跳转至

biweekly-contest-90

A

Statement

Metadata

给你一个字符串数组 words ,每一个字符串长度都相同,令所有字符串的长度都为 n 。

每个字符串 words[i] 可以被转化为一个长度为 n - 1 的 差值整数数组 difference[i] ,其中对于 0 <= j <= n - 2 有 difference[i][j] = words[i][j+1] - words[i][j] 。注意两个字母的差值定义为它们在字母表中 位置 之差,也就是说 'a' 的位置是 0 ,'b' 的位置是 1 ,'z' 的位置是 25 。

  • 比方说,字符串 "acb" 的差值整数数组是 [2 - 0, 1 - 2] = [2, -1] 。

words 中所有字符串 除了一个字符串以外 ,其他字符串的差值整数数组都相同。你需要找到那个不同的字符串。

请你返回 words中 差值整数数组 不同的字符串。

 

示例 1:

输入:words = ["adc","wzy","abc"]
输出:"abc"
解释:
- "adc" 的差值整数数组是 [3 - 0, 2 - 3] = [3, -1] 。
- "wzy" 的差值整数数组是 [25 - 22, 24 - 25]= [3, -1] 。
- "abc" 的差值整数数组是 [1 - 0, 2 - 1] = [1, 1] 。
不同的数组是 [1, 1],所以返回对应的字符串,"abc"。

示例 2:

输入:words = ["aaa","bob","ccc","ddd"]
输出:"bob"
解释:除了 "bob" 的差值整数数组是 [13, -13] 以外,其他字符串的差值整数数组都是 [0, 0] 。

 

提示:

  • 3 <= words.length <= 100
  • n == words[i].length
  • 2 <= n <= 20
  • words[i] 只含有小写英文字母。

Metadata

You are given an array of equal-length strings words. Assume that the length of each string is n.

Each string words[i] can be converted into a difference integer array difference[i] of length n - 1 where difference[i][j] = words[i][j+1] - words[i][j] where 0 <= j <= n - 2. Note that the difference between two letters is the difference between their positions in the alphabet i.e. the position of 'a' is 0, 'b' is 1, and 'z' is 25.

  • For example, for the string "acb", the difference integer array is [2 - 0, 1 - 2] = [2, -1].

All the strings in words have the same difference integer array, except one. You should find that string.

Return the string in words that has different difference integer array.

 

Example 1:

Input: words = ["adc","wzy","abc"]
Output: "abc"
Explanation: 
- The difference integer array of "adc" is [3 - 0, 2 - 3] = [3, -1].
- The difference integer array of "wzy" is [25 - 22, 24 - 25]= [3, -1].
- The difference integer array of "abc" is [1 - 0, 2 - 1] = [1, 1]. 
The odd array out is [1, 1], so we return the corresponding string, "abc".

Example 2:

Input: words = ["aaa","bob","ccc","ddd"]
Output: "bob"
Explanation: All the integer arrays are [0, 0] except for "bob", which corresponds to [13, -13].

 

Constraints:

  • 3 <= words.length <= 100
  • n == words[i].length
  • 2 <= n <= 20
  • words[i] consists of lowercase English letters.

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 oddString(vector<string>& words) {
        int m = int(words[0].size());

        const auto get_v = [&m](const string& s) {
            auto v = vector<int>();
            for (int i = 1; i < m; i++) {
                v.push_back(s[i] - s[i - 1]);
            }

            return v;
        };

        auto mp = map<vector<int>, int>();
        for (const auto& word : words) {
            mp[get_v(word)]++;
        }

        for (const auto& word : words) {
            auto v = get_v(word);
            if (mp[v] == 1) {
                return word;
            }
        }

        return "";
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

Metadata

给你两个字符串数组 queries 和 dictionary 。数组中所有单词都只包含小写英文字母,且长度都相同。

一次 编辑 中,你可以从 queries 中选择一个单词,将任意一个字母修改成任何其他字母。从 queries 中找到所有满足以下条件的字符串:不超过 两次编辑内,字符串与 dictionary 中某个字符串相同。

请你返回 queries 中的单词列表,这些单词距离 dictionary 中的单词 编辑次数 不超过 两次 。单词返回的顺序需要与 queries 中原本顺序相同。

 

示例 1:

输入:queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]
输出:["word","note","wood"]
解释:
- 将 "word" 中的 'r' 换成 'o' ,得到 dictionary 中的单词 "wood" 。
- 将 "note" 中的 'n' 换成 'j' 且将 't' 换成 'k' ,得到 "joke" 。
- "ants" 需要超过 2 次编辑才能得到 dictionary 中的单词。
- "wood" 不需要修改(0 次编辑),就得到 dictionary 中相同的单词。
所以我们返回 ["word","note","wood"] 。

示例 2:

输入:queries = ["yes"], dictionary = ["not"]
输出:[]
解释:
"yes" 需要超过 2 次编辑才能得到 "not" 。
所以我们返回空数组。

 

提示:

  • 1 <= queries.length, dictionary.length <= 100
  • n == queries[i].length == dictionary[j].length
  • 1 <= n <= 100
  • 所有 queries[i] 和 dictionary[j] 都只包含小写英文字母。

Metadata

You are given two string arrays, queries and dictionary. All words in each array comprise of lowercase English letters and have the same length.

In one edit you can take a word from queries, and change any letter in it to any other letter. Find all words from queries that, after a maximum of two edits, equal some word from dictionary.

Return a list of all words from queries, that match with some word from dictionary after a maximum of two edits. Return the words in the same order they appear in queries.

 

Example 1:

Input: queries = ["word","note","ants","wood"], dictionary = ["wood","joke","moat"]
Output: ["word","note","wood"]
Explanation:
- Changing the 'r' in "word" to 'o' allows it to equal the dictionary word "wood".
- Changing the 'n' to 'j' and the 't' to 'k' in "note" changes it to "joke".
- It would take more than 2 edits for "ants" to equal a dictionary word.
- "wood" can remain unchanged (0 edits) and match the corresponding dictionary word.
Thus, we return ["word","note","wood"].

Example 2:

Input: queries = ["yes"], dictionary = ["not"]
Output: []
Explanation:
Applying any two edits to "yes" cannot make it equal to "not". Thus, we return an empty array.

 

Constraints:

  • 1 <= queries.length, dictionary.length <= 100
  • n == queries[i].length == dictionary[j].length
  • 1 <= n <= 100
  • All queries[i] and dictionary[j] are composed of lowercase English letters.

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:
    vector<string> twoEditWords(vector<string>& queries, vector<string>& dictionary) {
        auto res = vector<string>();

        for (const auto& q : queries) {
            int len = int(q.size());
            for (const auto& d : dictionary) {
                int len_d = int(d.size());
                if (len != len_d) {
                    continue;
                }

                int cnt = 0;

                for (int i = 0; i < len; i++) {
                    cnt += (q[i] != d[i]);
                }

                if (cnt <= 2) {
                    res.push_back(q);
                    break;
                }
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

Metadata

给你一个下标从 0 开始的数组 nums ,它包含若干正整数,表示数轴上你需要摧毁的目标所在的位置。同时给你一个整数 space 。

你有一台机器可以摧毁目标。给机器 输入 nums[i] ,这台机器会摧毁所有位置在 nums[i] + c * space 的目标,其中 c 是任意非负整数。你想摧毁 nums 中 尽可能多 的目标。

请你返回在摧毁数目最多的前提下,nums[i] 的 最小值 。

 

示例 1:

输入:nums = [3,7,8,1,1,5], space = 2
输出:1
解释:如果我们输入 nums[3] ,我们可以摧毁位于 1,3,5,7,9,… 这些位置的目标。
这种情况下, 我们总共可以摧毁 5 个目标(除了 nums[2])。
没有办法摧毁多于 5 个目标,所以我们返回 nums[3] 。

示例 2:

输入:nums = [1,3,5,2,4,6], space = 2
输出:1
解释:输入 nums[0] 或者 nums[3] 都会摧毁 3 个目标。
没有办法摧毁多于 3 个目标。
由于 nums[0] 是最小的可以摧毁 3 个目标的整数,所以我们返回 1 。

示例 3:

输入:nums = [6,2,5], space = 100
输出:2
解释:无论我们输入哪个数字,都只能摧毁 1 个目标。输入的最小整数是 nums[1] 。

 

提示:

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

Metadata

You are given a 0-indexed array nums consisting of positive integers, representing targets on a number line. You are also given an integer space.

You have a machine which can destroy targets. Seeding the machine with some nums[i] allows it to destroy all targets with values that can be represented as nums[i] + c * space, where c is any non-negative integer. You want to destroy the maximum number of targets in nums.

Return the minimum value of nums[i] you can seed the machine with to destroy the maximum number of targets.

 

Example 1:

Input: nums = [3,7,8,1,1,5], space = 2
Output: 1
Explanation: If we seed the machine with nums[3], then we destroy all targets equal to 1,3,5,7,9,… 
In this case, we would destroy 5 total targets (all except for nums[2]). 
It is impossible to destroy more than 5 targets, so we return nums[3].

Example 2:

Input: nums = [1,3,5,2,4,6], space = 2
Output: 1
Explanation: Seeding the machine with nums[0], or nums[3] destroys 3 targets. 
It is not possible to destroy more than 3 targets.
Since nums[0] is the minimal integer that can destroy 3 targets, we return 1.

Example 3:

Input: nums = [6,2,5], space = 100
Output: 2
Explanation: Whatever initial seed we select, we can only destroy 1 target. The minimal seed is nums[1].

 

Constraints:

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

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:
    int destroyTargets(vector<int> &nums, int space) {
        sort(all(nums));
        reverse(all(nums));

        auto mp = map<int, int>();

        int mx = 0;
        int res = -1;

        for (const auto &a : nums) {
            int b = a % space;
            ++mp[b];
            if (mp[b] >= mx) {
                mx = mp[b];
                res = a;
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

Metadata

给你一个下标从 0 开始的非负整数数组 nums 。对于 nums 中每一个整数,你必须找到对应元素的 第二大 整数。

如果 nums[j] 满足以下条件,那么我们称它为 nums[i] 的 第二大 整数:

  • j > i
  • nums[j] > nums[i]
  • 恰好存在 一个 k 满足 i < k < j 且 nums[k] > nums[i] 。

如果不存在 nums[j] ,那么第二大整数为 -1 。

  • 比方说,数组 [1, 2, 4, 3] 中,1 的第二大整数是 4 ,2 的第二大整数是 3 ,3 和 4 的第二大整数是 -1 。

请你返回一个整数数组 answer ,其中 answer[i] nums[i] 的第二大整数。

 

示例 1:

输入:nums = [2,4,0,9,6]
输出:[9,6,6,-1,-1]
解释:
下标为 0 处:2 的右边,4 是大于 2 的第一个整数,9 是第二个大于 2 的整数。
下标为 1 处:4 的右边,9 是大于 4 的第一个整数,6 是第二个大于 4 的整数。
下标为 2 处:0 的右边,9 是大于 0 的第一个整数,6 是第二个大于 0 的整数。
下标为 3 处:右边不存在大于 9 的整数,所以第二大整数为 -1 。
下标为 4 处:右边不存在大于 6 的整数,所以第二大整数为 -1 。
所以我们返回 [9,6,6,-1,-1] 。

示例 2:

输入:nums = [3,3]
输出:[-1,-1]
解释:
由于每个数右边都没有更大的数,所以我们返回 [-1,-1] 。

 

提示:

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

Metadata

You are given a 0-indexed array of non-negative integers nums. For each integer in nums, you must find its respective second greater integer.

The second greater integer of nums[i] is nums[j] such that:

  • j > i
  • nums[j] > nums[i]
  • There exists exactly one index k such that nums[k] > nums[i] and i < k < j.

If there is no such nums[j], the second greater integer is considered to be -1.

  • For example, in the array [1, 2, 4, 3], the second greater integer of 1 is 4, 2 is 3, and that of 3 and 4 is -1.

Return an integer array answer, where answer[i] is the second greater integer of nums[i].

 

Example 1:

Input: nums = [2,4,0,9,6]
Output: [9,6,6,-1,-1]
Explanation:
0th index: 4 is the first integer greater than 2, and 9 is the second integer greater than 2, to the right of 2.
1st index: 9 is the first, and 6 is the second integer greater than 4, to the right of 4.
2nd index: 9 is the first, and 6 is the second integer greater than 0, to the right of 0.
3rd index: There is no integer greater than 9 to its right, so the second greater integer is considered to be -1.
4th index: There is no integer greater than 6 to its right, so the second greater integer is considered to be -1.
Thus, we return [9,6,6,-1,-1].

Example 2:

Input: nums = [3,3]
Output: [-1,-1]
Explanation:
We return [-1,-1] since neither integer has any integer greater than it.

 

Constraints:

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

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:
    vector<int> secondGreaterElement(vector<int> &nums) {
        int n = int(nums.size());

        auto se = set<int>();
        auto vec = vector<pair<int, int>>();
        for (int i = 0; i < n; i++) {
            vec.push_back(make_pair(nums[i], i));
        }

        sort(all(vec));
        reverse(all(vec));

        auto res = vector<int>(n, -1);

        for (int i = 0, j = 0; i < n; i++) {
            for (; j < i && vec[j].first > vec[i].first; j++) {
                se.insert(vec[j].second);
            }

            const auto &p = vec[i];
            auto pos = se.lower_bound(p.second);
            if (pos != se.end()) {
                ++pos;
                if (pos != se.end()) {
                    res[p.second] = nums[*pos];
                }
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: January 15, 2023
回到页面顶部