feat: add solutions to lc problem: No.3119 (doocs#2608)

No.3119.Maximum Number of Potholes That Can Be Fixed

Author: yanglbme

solution/3100-3199/3119.Maximum Number of Potholes That Can Be Fixed/README.md

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+# [3119. Maximum Number of Potholes That Can Be Fixed](https://leetcode.cn/problems/maximum-number-of-potholes-that-can-be-fixed)
+
+[English Version](/solution/3100-3199/3119.Maximum%20Number%20of%20Potholes%20That%20Can%20Be%20Fixed/README_EN.md)
+
+<!-- tags: -->
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>You are given a string <code>road</code>, consisting only of characters <code>&quot;x&quot;</code> and <code>&quot;.&quot;</code>, where each <code>&quot;x&quot;</code> denotes a <em>pothole</em> and each <code>&quot;.&quot;</code> denotes a smooth road, and an integer <code>budget</code>.</p>
+
+<p>In one repair operation, you can repair <code>n</code> <strong>consecutive</strong> potholes for a price of <code>n + 1</code>.</p>
+
+<p>Return the <strong>maximum</strong> number of potholes that can be fixed such that the sum of the prices of all of the fixes <strong>doesn&#39;t go over</strong> the given budget.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">road = &quot;..&quot;, budget = 5</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>There are no potholes to be fixed.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">road = &quot;..xxxxx&quot;, budget = 4</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">3</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We fix the first three potholes (they are consecutive). The budget needed for this task is <code>3 + 1 = 4</code>.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">road = &quot;x.x.xxx...x&quot;, budget = 14</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">6</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>We can fix all the potholes. The total cost would be <code>(1 + 1) + (1 + 1) + (3 + 1) + (1 + 1) = 10</code> which is within our budget of 14.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= road.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= budget &lt;= 10<sup>5</sup> + 1</code></li>
+	<li><code>road</code> consists only of characters <code>&#39;.&#39;</code> and <code>&#39;x&#39;</code>.</li>
+</ul>
+
+## 解法
+
+### 方法一：计数 + 贪心
+
+我们首先统计出每个连续的坑洼的数量，记录在数组 $cnt$ 中，即 $cnt[k]$ 表示有 $cnt[k]$ 个长度为 $k$ 的连续坑洼。
+
+由于我们要尽可能多地修补坑洼，而对于长度为 $k$ 的连续坑洼，我们需要花费 $k + 1$ 的代价，应该优先修补长度较长的坑洼，这样可以使得代价最小。
+
+因此，我们从最长的坑洼开始修补，对于长度为 $k$ 的坑洼，我们最多可以修补的个数为 $t = \min(\text{budget} / (k + 1), \text{cnt}[k])$，我们将修补的个数乘以长度 $k$ 加到答案中，然后更新剩余的预算。对于长度为 $k$ 的其余 $cnt[k] - t$ 个坑洼，我们将它们合并到长度为 $k - 1$ 的坑洼中。继续这个过程，直到遍历完所有的坑洼。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $road$ 的长度。
+
+<!-- tabs:start -->
+
+```python
+class Solution:
+    def maxPotholes(self, road: str, budget: int) -> int:
+        road += "."
+        n = len(road)
+        cnt = [0] * n
+        k = 0
+        for c in road:
+            if c == "x":
+                k += 1
+            elif k:
+                cnt[k] += 1
+                k = 0
+        ans = 0
+        for k in range(n - 1, 0, -1):
+            t = min(budget // (k + 1), cnt[k])
+            ans += t * k
+            budget -= t * (k + 1)
+            cnt[k - 1] += cnt[k] - t
+        return ans
+```
+
+```java
+class Solution {
+    public int maxPotholes(String road, int budget) {
+        road += ".";
+        int n = road.length();
+        int[] cnt = new int[n];
+        int k = 0;
+        for (char c : road.toCharArray()) {
+            if (c == 'x') {
+                ++k;
+            } else if (k > 0) {
+                ++cnt[k];
+                k = 0;
+            }
+        }
+        int ans = 0;
+        for (k = n - 1; k > 0; --k) {
+            int t = Math.min(budget / (k + 1), cnt[k]);
+            ans += t * k;
+            budget -= t * (k + 1);
+            cnt[k - 1] += cnt[k] - t;
+        }
+        return ans;
+    }
+}
+```
+
+```cpp
+class Solution {
+public:
+    int maxPotholes(string road, int budget) {
+        road.push_back('.');
+        int n = road.size();
+        vector<int> cnt(n);
+        int k = 0;
+        for (char& c : road) {
+            if (c == 'x') {
+                ++k;
+            } else if (k) {
+                ++cnt[k];
+                k = 0;
+            }
+        }
+        int ans = 0;
+        for (k = n - 1; k; --k) {
+            int t = min(budget / (k + 1), cnt[k]);
+            ans += t * k;
+            budget -= t * (k + 1);
+            cnt[k - 1] += cnt[k] - t;
+        }
+        return ans;
+    }
+};
+```
+
+```go
+func maxPotholes(road string, budget int) (ans int) {
+	road += "."
+	n := len(road)
+	cnt := make([]int, n)
+	k := 0
+	for _, c := range road {
+		if c == 'x' {
+			k++
+		} else if k > 0 {
+			cnt[k]++
+			k = 0
+		}
+	}
+	for k = n - 1; k > 0; k-- {
+		t := min(budget/(k+1), cnt[k])
+		ans += t * k
+		budget -= t * (k + 1)
+		cnt[k-1] += cnt[k] - t
+	}
+	return
+}
+```
+
+```ts
+function maxPotholes(road: string, budget: number): number {
+    road += '.';
+    const n = road.length;
+    const cnt: number[] = Array(n).fill(0);
+    let k = 0;
+    for (const c of road) {
+        if (c === 'x') {
+            ++k;
+        } else if (k) {
+            ++cnt[k];
+            k = 0;
+        }
+    }
+    let ans = 0;
+    for (k = n - 1; k; --k) {
+        const t = Math.min(Math.floor(budget / (k + 1)), cnt[k]);
+        ans += t * k;
+        budget -= t * (k + 1);
+        cnt[k - 1] += cnt[k] - t;
+    }
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- end -->