weekly-contest-316

A

Statement

简体中文English

Metadata

Link: 判断两个事件是否存在冲突
Difficulty: Easy
Tag:

给你两个字符串数组 event1 和 event2 ，表示发生在同一天的两个闭区间时间段事件，其中：

event1 = [startTime₁, endTime₁] 且
event2 = [startTime₂, endTime₂]

事件的时间为有效的 24 小时制且按 HH:MM 格式给出。

当两个事件存在某个非空的交集时（即，某些时刻是两个事件都包含的），则认为出现冲突。

如果两个事件之间存在冲突，返回 true ；否则，返回 false 。

示例 1：

输入：event1 = ["01:15","02:00"], event2 = ["02:00","03:00"]
输出：true
解释：两个事件在 2:00 出现交集。

示例 2：

输入：event1 = ["01:00","02:00"], event2 = ["01:20","03:00"]
输出：true
解释：两个事件的交集从 01:20 开始，到 02:00 结束。

示例 3：

输入：event1 = ["10:00","11:00"], event2 = ["14:00","15:00"]
输出：false
解释：两个事件不存在交集。

提示：

evnet1.length == event2.length == 2.
event1[i].length == event2[i].length == 5
startTime₁ <= endTime₁
startTime₂ <= endTime₂
所有事件的时间都按照 HH:MM 格式给出

Metadata

Link: Determine if Two Events Have Conflict
Difficulty: Easy
Tag:

You are given two arrays of strings that represent two inclusive events that happened on the same day, event1 and event2, where:

event1 = [startTime₁, endTime₁] and
event2 = [startTime₂, endTime₂].

Event times are valid 24 hours format in the form of HH:MM.

A conflict happens when two events have some non-empty intersection (i.e., some moment is common to both events).

Return true if there is a conflict between two events. Otherwise, return false.

Example 1:

Input: event1 = ["01:15","02:00"], event2 = ["02:00","03:00"]
Output: true
Explanation: The two events intersect at time 2:00.

Example 2:

Input: event1 = ["01:00","02:00"], event2 = ["01:20","03:00"]
Output: true
Explanation: The two events intersect starting from 01:20 to 02:00.

Example 3:

Input: event1 = ["10:00","11:00"], event2 = ["14:00","15:00"]
Output: false
Explanation: The two events do not intersect.

Constraints:

evnet1.length == event2.length == 2.
event1[i].length == event2[i].length == 5
startTime₁ <= endTime₁
startTime₂ <= endTime₂
All the event times follow the HH:MM format.

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:
    bool haveConflict(vector<string> &event1, vector<string> &event2) {
        if (event1[1] < event2[0] || event2[1] < event1[0]) {
            return false;
        }

        return true;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

简体中文English

Metadata

Link: 最大公因数等于 K 的子数组数目
Difficulty: Medium
Tag:

给你一个整数数组 nums 和一个整数 k ，请你统计并返回 nums 的子数组中元素的最大公因数等于 k 的子数组数目。

子数组 是数组中一个连续的非空序列。

数组的最大公因数 是能整除数组中所有元素的最大整数。

示例 1：

输入：nums = [9,3,1,2,6,3], k = 3
输出：4
解释：nums 的子数组中，以 3 作为最大公因数的子数组如下：
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]

示例 2：

输入：nums = [4], k = 7
输出：0
解释：不存在以 7 作为最大公因数的子数组。

提示：

1 <= nums.length <= 1000
1 <= nums[i], k <= 10⁹

Metadata

Link: Number of Subarrays With GCD Equal to K
Difficulty: Medium
Tag:

Given an integer array nums and an integer k, return the number of subarrays of nums where the greatest common divisor of the subarray's elements is k.

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

The greatest common divisor of an array is the largest integer that evenly divides all the array elements.

Example 1:

Input: nums = [9,3,1,2,6,3], k = 3
Output: 4
Explanation: The subarrays of nums where 3 is the greatest common divisor of all the subarray's elements are:
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]
- [9,3,1,2,6,3]

Example 2:

Input: nums = [4], k = 7
Output: 0
Explanation: There are no subarrays of nums where 7 is the greatest common divisor of all the subarray's elements.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i], k <= 10⁹

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 subarrayGCD(vector<int> &nums, int k) {
        int n = int(nums.size());

        int res = 0;

        for (int i = 0; i < n; i++) {
            int g = 0;
            for (int j = i; j < n; j++) {
                g = __gcd(g, nums[j]);
                if (g == k) {
                    ++res;
                }

                if (g < k) {
                    break;
                }
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

简体中文English

Metadata

Link: 使数组相等的最小开销
Difficulty: Hard
Tag:

给你两个下标从 0 开始的数组 nums 和 cost ，分别包含 n 个正整数。

你可以执行下面操作任意次：

将 nums 中任意元素增加或者减小 1 。

对第 i 个元素执行一次操作的开销是 cost[i] 。

请你返回使 nums 中所有元素相等的最少总开销。

示例 1：

输入：nums = [1,3,5,2], cost = [2,3,1,14]
输出：8
解释：我们可以执行以下操作使所有元素变为 2 ：
- 增加第 0 个元素 1 次，开销为 2 。
- 减小第 1 个元素 1 次，开销为 3 。
- 减小第 2 个元素 3 次，开销为 1 + 1 + 1 = 3 。
总开销为 2 + 3 + 3 = 8 。
这是最小开销。

示例 2：

输入：nums = [2,2,2,2,2], cost = [4,2,8,1,3]
输出：0
解释：数组中所有元素已经全部相等，不需要执行额外的操作。

提示：

n == nums.length == cost.length
1 <= n <= 10⁵
1 <= nums[i], cost[i] <= 10⁶

Metadata

Link: Minimum Cost to Make Array Equal
Difficulty: Hard
Tag:

You are given two 0-indexed arrays nums and cost consisting each of n positive integers.

You can do the following operation any number of times:

Increase or decrease any element of the array nums by 1.

The cost of doing one operation on the i^th element is cost[i].

Return the minimum total cost such that all the elements of the array nums become equal.

Example 1:

Input: nums = [1,3,5,2], cost = [2,3,1,14]
Output: 8
Explanation: We can make all the elements equal to 2 in the following way:
- Increase the 0^th element one time. The cost is 2.
- Decrease the 1^st element one time. The cost is 3.
- Decrease the 2^nd element three times. The cost is 1 + 1 + 1 = 3.
The total cost is 2 + 3 + 3 = 8.
It can be shown that we cannot make the array equal with a smaller cost.

Example 2:

Input: nums = [2,2,2,2,2], cost = [4,2,8,1,3]
Output: 0
Explanation: All the elements are already equal, so no operations are needed.

Constraints:

n == nums.length == cost.length
1 <= n <= 10⁵
1 <= nums[i], cost[i] <= 10⁶

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:
    long long minCost(vector<int> &nums, vector<int> &cost) {
        int n = int(nums.size());

        int mi = *min_element(all(nums));
        int mx = *max_element(all(nums));

        ll co = 0;
        ll res = 0x3f3f3f3f3f3f3f3f;
        ll left = 0;
        ll right = 0;

        auto f = vector<vector<int>>(mx + 5, vector<int>());

        for (int i = 0; i < n; i++) {
            f[nums[i]].push_back(cost[i]);
            co += 1ll * abs(nums[i] - (mi - 1)) * cost[i];
            right += cost[i];
        }

        for (int i = mi; i <= mx; i++) {
            co -= right;
            co += left;
            for (const auto &a : f[i]) {
                left += a;
                right -= a;
            }

            res = min(res, co);
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

简体中文English

Metadata

Link: 使数组相似的最少操作次数
Difficulty: Hard
Tag:

给你两个正整数数组 nums 和 target ，两个数组长度相等。

在一次操作中，你可以选择两个不同的下标 i 和 j ，其中 0 <= i, j < nums.length ，并且：

令 nums[i] = nums[i] + 2 且
令 nums[j] = nums[j] - 2 。

如果两个数组中每个元素出现的频率相等，我们称两个数组是相似的。

请你返回将 nums 变得与 target 相似的最少操作次数。测试数据保证 nums 一定能变得与 target 相似。

示例 1：

输入：nums = [8,12,6], target = [2,14,10]
输出：2
解释：可以用两步操作将 nums 变得与 target 相似：
- 选择 i = 0 和 j = 2 ，nums = [10,12,4] 。
- 选择 i = 1 和 j = 2 ，nums = [10,14,2] 。
2 次操作是最少需要的操作次数。

示例 2：

输入：nums = [1,2,5], target = [4,1,3]
输出：1
解释：一步操作可以使 nums 变得与 target 相似：
- 选择 i = 1 和 j = 2 ，nums = [1,4,3] 。

示例 3：

输入：nums = [1,1,1,1,1], target = [1,1,1,1,1]
输出：0
解释：数组 nums 已经与 target 相似。

提示：

n == nums.length == target.length
1 <= n <= 10⁵
1 <= nums[i], target[i] <= 10⁶
nums 一定可以变得与 target 相似。

Metadata

Link: Minimum Number of Operations to Make Arrays Similar
Difficulty: Hard
Tag:

You are given two positive integer arrays nums and target, of the same length.

In one operation, you can choose any two distinct indices i and j where 0 <= i, j < nums.length and:

set nums[i] = nums[i] + 2 and
set nums[j] = nums[j] - 2.

Two arrays are considered to be similar if the frequency of each element is the same.

Return the minimum number of operations required to make nums similar to target. The test cases are generated such that nums can always be similar to target.

Example 1:

Input: nums = [8,12,6], target = [2,14,10]
Output: 2
Explanation: It is possible to make nums similar to target in two operations:
- Choose i = 0 and j = 2, nums = [10,12,4].
- Choose i = 1 and j = 2, nums = [10,14,2].
It can be shown that 2 is the minimum number of operations needed.

Example 2:

Input: nums = [1,2,5], target = [4,1,3]
Output: 1
Explanation: We can make nums similar to target in one operation:
- Choose i = 1 and j = 2, nums = [1,4,3].

Example 3:

Input: nums = [1,1,1,1,1], target = [1,1,1,1,1]
Output: 0
Explanation: The array nums is already similiar to target.

Constraints:

n == nums.length == target.length
1 <= n <= 10⁵
1 <= nums[i], target[i] <= 10⁶
It is possible to make nums similar to target.

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

// 6 8 12
// 2 10 14
// 4 -2 -2
// 1 2 5
// 1 3 4
// 1 5 2
// 1 3 4
// 0 -2 2

class Solution {
public:
    long long makeSimilar(vector<int> &nums, vector<int> &target) {
        int n = int(nums.size());

        auto a = vector<vector<int>>(2, vector<int>());
        auto b = vector<vector<int>>(2, vector<int>());

        for (int i = 0; i < n; i++) {
            a[nums[i] % 2].push_back(nums[i]);
            b[target[i] % 2].push_back(target[i]);
        }

        for (int i = 0; i < 2; i++) {
            sort(all(a[i]));
            sort(all(b[i]));
        }

        ll res = 0;
        for (int i = 0; i < 2; i++) {
            int m = int(a[i].size());
            for (int j = 0; j < m; j++) {
                res += abs(a[i][j] - b[i][j]) / 2;
            }
        }

        res /= 2;

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: October 11, 2023