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biweekly-contest-74

A

Statement

Metadata

给你一个整数数组 nums ,它包含 2 * n 个整数。

你需要将 nums 划分成 n 个数对,满足:

  • 每个元素 只属于一个 数对。
  • 同一数对中的元素 相等 。

如果可以将 nums 划分成 n 个数对,请你返回 true ,否则返回 false 。

 

示例 1:

输入:nums = [3,2,3,2,2,2]
输出:true
解释:
nums 中总共有 6 个元素,所以它们应该被划分成 6 / 2 = 3 个数对。
nums 可以划分成 (2, 2) ,(3, 3) 和 (2, 2) ,满足所有要求。

示例 2:

输入:nums = [1,2,3,4]
输出:false
解释:
无法将 nums 划分成 4 / 2 = 2 个数对且满足所有要求。

 

提示:

  • nums.length == 2 * n
  • 1 <= n <= 500
  • 1 <= nums[i] <= 500

Metadata

You are given an integer array nums consisting of 2 * n integers.

You need to divide nums into n pairs such that:

  • Each element belongs to exactly one pair.
  • The elements present in a pair are equal.

Return true if nums can be divided into n pairs, otherwise return false.

 

Example 1:

Input: nums = [3,2,3,2,2,2]
Output: true
Explanation: 
There are 6 elements in nums, so they should be divided into 6 / 2 = 3 pairs.
If nums is divided into the pairs (2, 2), (3, 3), and (2, 2), it will satisfy all the conditions.

Example 2:

Input: nums = [1,2,3,4]
Output: false
Explanation: 
There is no way to divide nums into 4 / 2 = 2 pairs such that the pairs satisfy every condition.

 

Constraints:

  • nums.length == 2 * n
  • 1 <= n <= 500
  • 1 <= nums[i] <= 500

Solution

from typing import Counter, List


class Solution:
    def divideArray(self, nums: List[int]) -> bool:
        a = Counter(nums)
        return all((v & 1 == 0) for v in a.values()) and (len(nums) // 2) >= len(a.keys())

B

Statement

Metadata

给你一个下标从 0 开始的字符串 text 和另一个下标从 0 开始且长度为 2 的字符串 pattern ,两者都只包含小写英文字母。

你可以在 text 中任意位置插入 一个 字符,这个插入的字符必须是 pattern[0] 或者 pattern[1] 。注意,这个字符可以插入在 text 开头或者结尾的位置。

请你返回插入一个字符后,text 中最多包含多少个等于 pattern 的 子序列 。

子序列 指的是将一个字符串删除若干个字符后(也可以不删除),剩余字符保持原本顺序得到的字符串。

 

示例 1:

输入:text = "abdcdbc", pattern = "ac"
输出:4
解释:
如果我们在 text[1] 和 text[2] 之间添加 pattern[0] = 'a' ,那么我们得到 "abadcdbc" 。那么 "ac" 作为子序列出现 4 次。
其他得到 4 个 "ac" 子序列的方案还有 "aabdcdbc" 和 "abdacdbc" 。
但是,"abdcadbc" ,"abdccdbc" 和 "abdcdbcc" 这些字符串虽然是可行的插入方案,但是只出现了 3 次 "ac" 子序列,所以不是最优解。
可以证明插入一个字符后,无法得到超过 4 个 "ac" 子序列。

示例 2:

输入:text = "aabb", pattern = "ab"
输出:6
解释:
可以得到 6 个 "ab" 子序列的部分方案为 "aaabb" ,"aaabb" 和 "aabbb" 。

 

提示:

  • 1 <= text.length <= 105
  • pattern.length == 2
  • text 和 pattern 都只包含小写英文字母。

Metadata

You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both of which consist of only lowercase English letters.

You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the character can be added even at the beginning or at the end of text.

Return the maximum number of times pattern can occur as a subsequence of the modified text.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

 

Example 1:

Input: text = "abdcdbc", pattern = "ac"
Output: 4
Explanation:
If we add pattern[0] = 'a' in between text[1] and text[2], we get "abadcdbc". Now, the number of times "ac" occurs as a subsequence is 4.
Some other strings which have 4 subsequences "ac" after adding a character to text are "aabdcdbc" and "abdacdbc".
However, strings such as "abdcadbc", "abdccdbc", and "abdcdbcc", although obtainable, have only 3 subsequences "ac" and are thus suboptimal.
It can be shown that it is not possible to get more than 4 subsequences "ac" by adding only one character.

Example 2:

Input: text = "aabb", pattern = "ab"
Output: 6
Explanation:
Some of the strings which can be obtained from text and have 6 subsequences "ab" are "aaabb", "aaabb", and "aabbb".

 

Constraints:

  • 1 <= text.length <= 105
  • pattern.length == 2
  • text and pattern consist only 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>;
const ll mod = 1e9 + 7;

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 maximumSubsequenceCount(string text, string pattern) {
        char a = pattern[0];
        char b = pattern[1];

        long long ca = 0;
        long long cb = 0;
        long long res = 0;
        for (auto &c : text) {
            if (c == b) {
                ++cb;
                res += ca;
            }

            if (c == a) {
                ++ca;
            }
        }

        return res + max(ca, cb);
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

Metadata

给你一个正整数数组 nums 。每一次操作中,你可以从 nums 中选择 任意 一个数并将它减小到 恰好 一半。(注意,在后续操作中你可以对减半过的数继续执行操作)

请你返回将 nums 数组和 至少 减少一半的 最少 操作数。

 

示例 1:

输入:nums = [5,19,8,1]
输出:3
解释:初始 nums 的和为 5 + 19 + 8 + 1 = 33 。
以下是将数组和减少至少一半的一种方法:
选择数字 19 并减小为 9.5 。
选择数字 9.5 并减小为 4.75 。
选择数字 8 并减小为 4 。
最终数组为 [5, 4.75, 4, 1] ,和为 5 + 4.75 + 4 + 1 = 14.75 。
nums 的和减小了 33 - 14.75 = 18.25 ,减小的部分超过了初始数组和的一半,18.25 >= 33/2 = 16.5 。
我们需要 3 个操作实现题目要求,所以返回 3 。
可以证明,无法通过少于 3 个操作使数组和减少至少一半。

示例 2:

输入:nums = [3,8,20]
输出:3
解释:初始 nums 的和为 3 + 8 + 20 = 31 。
以下是将数组和减少至少一半的一种方法:
选择数字 20 并减小为 10 。
选择数字 10 并减小为 5 。
选择数字 3 并减小为 1.5 。
最终数组为 [1.5, 8, 5] ,和为 1.5 + 8 + 5 = 14.5 。
nums 的和减小了 31 - 14.5 = 16.5 ,减小的部分超过了初始数组和的一半, 16.5 >= 31/2 = 16.5 。
我们需要 3 个操作实现题目要求,所以返回 3 。
可以证明,无法通过少于 3 个操作使数组和减少至少一半。

 

提示:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 107

Metadata

You are given an array nums of positive integers. In one operation, you can choose any number from nums and reduce it to exactly half the number. (Note that you may choose this reduced number in future operations.)

Return the minimum number of operations to reduce the sum of nums by at least half.

 

Example 1:

Input: nums = [5,19,8,1]
Output: 3
Explanation: The initial sum of nums is equal to 5 + 19 + 8 + 1 = 33.
The following is one of the ways to reduce the sum by at least half:
Pick the number 19 and reduce it to 9.5.
Pick the number 9.5 and reduce it to 4.75.
Pick the number 8 and reduce it to 4.
The final array is [5, 4.75, 4, 1] with a total sum of 5 + 4.75 + 4 + 1 = 14.75. 
The sum of nums has been reduced by 33 - 14.75 = 18.25, which is at least half of the initial sum, 18.25 >= 33/2 = 16.5.
Overall, 3 operations were used so we return 3.
It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

Example 2:

Input: nums = [3,8,20]
Output: 3
Explanation: The initial sum of nums is equal to 3 + 8 + 20 = 31.
The following is one of the ways to reduce the sum by at least half:
Pick the number 20 and reduce it to 10.
Pick the number 10 and reduce it to 5.
Pick the number 3 and reduce it to 1.5.
The final array is [1.5, 8, 5] with a total sum of 1.5 + 8 + 5 = 14.5. 
The sum of nums has been reduced by 31 - 14.5 = 16.5, which is at least half of the initial sum, 16.5 >= 31/2 = 16.5.
Overall, 3 operations were used so we return 3.
It can be shown that we cannot reduce the sum by at least half in less than 3 operations.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 107

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>;
const ll mod = 1e9 + 7;

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 halveArray(vector<int> &nums) {
        long double res = 0;
        priority_queue<double, vector<double>, less<double>> pq;
        for (auto &a : nums) {
            res += a;
            pq.push(a);
        }

        long double need = res * 1.0 / 2;
        int cnt = 0;
        while (res > need) {
            long double a = pq.top();
            pq.pop();
            a /= 2;
            res -= a;
            pq.push(a);
            cnt += 1;
        }

        return cnt;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

Metadata

给你一个下标从 0 开始的 二进制 字符串 floor ,它表示地板上砖块的颜色。

  • floor[i] = '0' 表示地板上第 i 块砖块的颜色是 黑色 。
  • floor[i] = '1' 表示地板上第 i 块砖块的颜色是 白色 。

同时给你 numCarpets 和 carpetLen 。你有 numCarpets 条 黑色 的地毯,每一条 黑色 的地毯长度都为 carpetLen 块砖块。请你使用这些地毯去覆盖砖块,使得未被覆盖的剩余 白色 砖块的数目 最小 。地毯相互之间可以覆盖。

请你返回没被覆盖的白色砖块的 最少 数目。

 

示例 1:

输入:floor = "10110101", numCarpets = 2, carpetLen = 2
输出:2
解释:
上图展示了剩余 2 块白色砖块的方案。
没有其他方案可以使未被覆盖的白色砖块少于 2 块。

示例 2:

输入:floor = "11111", numCarpets = 2, carpetLen = 3
输出:0
解释:
上图展示了所有白色砖块都被覆盖的一种方案。
注意,地毯相互之间可以覆盖。

 

提示:

  • 1 <= carpetLen <= floor.length <= 1000
  • floor[i] 要么是 '0' ,要么是 '1' 。
  • 1 <= numCarpets <= 1000

Metadata

You are given a 0-indexed binary string floor, which represents the colors of tiles on a floor:

  • floor[i] = '0' denotes that the ith tile of the floor is colored black.
  • On the other hand, floor[i] = '1' denotes that the ith tile of the floor is colored white.

You are also given numCarpets and carpetLen. You have numCarpets black carpets, each of length carpetLen tiles. Cover the tiles with the given carpets such that the number of white tiles still visible is minimum. Carpets may overlap one another.

Return the minimum number of white tiles still visible.

 

Example 1:

Input: floor = "10110101", numCarpets = 2, carpetLen = 2
Output: 2
Explanation: 
The figure above shows one way of covering the tiles with the carpets such that only 2 white tiles are visible.
No other way of covering the tiles with the carpets can leave less than 2 white tiles visible.

Example 2:

Input: floor = "11111", numCarpets = 2, carpetLen = 3
Output: 0
Explanation: 
The figure above shows one way of covering the tiles with the carpets such that no white tiles are visible.
Note that the carpets are able to overlap one another.

 

Constraints:

  • 1 <= carpetLen <= floor.length <= 1000
  • floor[i] is either '0' or '1'.
  • 1 <= numCarpets <= 1000

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>;
const ll mod = 1e9 + 7;

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 minimumWhiteTiles(string floor, int numCarpets, int carpetLen) {
        int n = floor.length();
        int m = max(n, numCarpets) + 5;

        vector<vector<int>> f(n + 5, vector<int>(numCarpets + 5, 0x3f3f3f3f));
        for (int i = 0; i <= numCarpets; i++) {
            f[0][i] = 0;
        }

        for (int i = 1; i <= n; i++) {
            f[i][0] = f[i - 1][0] + (floor[i - 1] == '1');
        }

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= numCarpets; j++) {
                f[i][j] = min(f[i][j], f[max(0, i - carpetLen)][j - 1]);
                f[i][j] = min(f[i][j], f[i - 1][j] + (floor[i - 1] == '1'));
                f[i][j] = min(f[i][j], f[i][j - 1]);
            }
        }

        return f[n][numCarpets];
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: January 15, 2023
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