biweekly-contest-72

A

Statement

简体中文English

Metadata

给你一个下标从 0 开始长度为 n 的整数数组 nums 和一个整数 k ，请你返回满足 0 <= i < j < n ，nums[i] == nums[j] 且 (i * j) 能被 k 整除的数对 (i, j) 的数目。

示例 1：

输入：nums = [3,1,2,2,2,1,3], k = 2
输出：4
解释：
总共有 4 对数符合所有要求：
- nums[0] == nums[6] 且 0 * 6 == 0 ，能被 2 整除。
- nums[2] == nums[3] 且 2 * 3 == 6 ，能被 2 整除。
- nums[2] == nums[4] 且 2 * 4 == 8 ，能被 2 整除。
- nums[3] == nums[4] 且 3 * 4 == 12 ，能被 2 整除。

示例 2：

输入：nums = [1,2,3,4], k = 1
输出：0
解释：由于数组中没有重复数值，所以没有数对 (i,j) 符合所有要求。

提示：

1 <= nums.length <= 100
1 <= nums[i], k <= 100

Metadata

Link: Count Equal and Divisible Pairs in an Array
Difficulty: Easy
Tag:

Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i * j) is divisible by k.

Example 1:

Input: nums = [3,1,2,2,2,1,3], k = 2
Output: 4
Explanation:
There are 4 pairs that meet all the requirements:
- nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2.
- nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2.
- nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2.
- nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.

Example 2:

Input: nums = [1,2,3,4], k = 1
Output: 0
Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.

Constraints:

1 <= nums.length <= 100
1 <= nums[i], k <= 100

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 lowbit(x) ((x) & (-(x)))
#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 countPairs(vector<int> &nums, int k) {
        int res = 0;
        int n = nums.size();
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (nums[i] == nums[j] && (i * j) % k == 0) {
                    ++res;
                }
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

简体中文English

Metadata

Link: 找到和为给定整数的三个连续整数
Difficulty: Medium
Tag:

给你一个整数 num ，请你返回三个连续的整数，它们的和为 num 。如果 num 无法被表示成三个连续整数的和，请你返回一个空数组。

示例 1：

输入：num = 33
输出：[10,11,12]
解释：33 可以表示为 10 + 11 + 12 = 33 。
10, 11, 12 是 3 个连续整数，所以返回 [10, 11, 12] 。

示例 2：

输入：num = 4
输出：[]
解释：没有办法将 4 表示成 3 个连续整数的和。

提示：

0 <= num <= 10¹⁵

Metadata

Link: Find Three Consecutive Integers That Sum to a Given Number
Difficulty: Medium
Tag:

Given an integer num, return three consecutive integers (as a sorted array) that sum to num. If num cannot be expressed as the sum of three consecutive integers, return an empty array.

Example 1:

Input: num = 33
Output: [10,11,12]
Explanation: 33 can be expressed as 10 + 11 + 12 = 33.
10, 11, 12 are 3 consecutive integers, so we return [10, 11, 12].

Example 2:

Input: num = 4
Output: []
Explanation: There is no way to express 4 as the sum of 3 consecutive integers.

Constraints:

0 <= num <= 10¹⁵

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 lowbit(x) ((x) & (-(x)))
#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:
    vector<long long> sumOfThree(long long num) {
        if (num % 3) {
            return {};
        }

        return {num / 3 - 1, num / 3, num / 3 + 1};
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

简体中文English

Metadata

Link: 拆分成最多数目的偶整数之和
Difficulty: Medium
Tag:

给你一个整数 finalSum 。请你将它拆分成若干个 互不相同 的偶整数之和，且拆分出来的偶整数数目最多。

比方说，给你 finalSum = 12 ，那么这些拆分是 符合要求 的（互不相同的偶整数且和为 finalSum）：(2 + 10) ，(2 + 4 + 6) 和 (4 + 8) 。它们中，(2 + 4 + 6) 包含最多数目的整数。注意 finalSum 不能拆分成 (2 + 2 + 4 + 4) ，因为拆分出来的整数必须互不相同。

请你返回一个整数数组，表示将整数拆分成最多数目的偶整数数组。如果没有办法将 finalSum 进行拆分，请你返回一个空数组。你可以按任意顺序返回这些整数。

示例 1：

输入：finalSum = 12
输出：[2,4,6]
解释：以下是一些符合要求的拆分：(2 + 10)，(2 + 4 + 6) 和 (4 + 8) 。
(2 + 4 + 6) 为最多数目的整数，数目为 3 ，所以我们返回 [2,4,6] 。
[2,6,4] ，[6,2,4] 等等也都是可行的解。

示例 2：

输入：finalSum = 7
输出：[]
解释：没有办法将 finalSum 进行拆分。
所以返回空数组。

示例 3：

输入：finalSum = 28
输出：[6,8,2,12]
解释：以下是一些符合要求的拆分：(2 + 26)，(6 + 8 + 2 + 12) 和 (4 + 24) 。
(6 + 8 + 2 + 12) 有最多数目的整数，数目为 4 ，所以我们返回 [6,8,2,12] 。
[10,2,4,12] ，[6,2,4,16] 等等也都是可行的解。

提示：

1 <= finalSum <= 10¹⁰

Metadata

Link: Maximum Split of Positive Even Integers
Difficulty: Medium
Tag:

You are given an integer finalSum. Split it into a sum of a maximum number of unique positive even integers.

For example, given finalSum = 12, the following splits are valid (unique positive even integers summing up to finalSum): (2 + 10), (2 + 4 + 6), and (4 + 8). Among them, (2 + 4 + 6) contains the maximum number of integers. Note that finalSum cannot be split into (2 + 2 + 4 + 4) as all the numbers should be unique.

Return a list of integers that represent a valid split containing a maximum number of integers. If no valid split exists for finalSum, return an empty list. You may return the integers in any order.

Example 1:

Input: finalSum = 12
Output: [2,4,6]
Explanation: The following are some valid splits: (2 + 10), (2 + 4 + 6), and (4 + 8).
(2 + 4 + 6) has the maximum number of integers, which is 3. Thus, we return [2,4,6].
Note that [2,6,4], [6,2,4], etc. are also accepted.

Example 2:

Input: finalSum = 7
Output: []
Explanation: There are no valid splits for the given finalSum.
Thus, we return an empty array.

Example 3:

Input: finalSum = 28
Output: [6,8,2,12]
Explanation: The following are some valid splits: (2 + 26), (6 + 8 + 2 + 12), and (4 + 24). 
(6 + 8 + 2 + 12) has the maximum number of integers, which is 4. Thus, we return [6,8,2,12].
Note that [10,2,4,12], [6,2,4,16], etc. are also accepted.

Constraints:

1 <= finalSum <= 10¹⁰

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 lowbit(x) ((x) & (-(x)))
#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:
    vector<long long> maximumEvenSplit(long long finalSum) {
        if (finalSum & 1) {
            return {};
        }

        vector<long long> res;
        for (long long i = 2;; i += 2) {
            if (finalSum >= i) {
                finalSum -= i;
                res.push_back(i);
            }

            if (finalSum == 0) {
                break;
            }

            if (res.back() >= finalSum) {
                break;
            }
        }

        if (!res.empty() && res.back() >= finalSum) {
            res.back() += finalSum;
            finalSum = 0;
        }

        if (finalSum) {
            res.push_back(finalSum);
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

简体中文English

Metadata

Link: 统计数组中好三元组数目
Difficulty: Hard
Tag:

给你两个下标从 0 开始且长度为 n 的整数数组 nums1 和 nums2 ，两者都是 [0, 1, …, n - 1] 的排列。

好三元组 指的是 3 个 互不相同 的值，且它们在数组 nums1 和 nums2 中出现顺序保持一致。换句话说，如果我们将 pos1_v 记为值 v 在 nums1 中出现的位置，pos2_v 为值 v 在 nums2 中的位置，那么一个好三元组定义为 0 <= x, y, z <= n - 1 ，且 pos1_x < pos1_y < pos1_z 和 pos2_x < pos2_y < pos2_z 都成立的 (x, y, z) 。

请你返回好三元组的 总数目 。

示例 1：

输入：nums1 = [2,0,1,3], nums2 = [0,1,2,3]
输出：1
解释：
总共有 4 个三元组 (x,y,z) 满足 pos1_x < pos1_y < pos1_z，分别是 (2,0,1) ，(2,0,3) ，(2,1,3) 和 (0,1,3) 。
这些三元组中，只有 (0,1,3) 满足 pos2_x < pos2_y < pos2_z 。所以只有 1 个好三元组。

示例 2：

输入：nums1 = [4,0,1,3,2], nums2 = [4,1,0,2,3]
输出：4
解释：总共有 4 个好三元组 (4,0,3) ，(4,0,2) ，(4,1,3) 和 (4,1,2) 。

提示：

n == nums1.length == nums2.length
3 <= n <= 10⁵
0 <= nums1[i], nums2[i] <= n - 1
nums1 和 nums2 是 [0, 1, …, n - 1] 的排列。

Metadata

Link: Count Good Triplets in an Array
Difficulty: Hard
Tag:

You are given two 0-indexed arrays nums1 and nums2 of length n, both of which are permutations of [0, 1, …, n - 1].

A good triplet is a set of 3 distinct values which are present in increasing order by position both in nums1 and nums2. In other words, if we consider pos1_v as the index of the value v in nums1 and pos2_v as the index of the value v in nums2, then a good triplet will be a set (x, y, z) where 0 <= x, y, z <= n - 1, such that pos1_x < pos1_y < pos1_z and pos2_x < pos2_y < pos2_z.

Return the total number of good triplets.

Example 1:

Input: nums1 = [2,0,1,3], nums2 = [0,1,2,3]
Output: 1
Explanation: 
There are 4 triplets (x,y,z) such that pos1_x < pos1_y < pos1_z. They are (2,0,1), (2,0,3), (2,1,3), and (0,1,3). 
Out of those triplets, only the triplet (0,1,3) satisfies pos2_x < pos2_y < pos2_z. Hence, there is only 1 good triplet.

Example 2:

Input: nums1 = [4,0,1,3,2], nums2 = [4,1,0,2,3]
Output: 4
Explanation: The 4 good triplets are (4,0,3), (4,0,2), (4,1,3), and (4,1,2).

Constraints:

n == nums1.length == nums2.length
3 <= n <= 10⁵
0 <= nums1[i], nums2[i] <= n - 1
nums1 and nums2 are permutations of [0, 1, …, n - 1].

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>;
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

struct BIT {
    vector<ll> a;
    int n;
    void init(int n) {
        this->n = n + 10;
        a = vector<ll>(this->n, 0);
    }
    int lowbit(int x) {
        return x & -x;
    }
    void add(int x, ll v) {
        for (int i = x; i < n; i += lowbit(i)) a[i] += v;
    }
    ll query(int x) {
        ll ret = 0;
        for (int i = x; i > 0; i -= lowbit(i)) ret += a[i];
        return ret;
    }
    ll query(int l, int r) {
        if (l > r)
            return 0;
        return query(r) - query(l - 1);
    }
    void add(int l, int r, ll v) {
        if (l > r)
            return;
        add(l, v);
        add(r + 1, -v);
    }
} bf, bg;

class Solution {
public:
    long long goodTriplets(vector<int> &nums1, vector<int> &nums2) {
        int n = nums1.size();

        vector<int> pos(n + 1, 0);
        for (int i = 0; i < n; i++) {
            pos[nums2[i]] = i;
        }

        bf.init(n);
        bg.init(n);

        vector<ll> f(n + 1, 0), g(n + 1, 0);

        {
            for (int i = n - 1; i >= 0; i--) {
                int x = nums1[i], y = pos[x];
                f[x] = bf.query(y + 1);
                // dbg(i, f[x]);
                bf.add(1, y, 1);
            }
        }

        {
            for (int i = n - 1; i >= 0; i--) {
                int x = nums1[i], y = pos[x];
                g[x] = bg.query(y + 1);
                // dbg(i, g[x]);
                bg.add(1, y, f[x]);
            }
        }

        long long res = 0;
        for (int i = 0; i < n; i++) {
            res += g[i];
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    {
        auto s = new Solution();
        auto a = vector<int>{2, 0, 1, 3};
        auto b = vector<int>{0, 1, 2, 3};
        auto ans = s->goodTriplets(a, b);
        dbg(ans);
    }

    return 0;
}

#endif

最后更新: October 11, 2023