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Commit dc4d031

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feat: update solutions to lc problems: No.2526,2528 (doocs#3722)
* No.2526.Find Consecutive Integers from a Data Stream * No.2528.Maximize the Minimum Powered City
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‎solution/2500-2599/2526.Find Consecutive Integers from a Data Stream/README.md‎

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[null, false, false, true, false]
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<strong>解释:</strong>
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DataStream dataStream = new DataStream(4, 3); // value = 4, k = 3
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DataStream dataStream = new DataStream(4, 3); // value = 4, k = 3
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dataStream.consec(4); // 数据流中只有 1 个整数,所以返回 False 。
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dataStream.consec(4); // 数据流中只有 2 个整数
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// 由于 2 小于 k ,返回 False 。
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### 方法一:计数
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维护一个计数器 $cnt,ドル记录当前连续整数为 `value` 的个数。
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我们可以维护一个计数器 $\textit{cnt},ドル记录当前连续整数为 $\textit{value}$ 的个数。
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当 `num` 与 `value` 相等时,$cnt$ 自增 1,否则 $cnt$ 重置为 0。然后判断 $cnt$ 是否大于等于 `k` 即可。
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调用 `consec` 方法时,如果 $\textit{num}$ 与 $\textit{value}$ 相等,我们将 $\textit{cnt}$ 自增 1,否则将 $\textit{cnt}$ 重置为 0。然后判断 $\textit{cnt}$ 是否大于等于 $\textit{k}$ 即可。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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‎solution/2500-2599/2526.Find Consecutive Integers from a Data Stream/README_EN.md‎

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[null, false, false, true, false]
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<strong>Explanation</strong>
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DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
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dataStream.consec(4); // Only 1 integer is parsed, so returns False.
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DataStream dataStream = new DataStream(4, 3); //value = 4, k = 3
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dataStream.consec(4); // Only 1 integer is parsed, so returns False.
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dataStream.consec(4); // Only 2 integers are parsed.
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// Since 2 is less than k, returns False.
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dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
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// Since 2 is less than k, returns False.
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dataStream.consec(4); // The 3 integers parsed are all equal to value, so returns True.
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dataStream.consec(3); // The last k integers parsed in the stream are [4,4,3].
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// Since 3 is not equal to value, it returns False.
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</pre>
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Counting
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We can maintain a counter $\textit{cnt}$ to record the current number of consecutive integers equal to $\textit{value}$.
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When calling the `consec` method, if $\textit{num}$ is equal to $\textit{value},ドル we increment $\textit{cnt}$ by 1; otherwise, we reset $\textit{cnt}$ to 0. Then we check whether $\textit{cnt}$ is greater than or equal to $\textit{k}$.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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<!-- tabs:start -->
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‎solution/2500-2599/2528.Maximize the Minimum Powered City/README.md‎

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函数 $check(x, k)$ 的实现逻辑是:
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遍历每座城市,如果当前城市 $i$ 的供电站数目小于 $x,ドル此时我们可以贪心地在尽可能右边的位置上建造供电站,位置 $j = min(i + r, n - 1),ドル这样可以使得供电站覆盖尽可能多的城市。过程中我们可以借助差分数组,给一段连续的位置加上某个值。如果需要额外建造的供电站数量超过 $k,ドル那么 $x$ 不满足条件,返回 `false`。否则遍历结束后,返回 `true`。
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遍历每座城市,如果当前城市 $i$ 的供电站数目小于 $x,ドル此时我们可以贪心地在尽可能右边的位置上建造供电站,位置 $j = \min(i + r, n - 1),ドル这样可以使得供电站覆盖尽可能多的城市。过程中我们可以借助差分数组,给一段连续的位置加上某个值。如果需要额外建造的供电站数量超过 $k,ドル那么 $x$ 不满足条件,返回 `false`。否则遍历结束后,返回 `true`。
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时间复杂度 $O(n \times \log M),ドル空间复杂度 $O(n)$。其中 $n$ 为城市数量,而 $M$ 我们固定取 2ドル^{40}$。
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‎solution/2500-2599/2528.Maximize the Minimum Powered City/README_EN.md‎

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<pre>
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<strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2
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<strong>Output:</strong> 5
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<strong>Explanation:</strong>
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One of the optimal ways is to install both the power stations at city 1.
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<strong>Explanation:</strong>
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One of the optimal ways is to install both the power stations at city 1.
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So stations will become [1,4,4,5,0].
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- City 0 is provided by 1 + 4 = 5 power stations.
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- City 1 is provided by 1 + 4 + 4 = 9 power stations.
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<pre>
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<strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3
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<strong>Output:</strong> 4
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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It can be proved that we cannot make the minimum power of a city greater than 4.
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</pre>
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Binary Search + Difference Array + Greedy
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According to the problem description, the minimum number of power stations increases as the value of $k$ increases. Therefore, we can use binary search to find the largest minimum number of power stations, ensuring that the additional power stations needed do not exceed $k$.
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First, we use a difference array and prefix sum to calculate the initial number of power stations in each city, recording it in the array $s,ドル where $s[i]$ represents the number of power stations in the $i$-th city.
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Next, we define the left boundary of the binary search as 0ドル$ and the right boundary as 2ドル^{40}$. Then, we implement a function $check(x, k)$ to determine whether the minimum number of power stations in the cities can be $x,ドル ensuring that the additional power stations needed do not exceed $k$.
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The implementation logic of the function $check(x, k)$ is as follows:
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Traverse each city. If the number of power stations in the current city $i$ is less than $x,ドル we can greedily build a power station at the rightmost possible position, $j = \min(i + r, n - 1),ドル to cover as many cities as possible. During this process, we can use the difference array to add a certain value to a continuous segment. If the number of additional power stations needed exceeds $k,ドル then $x$ does not meet the condition, and we return `false`. Otherwise, after the traversal, return `true`.
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The time complexity is $O(n \times \log M),ドル and the space complexity is $O(n)$. Here, $n$ is the number of cities, and $M$ is fixed at 2ドル^{40}$.
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