weekly-contest-315

A

Statement

简体中文English

Metadata

Link: 与对应负数同时存在的最大正整数
Difficulty: Easy
Tag:

给你一个 不包含 任何零的整数数组 nums ，找出自身与对应的负数都在数组中存在的最大正整数 k 。

返回正整数 k ，如果不存在这样的整数，返回 -1 。

示例 1：

输入：nums = [-1,2,-3,3]
输出：3
解释：3 是数组中唯一一个满足题目要求的 k 。

示例 2：

输入：nums = [-1,10,6,7,-7,1]
输出：7
解释：数组中存在 1 和 7 对应的负数，7 的值更大。

示例 3：

输入：nums = [-10,8,6,7,-2,-3]
输出：-1
解释：不存在满足题目要求的 k ，返回 -1 。

提示：

1 <= nums.length <= 1000
-1000 <= nums[i] <= 1000
nums[i] != 0

Metadata

Link: Largest Positive Integer That Exists With Its Negative
Difficulty: Easy
Tag:

Given an integer array nums that does not contain any zeros, find the largest positive integer k such that -k also exists in the array.

Return the positive integer k. If there is no such integer, return -1.

Example 1:

Input: nums = [-1,2,-3,3]
Output: 3
Explanation: 3 is the only valid k we can find in the array.

Example 2:

Input: nums = [-1,10,6,7,-7,1]
Output: 7
Explanation: Both 1 and 7 have their corresponding negative values in the array. 7 has a larger value.

Example 3:

Input: nums = [-10,8,6,7,-2,-3]
Output: -1
Explanation: There is no a single valid k, we return -1.

Constraints:

1 <= nums.length <= 1000
-1000 <= nums[i] <= 1000
nums[i] != 0

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 findMaxK(vector<int> &nums) {
        auto mp = map<int, bool>();
        for (const auto &a : nums) {
            mp[a] = true;
        }

        int res = -1;
        for (const auto &a : nums) {
            if (a > 0 && mp.count(-a)) {
                res = max(res, a);
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

B

Statement

简体中文English

Metadata

Link: 反转之后不同整数的数目
Difficulty: Medium
Tag:

给你一个由正整数组成的数组 nums 。

你必须取出数组中的每个整数，反转其中每个数位，并将反转后得到的数字添加到数组的末尾。这一操作只针对 nums 中原有的整数执行。

返回结果数组中不同整数的数目。

示例 1：

输入：nums = [1,13,10,12,31]
输出：6
解释：反转每个数字后，结果数组是 [1,13,10,12,31,1,31,1,21,13] 。
反转后得到的数字添加到数组的末尾并按斜体加粗表示。注意对于整数 10 ，反转之后会变成 01 ，即 1 。
数组中不同整数的数目为 6（数字 1、10、12、13、21 和 31）。

示例 2：

输入：nums = [2,2,2]
输出：1
解释：反转每个数字后，结果数组是 [2,2,2,2,2,2] 。
数组中不同整数的数目为 1（数字 2）。

提示：

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Metadata

Link: Count Number of Distinct Integers After Reverse Operations
Difficulty: Medium
Tag:

You are given an array nums consisting of positive integers.

You have to take each integer in the array, reverse its digits, and add it to the end of the array. You should apply this operation to the original integers in nums.

Return the number of distinct integers in the final array.

Example 1:

Input: nums = [1,13,10,12,31]
Output: 6
Explanation: After including the reverse of each number, the resulting array is [1,13,10,12,31,1,31,1,21,13].
The reversed integers that were added to the end of the array are underlined. Note that for the integer 10, after reversing it, it becomes 01 which is just 1.
The number of distinct integers in this array is 6 (The numbers 1, 10, 12, 13, 21, and 31).

Example 2:

Input: nums = [2,2,2]
Output: 1
Explanation: After including the reverse of each number, the resulting array is [2,2,2,2,2,2].
The number of distinct integers in this array is 1 (The number 2).

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[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:
    int f(int x) {
        int res = 0;
        while (x) {
            res = res * 10 + (x % 10);
            x /= 10;
        }

        return res;
    }

    int countDistinctIntegers(vector<int> &nums) {
        auto se = set<int>();
        for (const auto &a : nums) {
            se.insert(a);
            se.insert(f(a));
        }

        return int(se.size());
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

C

Statement

简体中文English

Metadata

Link: 反转之后的数字和
Difficulty: Medium
Tag:

给你一个非负整数 num 。如果存在某个非负整数 k 满足 k + reverse(k) = num ，则返回 true ；否则，返回 false 。

reverse(k) 表示 k 反转每个数位后得到的数字。

示例 1：

输入：num = 443
输出：true
解释：172 + 271 = 443 ，所以返回 true 。

示例 2：

输入：num = 63
输出：false
解释：63 不能表示为非负整数及其反转后数字之和，返回 false 。

示例 3：

输入：num = 181
输出：true
解释：140 + 041 = 181 ，所以返回 true 。注意，反转后的数字可能包含前导零。

提示：

0 <= num <= 10⁵

Metadata

Link: Sum of Number and Its Reverse
Difficulty: Medium
Tag:

Given a non-negative integer num, return true if num can be expressed as the sum of any non-negative integer and its reverse, or false otherwise.

Example 1:

Input: num = 443
Output: true
Explanation: 172 + 271 = 443 so we return true.

Example 2:

Input: num = 63
Output: false
Explanation: 63 cannot be expressed as the sum of a non-negative integer and its reverse so we return false.

Example 3:

Input: num = 181
Output: true
Explanation: 140 + 041 = 181 so we return true. Note that when a number is reversed, there may be leading zeros.

Constraints:

0 <= num <= 10⁵

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

constexpr int N = 1e5 + 10;

class Solution {
public:
    vector<bool> vis{};

    int f(int x) {
        int res = 0;
        while (x) {
            res = res * 10 + (x % 10);
            x /= 10;
        }

        return res;
    }

    void init() {
        vis = vector<bool>(N, false);
        for (int i = 0; i < N; i++) {
            int x = i + f(i);
            if (x < N) {
                vis[x] = true;
            }
        }
    }

    bool sumOfNumberAndReverse(int num) {
        if (vis.empty()) {
            init();
        }

        return vis[num];
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

简体中文English

Metadata

Link: 统计定界子数组的数目
Difficulty: Hard
Tag:

给你一个整数数组 nums 和两个整数 minK 以及 maxK 。

nums 的定界子数组是满足下述条件的一个子数组：

子数组中的 最小值 等于 minK 。
子数组中的 最大值 等于 maxK 。

返回定界子数组的数目。

子数组是数组中的一个连续部分。

示例 1：

输入：nums = [1,3,5,2,7,5], minK = 1, maxK = 5
输出：2
解释：定界子数组是 [1,3,5] 和 [1,3,5,2] 。

示例 2：

输入：nums = [1,1,1,1], minK = 1, maxK = 1
输出：10
解释：nums 的每个子数组都是一个定界子数组。共有 10 个子数组。

提示：

2 <= nums.length <= 10⁵
1 <= nums[i], minK, maxK <= 10⁶

Metadata

Link: Count Subarrays With Fixed Bounds
Difficulty: Hard
Tag:

You are given an integer array nums and two integers minK and maxK.

A fixed-bound subarray of nums is a subarray that satisfies the following conditions:

The minimum value in the subarray is equal to minK.
The maximum value in the subarray is equal to maxK.

Return the number of fixed-bound subarrays.

A subarray is a contiguous part of an array.

Example 1:

Input: nums = [1,3,5,2,7,5], minK = 1, maxK = 5
Output: 2
Explanation: The fixed-bound subarrays are [1,3,5] and [1,3,5,2].

Example 2:

Input: nums = [1,1,1,1], minK = 1, maxK = 1
Output: 10
Explanation: Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i], minK, maxK <= 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 countSubarrays(vector<int> &nums, int minK, int maxK) {
        ll res = 0;

        int n = int(nums.size());

        int l1 = 0;
        int l2 = 0;
        auto se = multiset<int>();
        auto se2 = multiset<int>();

        for (int i = 0; i < n; i++) {
            int a = nums[i];
            se.insert(a);
            se2.insert(a);

            while (l1 <= i) {
                int mi = *se.begin();
                int mx = *(--(se.end()));

                if (mi < minK || mx > maxK) {
                    se.erase(se.lower_bound(nums[l1]));
                    ++l1;
                } else {
                    break;
                }
            }

            while (l2 < l1) {
                se2.erase(se2.lower_bound(nums[l2]));
                ++l2;
            }

            while (l2 <= i) {
                int mi = *se2.begin();
                int mx = *(--(se2.end()));

                if (mi == minK && mx == maxK) {
                    se2.erase(se2.lower_bound(nums[l2]));
                    ++l2;
                } else {
                    break;
                }
            }

            if (l1 <= i) {
                int mi = *se.begin();
                int mx = *(--(se.end()));

                if (mi == minK && mx == maxK) {
                    // cout << i << " " << l << " " << mi << " " << mx << endl;

                    res += l2 - l1;
                }
            }
        }

        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: October 11, 2023