weekly-contest-311

A

Statement

简体中文English

Metadata

Link: 最小偶倍数
Difficulty: Easy
Tag:

给你一个正整数 n ，返回 2 和 n 的最小公倍数（正整数）。

示例 1：

输入：n = 5
输出：10
解释：5 和 2 的最小公倍数是 10 。

示例 2：

输入：n = 6
输出：6
解释：6 和 2 的最小公倍数是 6 。注意数字会是它自身的倍数。

提示：

1 <= n <= 150

Metadata

Link: Smallest Even Multiple
Difficulty: Easy
Tag:

Given a positive integer n, return the smallest positive integer that is a multiple of both 2 and n.

Example 1:

Input: n = 5
Output: 10
Explanation: The smallest multiple of both 5 and 2 is 10.

Example 2:

Input: n = 6
Output: 6
Explanation: The smallest multiple of both 6 and 2 is 6. Note that a number is a multiple of itself.

Constraints:

1 <= n <= 150

Solution

C++

#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:
    int smallestEvenMultiple(int n) {
        return n * 2 / __gcd(n, 2);
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

简体中文English

Metadata

Link: 最长的字母序连续子字符串的长度
Difficulty: Medium
Tag:

字母序连续字符串 是由字母表中连续字母组成的字符串。换句话说，字符串 "abcdefghijklmnopqrstuvwxyz" 的任意子字符串都是 字母序连续字符串 。

例如，"abc" 是一个字母序连续字符串，而 "acb" 和 "za" 不是。

给你一个仅由小写英文字母组成的字符串 s ，返回其最长的字母序连续子字符串的长度。

示例 1：

输入：s = "abacaba"
输出：2
解释：共有 4 个不同的字母序连续子字符串 "a"、"b"、"c" 和 "ab" 。
"ab" 是最长的字母序连续子字符串。

示例 2：

输入：s = "abcde"
输出：5
解释："abcde" 是最长的字母序连续子字符串。

提示：

1 <= s.length <= 10⁵
s 由小写英文字母组成

Metadata

Link: Length of the Longest Alphabetical Continuous Substring
Difficulty: Medium
Tag:

An alphabetical continuous string is a string consisting of consecutive letters in the alphabet. In other words, it is any substring of the string "abcdefghijklmnopqrstuvwxyz".

For example, "abc" is an alphabetical continuous string, while "acb" and "za" are not.

Given a string s consisting of lowercase letters only, return the length of the longest alphabetical continuous substring.

Example 1:

Input: s = "abacaba"
Output: 2
Explanation: There are 4 distinct continuous substrings: "a", "b", "c" and "ab".
"ab" is the longest continuous substring.

Example 2:

Input: s = "abcde"
Output: 5
Explanation: "abcde" is the longest continuous substring.

Constraints:

1 <= s.length <= 10⁵
s consists of only English lowercase letters.

Solution

C++

#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:
    int longestContinuousSubstring(string s) {
        int n = int(s.size());
        int res = 1;
        int cur = 1;

        for (int i = 1; i < n; i++) {
            if (s[i] == s[i - 1] + 1) {
                ++cur;
                res = max(res, cur);
            } else {
                cur = 1;
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

简体中文English

Metadata

Link: 反转二叉树的奇数层
Difficulty: Medium
Tag:

给你一棵完美二叉树的根节点 root ，请你反转这棵树中每个奇数层的节点值。

例如，假设第 3 层的节点值是 [2,1,3,4,7,11,29,18] ，那么反转后它应该变成 [18,29,11,7,4,3,1,2] 。

反转后，返回树的根节点。

完美二叉树需满足：二叉树的所有父节点都有两个子节点，且所有叶子节点都在同一层。

节点的层数等于该节点到根节点之间的边数。

示例 1：

输入：root = [2,3,5,8,13,21,34]
输出：[2,5,3,8,13,21,34]
解释：
这棵树只有一个奇数层。
在第 1 层的节点分别是 3、5 ，反转后为 5、3 。

示例 2：

输入：root = [7,13,11]
输出：[7,11,13]
解释： 
在第 1 层的节点分别是 13、11 ，反转后为 11、13 。

示例 3：

输入：root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
输出：[0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
解释：奇数层由非零值组成。
在第 1 层的节点分别是 1、2 ，反转后为 2、1 。
在第 3 层的节点分别是 1、1、1、1、2、2、2、2 ，反转后为 2、2、2、2、1、1、1、1 。

提示：

树中的节点数目在范围 [1, 2¹⁴] 内
0 <= Node.val <= 10⁵
root 是一棵完美二叉树

Metadata

Link: Reverse Odd Levels of Binary Tree
Difficulty: Medium
Tag:

Given the root of a perfect binary tree, reverse the node values at each odd level of the tree.

For example, suppose the node values at level 3 are [2,1,3,4,7,11,29,18], then it should become [18,29,11,7,4,3,1,2].

Return the root of the reversed tree.

A binary tree is perfect if all parent nodes have two children and all leaves are on the same level.

The level of a node is the number of edges along the path between it and the root node.

Example 1:

Input: root = [2,3,5,8,13,21,34]
Output: [2,5,3,8,13,21,34]
Explanation: 
The tree has only one odd level.
The nodes at level 1 are 3, 5 respectively, which are reversed and become 5, 3.

Example 2:

Input: root = [7,13,11]
Output: [7,11,13]
Explanation: 
The nodes at level 1 are 13, 11, which are reversed and become 11, 13.

Example 3:

Input: root = [0,1,2,0,0,0,0,1,1,1,1,2,2,2,2]
Output: [0,2,1,0,0,0,0,2,2,2,2,1,1,1,1]
Explanation: 
The odd levels have non-zero values.
The nodes at level 1 were 1, 2, and are 2, 1 after the reversal.
The nodes at level 3 were 1, 1, 1, 1, 2, 2, 2, 2, and are 2, 2, 2, 2, 1, 1, 1, 1 after the reversal.

Constraints:

The number of nodes in the tree is in the range [1, 2¹⁴].
0 <= Node.val <= 10⁵
root is a perfect binary tree.

Solution

C++

#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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */

#ifdef LOCAL

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

#endif

struct node {
    TreeNode *t{nullptr};
    int dep{0};
};

class Solution {
public:
    TreeNode *reverseOddLevels(TreeNode *root) {
        queue<node> q;
        q.push(node{
                .t = root,
                .dep = 0,
        });

        auto vec = vector<vector<node>>(20, vector<node>{});

        while (!q.empty()) {
            auto front = q.front();
            q.pop();

            vec[front.dep].push_back(front);
            if (front.t->left) {
                q.push(node{
                        .t = front.t->left,
                        .dep = front.dep + 1,
                });
            }

            if (front.t->right) {
                q.push(node{
                        .t = front.t->right,
                        .dep = front.dep + 1,
                });
            }
        }

        for (int i = 1; i < 20; i += 2) {
            int n = int(vec[i].size());
            for (int j = 0, k = n - 1; j < k; j++, k--) {
                swap(vec[i][j].t->val, vec[i][k].t->val);
            }
        }

        return root;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

简体中文English

Metadata

Link: 字符串的前缀分数和
Difficulty: Hard
Tag:

给你一个长度为 n 的数组 words ，该数组由非空字符串组成。

定义字符串 word 的分数等于以 word 作为前缀的 words[i] 的数目。

例如，如果 words = ["a", "ab", "abc", "cab"] ，那么 "ab" 的分数是 2 ，因为 "ab" 是 "ab" 和 "abc" 的一个前缀。

返回一个长度为 n 的数组 answer ，其中 answer[i] 是 words[i] 的每个非空前缀的分数总和。

注意：字符串视作它自身的一个前缀。

示例 1：

输入：words = ["abc","ab","bc","b"]
输出：[5,4,3,2]
解释：对应每个字符串的答案如下：
- "abc" 有 3 个前缀："a"、"ab" 和 "abc" 。
- 2 个字符串的前缀为 "a" ，2 个字符串的前缀为 "ab" ，1 个字符串的前缀为 "abc" 。
总计 answer[0] = 2 + 2 + 1 = 5 。
- "ab" 有 2 个前缀："a" 和 "ab" 。
- 2 个字符串的前缀为 "a" ，2 个字符串的前缀为 "ab" 。
总计 answer[1] = 2 + 2 = 4 。
- "bc" 有 2 个前缀："b" 和 "bc" 。
- 2 个字符串的前缀为 "b" ，1 个字符串的前缀为 "bc" 。 
总计 answer[2] = 2 + 1 = 3 。
- "b" 有 1 个前缀："b"。
- 2 个字符串的前缀为 "b" 。
总计 answer[3] = 2 。

示例 2：

输入：words = ["abcd"]
输出：[4]
解释：
"abcd" 有 4 个前缀 "a"、"ab"、"abc" 和 "abcd"。
每个前缀的分数都是 1 ，总计 answer[0] = 1 + 1 + 1 + 1 = 4 。

提示：

1 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i] 由小写英文字母组成

Metadata

Link: Sum of Prefix Scores of Strings
Difficulty: Hard
Tag:

You are given an array words of size n consisting of non-empty strings.

We define the score of a string word as the number of strings words[i] such that word is a prefix of words[i].

For example, if words = ["a", "ab", "abc", "cab"], then the score of "ab" is 2, since "ab" is a prefix of both "ab" and "abc".

Return an array answer of size n where answer[i] is the sum of scores of every non-empty prefix of words[i].

Note that a string is considered as a prefix of itself.

Example 1:

Input: words = ["abc","ab","bc","b"]
Output: [5,4,3,2]
Explanation: The answer for each string is the following:
- "abc" has 3 prefixes: "a", "ab", and "abc".
- There are 2 strings with the prefix "a", 2 strings with the prefix "ab", and 1 string with the prefix "abc".
The total is answer[0] = 2 + 2 + 1 = 5.
- "ab" has 2 prefixes: "a" and "ab".
- There are 2 strings with the prefix "a", and 2 strings with the prefix "ab".
The total is answer[1] = 2 + 2 = 4.
- "bc" has 2 prefixes: "b" and "bc".
- There are 2 strings with the prefix "b", and 1 string with the prefix "bc".
The total is answer[2] = 2 + 1 = 3.
- "b" has 1 prefix: "b".
- There are 2 strings with the prefix "b".
The total is answer[3] = 2.

Example 2:

Input: words = ["abcd"]
Output: [4]
Explanation:
"abcd" has 4 prefixes: "a", "ab", "abc", and "abcd".
Each prefix has a score of one, so the total is answer[0] = 1 + 1 + 1 + 1 = 4.

Constraints:

1 <= words.length <= 1000
1 <= words[i].length <= 1000
words[i] consists of lowercase English letters.

Solution

C++

#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

constexpr int N = 1e6 + 10;

struct TRIE {
    struct node {
        int son[26];
        int cnt;
        node() {
            memset(son, -1, sizeof son);
            cnt = 0;
        }
    } t[N];
    int rt, cnt;

    void init() {
        rt = 0;
        t[0] = node();
        cnt = 0;
    }

    void insert(const std::string &s) {
        int cur_rt = rt;
        for (const auto &c : s) {
            if (t[cur_rt].son[c - 'a'] == -1) {
                ++cnt;
                t[cnt] = node();
                t[cur_rt].son[c - 'a'] = cnt;
            }

            cur_rt = t[cur_rt].son[c - 'a'];
            ++t[cur_rt].cnt;
        }
    }

    int query(const std::string &s) {
        int res = 0;
        int cur_rt = rt;
        for (const auto &c : s) {
            cur_rt = t[cur_rt].son[c - 'a'];
            res += t[cur_rt].cnt;
        }

        return res;
    }
} tr;

class Solution {
public:
    vector<int> sumPrefixScores(vector<string> &words) {
        tr.init();

        for (const auto &w : words) {
            tr.insert(w);
        }

        auto res = vector<int>();
        for (const auto &w : words) {
            res.push_back(tr.query(w));
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: October 11, 2023