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Commit 93169b2

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✨feat: add 878、1742
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‎Index/哈希表.md‎

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| [1640. 能否连接形成数组](https://leetcode.cn/problems/check-array-formation-through-concatenation/) | [LeetCode 题解链接](https://leetcode.cn/problems/check-array-formation-through-concatenation/solution/by-ac_oier-3jqf/) | 简单 | 🤩🤩🤩🤩🤩 |
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| [1700. 无法吃午餐的学生数量](https://leetcode.cn/problems/number-of-students-unable-to-eat-lunch/) | [LeetCode 题解链接](https://leetcode.cn/problems/number-of-students-unable-to-eat-lunch/solution/by-ac_oier-rvc3/) | 简单 | 🤩🤩🤩🤩🤩 |
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| [1711. 大餐计数](https://leetcode-cn.com/problems/count-good-meals/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/count-good-meals/solution/gong-shui-san-xie-xiang-jie-san-chong-gu-nn4f/) | 中等 | 🤩🤩🤩 |
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| [1742. 盒子中小球的最大数量](https://leetcode.cn/problems/maximum-number-of-balls-in-a-box/) | [LeetCode 题解链接](https://leetcode.cn/problems/maximum-number-of-balls-in-a-box/solution/by-ac_oier-3mxf/) | 简单 | 🤩🤩🤩 |
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| [1743. 从相邻元素对还原数组](https://leetcode-cn.com/problems/restore-the-array-from-adjacent-pairs/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/restore-the-array-from-adjacent-pairs/solution/gong-shui-san-xie-yi-ti-shuang-jie-dan-x-elpx/) | 中等 | 🤩🤩🤩🤩 |
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| [1748. 唯一元素的和](https://leetcode-cn.com/problems/sum-of-unique-elements/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/sum-of-unique-elements/solution/gong-shui-san-xie-yi-ti-shuang-jie-pai-x-atnd/) | 简单 | 🤩🤩🤩🤩 |
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| [1838. 最高频元素的频数](https://leetcode-cn.com/problems/frequency-of-the-most-frequent-element/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/frequency-of-the-most-frequent-element/solution/gong-shui-san-xie-cong-mei-ju-dao-pai-xu-kxnk/) | 中等 | 🤩🤩🤩 |

‎Index/模拟.md‎

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| [1732. 找到最高海拔](https://leetcode.cn/problems/find-the-highest-altitude/) | [LeetCode 题解链接](https://leetcode.cn/problems/find-the-highest-altitude/solution/by-ac_oier-a0j8/) | 简单 | 🤩🤩🤩🤩 |
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| [1736. 替换隐藏数字得到的最晚时间](https://leetcode-cn.com/problems/latest-time-by-replacing-hidden-digits/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/latest-time-by-replacing-hidden-digits/solution/gong-shui-san-xie-ti-huan-yin-cang-shu-z-2l1h/) | 简单 | 🤩🤩🤩🤩 |
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| [1737. 满足三条件之一需改变的最少字符数](https://leetcode.cn/problems/change-minimum-characters-to-satisfy-one-of-three-conditions/) | [LeetCode 题解链接](https://leetcode.cn/problems/change-minimum-characters-to-satisfy-one-of-three-conditions/solution/by-ac_oier-vs5u/) | 中等 | 🤩🤩🤩🤩 |
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| [1742. 盒子中小球的最大数量](https://leetcode.cn/problems/maximum-number-of-balls-in-a-box/) | [LeetCode 题解链接](https://leetcode.cn/problems/maximum-number-of-balls-in-a-box/solution/by-ac_oier-3mxf/) | 简单 | 🤩🤩🤩 |
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| [1743. 从相邻元素对还原数组](https://leetcode-cn.com/problems/restore-the-array-from-adjacent-pairs/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/restore-the-array-from-adjacent-pairs/solution/gong-shui-san-xie-yi-ti-shuang-jie-dan-x-elpx/) | 中等 | 🤩🤩🤩🤩 |
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| [1748. 唯一元素的和](https://leetcode-cn.com/problems/sum-of-unique-elements/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/sum-of-unique-elements/solution/gong-shui-san-xie-yi-ti-shuang-jie-pai-x-atnd/) | 简单 | 🤩🤩🤩🤩 |
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| [1763. 最长的美好子字符串](https://leetcode-cn.com/problems/longest-nice-substring/) | [LeetCode 题解链接](https://leetcode-cn.com/problems/longest-nice-substring/solution/cong-shu-ju-fan-wei-xuan-ze-he-gua-suan-n3y2a/) | 简单 | 🤩🤩🤩 |
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### 题目描述
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这是 LeetCode 上的 **[1742. 盒子中小球的最大数量](https://leetcode.cn/problems/maximum-number-of-balls-in-a-box/solution/by-ac_oier-3mxf/)** ,难度为 **简单**。
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Tag : 「哈希表」、「模拟」
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你在一家生产小球的玩具厂工作,有 `n` 个小球,编号从 `lowLimit` 开始,到 `highLimit` 结束(包括 `lowLimit` 和 `highLimit` ,即 `n == highLimit - lowLimit + 1`)。另有无限数量的盒子,编号从 `1` 到 `infinity` 。
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你的工作是将每个小球放入盒子中,其中盒子的编号应当等于小球编号上每位数字的和。例如,编号 `321` 的小球应当放入编号 `3 + 2 + 1 = 6` 的盒子,而编号 `10` 的小球应当放入编号 `1 + 0 = 1` 的盒子。
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给你两个整数 `lowLimit` 和 `highLimit` ,返回放有最多小球的盒子中的小球数量。如果有多个盒子都满足放有最多小球,只需返回其中任一盒子的小球数量。
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示例 1:
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```
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输入:lowLimit = 1, highLimit = 10
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输出:2
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解释:
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盒子编号:1 2 3 4 5 6 7 8 9 10 11 ...
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小球数量:2 1 1 1 1 1 1 1 1 0 0 ...
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编号 1 的盒子放有最多小球,小球数量为 2 。
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```
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示例 2:
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```
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输入:lowLimit = 5, highLimit = 15
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输出:2
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解释:
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盒子编号:1 2 3 4 5 6 7 8 9 10 11 ...
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小球数量:1 1 1 1 2 2 1 1 1 0 0 ...
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编号 5 和 6 的盒子放有最多小球,每个盒子中的小球数量都是 2 。
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```
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示例 3:
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```
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输入:lowLimit = 19, highLimit = 28
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输出:2
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解释:
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盒子编号:1 2 3 4 5 6 7 8 9 10 11 12 ...
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小球数量:0 1 1 1 1 1 1 1 1 2 0 0 ...
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编号 10 的盒子放有最多小球,小球数量为 2 。
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```
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提示:
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* 1ドル <= lowLimit <= highLimit <= 10^5$
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---
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### 模拟
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数据范围 $n = 1e5,ドル因此最大盒子编号 `99999 = 5 * 9 = 45`,我们可以用一个大小为 50ドル$ 的数组 `cnts` 来统计每个编号盒子中小球的数量,$cnts[idx] = x$ 含义为编号为 $idx$ 的盒子有 $x$ 个小球。
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Java 代码:
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```Java
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class Solution {
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public int countBalls(int l, int r) {
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int ans = 0;
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int[] cnts = new int[50];
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for (int i = l; i <= r; i++) {
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int j = i, cur = 0;
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while (j != 0) {
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cur += j % 10; j /= 10;
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}
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if (++cnts[cur] > ans) ans = cnts[cur];
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}
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return ans;
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}
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}
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```
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TypeScript 代码:
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```TypeScript
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function countBalls(l: number, r: number): number {
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let ans = 0
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const cnts = new Array<number>(50).fill(0)
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for (let i = l; i <= r; i++) {
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let j = i, cur = 0
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while (j != 0) {
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cur += j % 10
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j = Math.floor(j / 10)
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}
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if (++cnts[cur] > ans) ans = cnts[cur]
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}
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return ans
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}
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```
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Python 代码:
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```Python
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class Solution:
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def countBalls(self, l: int, r: int) -> int:
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ans = 0
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cnts = [0] * 50
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for i in range(l, r + 1):
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j, cur = i, 0
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while j != 0:
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j, cur = j // 10, cur + j % 10
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cnts[cur] += 1
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if cnts[cur] > ans:
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ans = cnts[cur]
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return ans
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```
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* 时间复杂度:$O(n\log{r})$
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* 空间复杂度:$O(C),ドル其中 $C = 50$ 为最大盒子编号
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---
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### 最后
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这是我们「刷穿 LeetCode」系列文章的第 `No.1742` 篇,系列开始于 2021年01月01日,截止于起始日 LeetCode 上共有 1916 道题目,部分是有锁题,我们将先把所有不带锁的题目刷完。
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在这个系列文章里面,除了讲解解题思路以外,还会尽可能给出最为简洁的代码。如果涉及通解还会相应的代码模板。
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为了方便各位同学能够电脑上进行调试和提交代码,我建立了相关的仓库:https://github.com/SharingSource/LogicStack-LeetCode 。
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在仓库地址里,你可以看到系列文章的题解链接、系列文章的相应代码、LeetCode 原题链接和其他优选题解。
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‎LeetCode/871-880/878. 第 N 个神奇数字(困难).md‎

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#### 提示四 : 如何确定值域
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一个合格的值域只需要确定答案在值域范围即可,因此我们可以直接定值域大小为 $1e20$。
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一个合格的值域只需要确定答案在值域范围即可,因此我们可以直接定值域大小为 $1e18$。
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或是根据 `a` 和 `b` 的取值来大致确定:假设两者中的较大值为 $m,ドル此时第 $n$ 个符合要求的数最大不会超过 $n \times m,ドル因此也可以设定值域大小为 $[0, 40000n]$。
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}
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public int nthMagicalNumber(int _n, int _a, int _b) {
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n = _n; a = _a; b = _b; c = a * b / gcd(a, b);
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long l = 0, r = (long)1e20;
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long l = 0, r = (long)1e18;
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while (l < r) {
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long mid = l + r >> 1;
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if (check(mid) >= n) r = mid;
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return Math.floor(x / a) + Math.floor(x / b) - Math.floor(x / c)
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}
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const c = Math.floor(a * b / gcd(a, b))
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let l = 0, r = 1e19
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let l = 0, r = 1e18
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while (l < r) {
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const mid = Math.floor((l + r) / 2)
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if (check(mid) >= n) r = mid
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def check(x):
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return x // a + x // b - x // c
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c = a * b // gcd(a, b)
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l, r = 0, 1e20
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l, r = 0, 1e18
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while l < r:
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mid = (l + r) // 2
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if check(mid) >= n:
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l = mid + 1
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return int(r % 1000000007)
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```
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* 时间复杂度:$O(\log{N}),ドル其中 $N = 1e20$ 为值域大小
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* 时间复杂度:$O(\log{N}),ドル其中 $N = 1e18$ 为值域大小
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* 空间复杂度:$O(1)$
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---

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