跳转至

biweekly-contest-84

A

Statement

Metadata

给你两个二维整数数组 items1 和 items2 ,表示两个物品集合。每个数组 items 有以下特质:

  • items[i] = [valuei, weighti] 其中 valuei 表示第 i 件物品的 价值 ,weighti 表示第 i 件物品的 重量 。
  • items 中每件物品的价值都是 唯一的 。

请你返回一个二维数组 ret,其中 ret[i] = [valuei, weighti], weighti 是所有价值为 valuei 物品的 重量之和 。

注意:ret 应该按价值 升序 排序后返回。

 

示例 1:

输入:items1 = [[1,1],[4,5],[3,8]], items2 = [[3,1],[1,5]]
输出:[[1,6],[3,9],[4,5]]
解释:
value = 1 的物品在 items1 中 weight = 1 ,在 items2 中 weight = 5 ,总重量为 1 + 5 = 6 。
value = 3 的物品再 items1 中 weight = 8 ,在 items2 中 weight = 1 ,总重量为 8 + 1 = 9 。
value = 4 的物品在 items1 中 weight = 5 ,总重量为 5 。
所以,我们返回 [[1,6],[3,9],[4,5]] 。

示例 2:

输入:items1 = [[1,1],[3,2],[2,3]], items2 = [[2,1],[3,2],[1,3]]
输出:[[1,4],[2,4],[3,4]]
解释:
value = 1 的物品在 items1 中 weight = 1 ,在 items2 中 weight = 3 ,总重量为 1 + 3 = 4 。
value = 2 的物品在 items1 中 weight = 3 ,在 items2 中 weight = 1 ,总重量为 3 + 1 = 4 。
value = 3 的物品在 items1 中 weight = 2 ,在 items2 中 weight = 2 ,总重量为 2 + 2 = 4 。
所以,我们返回 [[1,4],[2,4],[3,4]] 。

示例 3:

输入:items1 = [[1,3],[2,2]], items2 = [[7,1],[2,2],[1,4]]
输出:[[1,7],[2,4],[7,1]]
解释:
value = 1 的物品在 items1 中 weight = 3 ,在 items2 中 weight = 4 ,总重量为 3 + 4 = 7 。
value = 2 的物品在 items1 中 weight = 2 ,在 items2 中 weight = 2 ,总重量为 2 + 2 = 4 。
value = 7 的物品在 items2 中 weight = 1 ,总重量为 1 。
所以,我们返回 [[1,7],[2,4],[7,1]] 。

 

提示:

  • 1 <= items1.length, items2.length <= 1000
  • items1[i].length == items2[i].length == 2
  • 1 <= valuei, weighti <= 1000
  • items1 中每个 valuei 都是 唯一的 。
  • items2 中每个 valuei 都是 唯一的 。

Metadata

You are given two 2D integer arrays, items1 and items2, representing two sets of items. Each array items has the following properties:

  • items[i] = [valuei, weighti] where valuei represents the value and weighti represents the weight of the ith item.
  • The value of each item in items is unique.

Return a 2D integer array ret where ret[i] = [valuei, weighti], with weighti being the sum of weights of all items with value valuei.

Note: ret should be returned in ascending order by value.

 

Example 1:

Input: items1 = [[1,1],[4,5],[3,8]], items2 = [[3,1],[1,5]]
Output: [[1,6],[3,9],[4,5]]
Explanation: 
The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 5, total weight = 1 + 5 = 6.
The item with value = 3 occurs in items1 with weight = 8 and in items2 with weight = 1, total weight = 8 + 1 = 9.
The item with value = 4 occurs in items1 with weight = 5, total weight = 5.
Therefore, we return [[1,6],[3,9],[4,5]].

Example 2:

Input: items1 = [[1,1],[3,2],[2,3]], items2 = [[2,1],[3,2],[1,3]]
Output: [[1,4],[2,4],[3,4]]
Explanation: 
The item with value = 1 occurs in items1 with weight = 1 and in items2 with weight = 3, total weight = 1 + 3 = 4.
The item with value = 2 occurs in items1 with weight = 3 and in items2 with weight = 1, total weight = 3 + 1 = 4.
The item with value = 3 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4.
Therefore, we return [[1,4],[2,4],[3,4]].

Example 3:

Input: items1 = [[1,3],[2,2]], items2 = [[7,1],[2,2],[1,4]]
Output: [[1,7],[2,4],[7,1]]
Explanation:
The item with value = 1 occurs in items1 with weight = 3 and in items2 with weight = 4, total weight = 3 + 4 = 7. 
The item with value = 2 occurs in items1 with weight = 2 and in items2 with weight = 2, total weight = 2 + 2 = 4. 
The item with value = 7 occurs in items2 with weight = 1, total weight = 1.
Therefore, we return [[1,7],[2,4],[7,1]].

 

Constraints:

  • 1 <= items1.length, items2.length <= 1000
  • items1[i].length == items2[i].length == 2
  • 1 <= valuei, weighti <= 1000
  • Each valuei in items1 is unique.
  • Each valuei in items2 is unique.

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<vector<int>> mergeSimilarItems(vector<vector<int>>& items1, vector<vector<int>>& items2) {
        auto mp = map<int, int>();
        for (const auto& it : items1) {
            mp[it[0]] += it[1];
        }

        for (const auto& it : items2) {
            mp[it[0]] += it[1];
        }

        vector<vector<int>> res;
        res.reserve(mp.size());
        for (const auto& it : mp) {
            res.emplace_back(vector<int>({it.first, it.second}));
        }
        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

Metadata

给你一个下标从 0 开始的整数数组 nums 。如果 i < j 且 j - i != nums[j] - nums[i] ,那么我们称 (i, j) 是一个 数对 。

请你返回 nums 中 坏数对 的总数目。

 

示例 1:

输入:nums = [4,1,3,3]
输出:5
解释:数对 (0, 1) 是坏数对,因为 1 - 0 != 1 - 4 。
数对 (0, 2) 是坏数对,因为 2 - 0 != 3 - 4, 2 != -1 。
数对 (0, 3) 是坏数对,因为 3 - 0 != 3 - 4, 3 != -1 。
数对 (1, 2) 是坏数对,因为 2 - 1 != 3 - 1, 1 != 2 。
数对 (2, 3) 是坏数对,因为 3 - 2 != 3 - 3, 1 != 0 。
总共有 5 个坏数对,所以我们返回 5 。

示例 2:

输入:nums = [1,2,3,4,5]
输出:0
解释:没有坏数对。

 

提示:

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

Metadata

You are given a 0-indexed integer array nums. A pair of indices (i, j) is a bad pair if i < j and j - i != nums[j] - nums[i].

Return the total number of bad pairs in nums.

 

Example 1:

Input: nums = [4,1,3,3]
Output: 5
Explanation: The pair (0, 1) is a bad pair since 1 - 0 != 1 - 4.
The pair (0, 2) is a bad pair since 2 - 0 != 3 - 4, 2 != -1.
The pair (0, 3) is a bad pair since 3 - 0 != 3 - 4, 3 != -1.
The pair (1, 2) is a bad pair since 2 - 1 != 3 - 1, 1 != 2.
The pair (2, 3) is a bad pair since 3 - 2 != 3 - 3, 1 != 0.
There are a total of 5 bad pairs, so we return 5.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 0
Explanation: There are no bad pairs.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= 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:
    long long countBadPairs(vector<int> &nums) {
        int n = int(nums.size());

        ll res = 0;
        auto mp = map<int, int>();
        for (int i = 0; i < n; i++) {
            auto ix = nums[i] - i;
            res += i - mp[ix];
            ++mp[ix];
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

Metadata

给你一个下标从 0 开始的正整数数组 tasks ,表示需要 按顺序 完成的任务,其中 tasks[i] 表示第 i 件任务的 类型 。

同时给你一个正整数 space ,表示一个任务完成  ,另一个 相同 类型任务完成前需要间隔的 最少 天数。

在所有任务完成前的每一天,你都必须进行以下两种操作中的一种:

  • 完成 tasks 中的下一个任务
  • 休息一天

请你返回完成所有任务所需的 最少 天数。

 

示例 1:

输入:tasks = [1,2,1,2,3,1], space = 3
输出:9
解释:
9 天完成所有任务的一种方法是:
第 1 天:完成任务 0 。
第 2 天:完成任务 1 。
第 3 天:休息。
第 4 天:休息。
第 5 天:完成任务 2 。
第 6 天:完成任务 3 。
第 7 天:休息。
第 8 天:完成任务 4 。
第 9 天:完成任务 5 。
可以证明无法少于 9 天完成所有任务。

示例 2:

输入:tasks = [5,8,8,5], space = 2
输出:6
解释:
6 天完成所有任务的一种方法是:
第 1 天:完成任务 0 。
第 2 天:完成任务 1 。
第 3 天:休息。
第 4 天:休息。
第 5 天:完成任务 2 。
第 6 天:完成任务 3 。
可以证明无法少于 6 天完成所有任务。

 

提示:

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

Metadata

You are given a 0-indexed array of positive integers tasks, representing tasks that need to be completed in order, where tasks[i] represents the type of the ith task.

You are also given a positive integer space, which represents the minimum number of days that must pass after the completion of a task before another task of the same type can be performed.

Each day, until all tasks have been completed, you must either:

  • Complete the next task from tasks, or
  • Take a break.

Return the minimum number of days needed to complete all tasks.

 

Example 1:

Input: tasks = [1,2,1,2,3,1], space = 3
Output: 9
Explanation:
One way to complete all tasks in 9 days is as follows:
Day 1: Complete the 0th task.
Day 2: Complete the 1st task.
Day 3: Take a break.
Day 4: Take a break.
Day 5: Complete the 2nd task.
Day 6: Complete the 3rd task.
Day 7: Take a break.
Day 8: Complete the 4th task.
Day 9: Complete the 5th task.
It can be shown that the tasks cannot be completed in less than 9 days.

Example 2:

Input: tasks = [5,8,8,5], space = 2
Output: 6
Explanation:
One way to complete all tasks in 6 days is as follows:
Day 1: Complete the 0th task.
Day 2: Complete the 1st task.
Day 3: Take a break.
Day 4: Take a break.
Day 5: Complete the 2nd task.
Day 6: Complete the 3rd task.
It can be shown that the tasks cannot be completed in less than 6 days.

 

Constraints:

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

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 taskSchedulerII(vector<int> &tasks, int space) {
        auto mp = unordered_map<int, ll>();

        ll res = 0;
        for (const auto &t : tasks) {
            if (mp.count(t) == 0) {
                ++res;
            } else {
                res = max(res, mp[t] + space);
                ++res;
            }
            mp[t] = res;
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

Metadata

给你一个下表从 0 开始的整数数组 nums 。每次操作中,你可以将数组中任何一个元素替换为 任意两个 和为该元素的数字。

  • 比方说,nums = [5,6,7] 。一次操作中,我们可以将 nums[1] 替换成 2 和 4 ,将 nums 转变成 [5,2,4,7] 。

请你执行上述操作,将数组变成元素按 非递减 顺序排列的数组,并返回所需的最少操作次数。

 

示例 1:

输入:nums = [3,9,3]
输出:2
解释:以下是将数组变成非递减顺序的步骤:
- [3,9,3] ,将9 变成 3 和 6 ,得到数组 [3,3,6,3] 
- [3,3,6,3] ,将 6 变成 3 和 3 ,得到数组 [3,3,3,3,3] 
总共需要 2 步将数组变成非递减有序,所以我们返回 2 。

示例 2:

输入:nums = [1,2,3,4,5]
输出:0
解释:数组已经是非递减顺序,所以我们返回 0 。

 

提示:

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

Metadata

You are given a 0-indexed integer array nums. In one operation you can replace any element of the array with any two elements that sum to it.

  • For example, consider nums = [5,6,7]. In one operation, we can replace nums[1] with 2 and 4 and convert nums to [5,2,4,7].

Return the minimum number of operations to make an array that is sorted in non-decreasing order.

 

Example 1:

Input: nums = [3,9,3]
Output: 2
Explanation: Here are the steps to sort the array in non-decreasing order:
- From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3]
- From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3]
There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.

Example 2:

Input: nums = [1,2,3,4,5]
Output: 0
Explanation: The array is already in non-decreasing order. Therefore, we return 0. 

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= 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:
    long long minimumReplacement(vector<int> &nums) {
        int n = int(nums.size());

        int r = nums[n - 1];
        ll res = 0;

        for (int i = n - 2; i >= 0; i--) {
            if (nums[i] <= r) {
                r = nums[i];
            } else {
                int x = (nums[i] + r - 1) / r;
                int y = nums[i] / x;
                res += (x - 1);
                r = y;
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: January 15, 2023
回到页面顶部