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biweekly-contest-80

A

Statement

Metadata

如果一个密码满足以下所有条件，我们称它是一个 强 密码：

  • 它有至少 8 个字符。
  • 至少包含 一个小写英文 字母。
  • 至少包含 一个大写英文 字母。
  • 至少包含 一个数字 。
  • 至少包含 一个特殊字符 。特殊字符为："!@#$%^&*()-+" 中的一个。
  • 它 不 包含 2 个连续相同的字符（比方说 "aab" 不符合该条件，但是 "aba" 符合该条件）。

给你一个字符串 password ，如果它是一个 强 密码，返回 true，否则返回 false 。

 

示例 1：

输入：password = "IloveLe3tcode!"
输出：true
解释：密码满足所有的要求，所以我们返回 true 。

示例 2：

输入：password = "Me+You–IsMyDream"
输出：false
解释：密码不包含数字，且包含 2 个连续相同的字符。所以我们返回 false 。

示例 3：

输入：password = "1aB!"
输出：false
解释：密码不符合长度要求。所以我们返回 false 。

 

提示：

  • 1 <= password.length <= 100
  • password 包含字母，数字和 "!@#$%^&*()-+" 这些特殊字符。

Metadata

A password is said to be strong if it satisfies all the following criteria:

  • It has at least 8 characters.
  • It contains at least one lowercase letter.
  • It contains at least one uppercase letter.
  • It contains at least one digit.
  • It contains at least one special character. The special characters are the characters in the following string: "!@#$%^&*()-+".
  • It does not contain 2 of the same character in adjacent positions (i.e., "aab" violates this condition, but "aba" does not).

Given a string password, return true if it is a strong password. Otherwise, return false.

 

Example 1:

Input: password = "IloveLe3tcode!"
Output: true
Explanation: The password meets all the requirements. Therefore, we return true.

Example 2:

Input: password = "Me+You–IsMyDream"
Output: false
Explanation: The password does not contain a digit and also contains 2 of the same character in adjacent positions. Therefore, we return false.

Example 3:

Input: password = "1aB!"
Output: false
Explanation: The password does not meet the length requirement. Therefore, we return false.

 

Constraints:

  • 1 <= password.length <= 100
  • password consists of letters, digits, and special characters: "!@#$%^&*()-+".

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:
    bool strongPasswordCheckerII(string s) {
        int l = s.length();
        if (l < 8) {
            return false;
        }

        bool a = 0, b = 0, c = 0, d = 0;

        for (int i = 0; i < l; i++) {
            if (i + 1 < l && s[i] == s[i + 1]) {
                return false;
            }

            if (s[i] >= 'a' && s[i] <= 'z') {
                a = 1;
            }

            if (s[i] >= 'A' && s[i] <= 'Z') {
                b = 1;
            }

            if (s[i] >= '0' && s[i] <= '9') {
                c = 1;
            }

            if (string("!@#$%^&*()-+").find(s[i]) != string::npos) {
                d = 1;
            }
        }

        return a && b && c && d;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

Metadata

给你两个正整数数组 spells 和 potions ，长度分别为 n 和 m ，其中 spells[i] 表示第 i 个咒语的能量强度，potions[j] 表示第 j 瓶药水的能量强度。

同时给你一个整数 success 。一个咒语和药水的能量强度 相乘 如果 大于等于 success ，那么它们视为一对 成功 的组合。

请你返回一个长度为 n 的整数数组 pairs，其中 pairs[i] 是能跟第 i 个咒语成功组合的 药水 数目。

 

示例 1：

输入：spells = [5,1,3], potions = [1,2,3,4,5], success = 7
输出：[4,0,3]
解释：
- 第 0 个咒语：5 * [1,2,3,4,5] = [5,10,15,20,25] 。总共 4 个成功组合。
- 第 1 个咒语：1 * [1,2,3,4,5] = [1,2,3,4,5] 。总共 0 个成功组合。
- 第 2 个咒语：3 * [1,2,3,4,5] = [3,6,9,12,15] 。总共 3 个成功组合。
所以返回 [4,0,3] 。

示例 2：

输入：spells = [3,1,2], potions = [8,5,8], success = 16
输出：[2,0,2]
解释：
- 第 0 个咒语：3 * [8,5,8] = [24,15,24] 。总共 2 个成功组合。
- 第 1 个咒语：1 * [8,5,8] = [8,5,8] 。总共 0 个成功组合。
- 第 2 个咒语：2 * [8,5,8] = [16,10,16] 。总共 2 个成功组合。
所以返回 [2,0,2] 。

 

提示：

  • n == spells.length
  • m == potions.length
  • 1 <= n, m <= 105
  • 1 <= spells[i], potions[i] <= 105
  • 1 <= success <= 1010

Metadata

You are given two positive integer arrays spells and potions, of length n and m respectively, where spells[i] represents the strength of the ith spell and potions[j] represents the strength of the jth potion.

You are also given an integer success. A spell and potion pair is considered successful if the product of their strengths is at least success.

Return an integer array pairs of length n where pairs[i] is the number of potions that will form a successful pair with the ith spell.

 

Example 1:

Input: spells = [5,1,3], potions = [1,2,3,4,5], success = 7
Output: [4,0,3]
Explanation:
- 0th spell: 5 * [1,2,3,4,5] = [5,10,15,20,25]. 4 pairs are successful.
- 1st spell: 1 * [1,2,3,4,5] = [1,2,3,4,5]. 0 pairs are successful.
- 2nd spell: 3 * [1,2,3,4,5] = [3,6,9,12,15]. 3 pairs are successful.
Thus, [4,0,3] is returned.

Example 2:

Input: spells = [3,1,2], potions = [8,5,8], success = 16
Output: [2,0,2]
Explanation:
- 0th spell: 3 * [8,5,8] = [24,15,24]. 2 pairs are successful.
- 1st spell: 1 * [8,5,8] = [8,5,8]. 0 pairs are successful. 
- 2nd spell: 2 * [8,5,8] = [16,10,16]. 2 pairs are successful. 
Thus, [2,0,2] is returned.

 

Constraints:

  • n == spells.length
  • m == potions.length
  • 1 <= n, m <= 105
  • 1 <= spells[i], potions[i] <= 105
  • 1 <= success <= 1010

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:
    vector<int> successfulPairs(vector<int> &s, vector<int> &p, long long success) {
        sort(p.begin(), p.end());
        int n = p.size();

        auto res = vector<int>();

        for (const auto &_s : s) {
            auto it = lower_bound(p.begin(), p.end(), (success + _s - 1) / _s) - p.begin();
            res.push_back(n - it);
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

Metadata

给你两个字符串 s 和 sub 。同时给你一个二维字符数组 mappings ，其中 mappings[i] = [oldi, newi] 表示你可以替换 sub 中任意数目的 oldi 个字符，替换成 newi 。sub 中每个字符 不能 被替换超过一次。

如果使用 mappings 替换 0 个或者若干个字符，可以将 sub 变成 s 的一个子字符串，请你返回 true，否则返回 false 。

一个 子字符串 是字符串中连续非空的字符序列。

 

示例 1：

输入：s = "fool3e7bar", sub = "leet", mappings = [["e","3"],["t","7"],["t","8"]]
输出：true
解释：将 sub 中第一个 'e' 用 '3' 替换，将 't' 用 '7' 替换。
现在 sub = "l3e7" ，它是 s 的子字符串，所以我们返回 true 。

示例 2：

输入：s = "fooleetbar", sub = "f00l", mappings = [["o","0"]]
输出：false
解释：字符串 "f00l" 不是 s 的子串且没有可以进行的修改。
注意我们不能用 'o' 替换 '0' 。

示例 3：

输入：s = "Fool33tbaR", sub = "leetd", mappings = [["e","3"],["t","7"],["t","8"],["d","b"],["p","b"]]
输出：true
解释：将 sub 里第一个和第二个 'e' 用 '3' 替换，用 'b' 替换 sub 里的 'd' 。
得到 sub = "l33tb" ，它是 s 的子字符串，所以我们返回 true 。

 

提示：

  • 1 <= sub.length <= s.length <= 5000
  • 0 <= mappings.length <= 1000
  • mappings[i].length == 2
  • oldi != newi
  • s 和 sub 只包含大写和小写英文字母和数字。
  • oldi 和 newi 是大写、小写字母或者是个数字。

Metadata

You are given two strings s and sub. You are also given a 2D character array mappings where mappings[i] = [oldi, newi] indicates that you may replace any number of oldi characters of sub with newi. Each character in sub cannot be replaced more than once.

Return true if it is possible to make sub a substring of s by replacing zero or more characters according to mappings. Otherwise, return false.

A substring is a contiguous non-empty sequence of characters within a string.

 

Example 1:

Input: s = "fool3e7bar", sub = "leet", mappings = [["e","3"],["t","7"],["t","8"]]
Output: true
Explanation: Replace the first 'e' in sub with '3' and 't' in sub with '7'.
Now sub = "l3e7" is a substring of s, so we return true.

Example 2:

Input: s = "fooleetbar", sub = "f00l", mappings = [["o","0"]]
Output: false
Explanation: The string "f00l" is not a substring of s and no replacements can be made.
Note that we cannot replace '0' with 'o'.

Example 3:

Input: s = "Fool33tbaR", sub = "leetd", mappings = [["e","3"],["t","7"],["t","8"],["d","b"],["p","b"]]
Output: true
Explanation: Replace the first and second 'e' in sub with '3' and 'd' in sub with 'b'.
Now sub = "l33tb" is a substring of s, so we return true.

 

Constraints:

  • 1 <= sub.length <= s.length <= 5000
  • 0 <= mappings.length <= 1000
  • mappings[i].length == 2
  • oldi != newi
  • s and sub consist of uppercase and lowercase English letters and digits.
  • oldi and newi are either uppercase or lowercase English letters or digits.

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:
    bool matchReplacement(string s, string sub, vector<vector<char>> &mappings) {
        map<char, std::string> mp;

        for (int i = 'a'; i <= 'z'; i++) {
            mp[char(i)] = string(1, char(i));
        }

        for (int i = 'A'; i <= 'Z'; i++) {
            mp[char(i)] = string(1, char(i));
        }

        for (int i = '0'; i <= '9'; i++) {
            mp[char(i)] = string(1, char(i));
        }

        for (auto &mapping : mappings) {
            mp[mapping[0]] += mapping[1];
        }

        for (auto &[k, v] : mp) {
            sort(v.begin(), v.end());
            v.erase(unique(v.begin(), v.end()), v.end());
        }

        int n = int(s.length());
        int m = int(sub.length());

        for (int i = 0; i < n; i++) {
            if (i + m > n) {
                break;
            }

            bool ok = 1;

            for (int j = 0; j < m; j++) {
                if (mp[sub[j]].find(s[i + j]) == string::npos) {
                    ok = 0;
                    break;
                }
            }

            if (ok) {
                return true;
            }
        }

        return false;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

Metadata

一个数字的 分数 定义为数组之和 乘以 数组的长度。

  • 比方说，[1, 2, 3, 4, 5] 的分数为 (1 + 2 + 3 + 4 + 5) * 5 = 75 。

给你一个正整数数组 nums 和一个整数 k ，请你返回 nums 中分数 严格小于 k 的 非空整数子数组数目。

子数组 是数组中的一个连续元素序列。

 

示例 1：

输入：nums = [2,1,4,3,5], k = 10
输出：6
解释：
有 6 个子数组的分数小于 10 ：
- [2] 分数为 2 * 1 = 2 。
- [1] 分数为 1 * 1 = 1 。
- [4] 分数为 4 * 1 = 4 。
- [3] 分数为 3 * 1 = 3 。 
- [5] 分数为 5 * 1 = 5 。
- [2,1] 分数为 (2 + 1) * 2 = 6 。
注意，子数组 [1,4] 和 [4,3,5] 不符合要求，因为它们的分数分别为 10 和 36，但我们要求子数组的分数严格小于 10 。

示例 2：

输入：nums = [1,1,1], k = 5
输出：5
解释：
除了 [1,1,1] 以外每个子数组分数都小于 5 。
[1,1,1] 分数为 (1 + 1 + 1) * 3 = 9 ，大于 5 。
所以总共有 5 个子数组得分小于 5 。

 

提示：

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 105
  • 1 <= k <= 1015

Metadata

The score of an array is defined as the product of its sum and its length.

  • For example, the score of [1, 2, 3, 4, 5] is (1 + 2 + 3 + 4 + 5) * 5 = 75.

Given a positive integer array nums and an integer k, return the number of non-empty subarrays of nums whose score is strictly less than k.

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

 

Example 1:

Input: nums = [2,1,4,3,5], k = 10
Output: 6
Explanation:
The 6 subarrays having scores less than 10 are:
- [2] with score 2 * 1 = 2.
- [1] with score 1 * 1 = 1.
- [4] with score 4 * 1 = 4.
- [3] with score 3 * 1 = 3. 
- [5] with score 5 * 1 = 5.
- [2,1] with score (2 + 1) * 2 = 6.
Note that subarrays such as [1,4] and [4,3,5] are not considered because their scores are 10 and 36 respectively, while we need scores strictly less than 10.

Example 2:

Input: nums = [1,1,1], k = 5
Output: 5
Explanation:
Every subarray except [1,1,1] has a score less than 5.
[1,1,1] has a score (1 + 1 + 1) * 3 = 9, which is greater than 5.
Thus, there are 5 subarrays having scores less than 5.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 105
  • 1 <= k <= 1015

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:
    long long countSubarrays(vector<int> &nums, long long k) {
        ll res = 0;
        int r = -1;
        int n = int(nums.size());
        ll sum = 0;

        for (int l = 0; l < n; l++) {
            while (r + 1 < n) {
                ll nx_sum = sum + nums[r + 1];
                ll len = r + 1 - l + 1;
                if (nx_sum * len >= k) {
                    break;
                }

                sum = nx_sum;
                ++r;
            }

            res += r - l + 1;
            if (r < l) {
                r = l;
            } else {
                sum -= nums[l];
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif
最后更新: October 11, 2023
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