= n <= 105
  • 1 <= rolls[i] <= k <= 105

Solution

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <vector>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
    return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
    return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif
// head

class Solution {
public:
    int shortestSequence(vector<int> &rolls, int k) {
        int res = 0;
        auto f = vector<int>(k + 1, 0);
        int cnt = 0;
        for (const auto &r : rolls) {
            if (f[r] == 0) {
                ++cnt;
                f[r] = 1;
                if (cnt == k) {
                    ++res;
                    cnt = 0;
                    auto g = vector<int>(k + 1, 0);
                    f.swap(g);
                }
            }
        }

        return res + 1;
    }
};

#ifdef LOCAL

int main() {
    auto s = Solution();

    {
        auto r = vector<int>({4, 2, 1, 2, 3, 3, 2, 4, 1});
        dbg(s.shortestSequence(r, 4));
    }

    {
        auto r = vector<int>({1, 1, 2, 2});
        dbg(s.shortestSequence(r, 2));
    }

    {
        auto r = vector<int>({1, 1, 3, 2, 2, 2, 3, 3});
        dbg(s.shortestSequence(r, 4));
    }

    {
        auto r = vector<int>({1, 1});
        dbg(s.shortestSequence(r, 2));
    }

    return 0;
}

#endif
最后更新: October 11, 2023
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