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feat: add solutions to lc problem: No.1326
No.1326.Minimum Number of Taps to Open to Water a Garden
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‎solution/1000-1099/1024.Video Stitching/README.md

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时间复杂度 $O(n+m)$，空间复杂度 $O(m)$。其中 $n$ 和 $m$ 分别是数组 `clips` 的长度和 `time` 的值。
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相似题目：[1326. 灌溉花园的最少水龙头数目](/solution/1300-1399/1326.Minimum%20Number%20of%20Taps%20to%20Open%20to%20Water%20a%20Garden/README.md)
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<!-- tabs:start -->
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### **Python3**

‎solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心**
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注意到，对于所有能覆盖某个左端点的水龙头，选择能覆盖最远右端点的那个水龙头是最优的。
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因此，我们可以预处理 `ranges`，对于每一个位置 $i$，左右端点分别为 `l = max(0, i-ranges[i])` 和 `r = min(n, i+ranges[i])`，我们算出所有能覆盖左端点 $l$ 的水龙头中，右端点最大的那个位置，记录在数组 $last[i]$ 中。
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我们定义变量 `mx` 表示当前能够到达的最远位置，变量 `ans` 表示当前需要的最少子区间数，变量 `pre` 表示上一个被使用的子区间的右端点。
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接下来，我们从 $0$ 开始枚举所有位置 $i$，用 $last[i]$ 来更新 `mx`。如果更新后 $mx = i$，说明无法覆盖下一个位置，返回 $-1$。
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同时我们记录上一个被使用的子区间的右端点 `pre`，如果 $pre = i$，说明需要使用一个新的子区间，因此我们将 `ans` 加 $1$，并将 `pre` 更新为 `mx`。
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遍历结束后，返回 `ans` 即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为花园的长度。
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相似题目：[1024. 视频拼接](/solution/1000-1099/1024.Video%20Stitching/README.md)
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def minTaps(self, n: int, ranges: List[int]) -> int:
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last = [0] * (n + 1)
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for i, v in enumerate(ranges):
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l, r = max(0, i - v), min(n, i + v)
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last[l] = max(last[l], r)
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ans = mx = pre = 0
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for i in range(n):
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mx = max(mx, last[i])
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if mx <= i:
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return -1
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if pre == i:
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ans += 1
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pre = mx
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int minTaps(int n, int[] ranges) {
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int[] last = new int[n + 1];
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for (int i = 0; i < n + 1; ++i) {
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int v = ranges[i];
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int l = Math.max(0, i - v), r = Math.min(n, i + v);
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last[l] = Math.max(last[l], r);
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}
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int ans = 0, mx = 0, pre = 0;
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for (int i = 0; i < n; ++i) {
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mx = Math.max(mx, last[i]);
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if (mx <= i) {
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return -1;
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}
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if (pre == i) {
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++ans;
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pre = mx;
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}
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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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### **C++**
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```cpp
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class Solution {
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public:
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int minTaps(int n, vector<int>& ranges) {
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vector<int> last(n + 1);
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for (int i = 0; i < n + 1; ++i) {
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int v = ranges[i];
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int l = max(0, i - v), r = min(n, i + v);
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last[l] = max(last[l], r);
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}
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int ans = 0, mx = 0, pre = 0;
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for (int i = 0; i < n; ++i) {
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mx = max(mx, last[i]);
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if (mx <= i) {
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return -1;
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}
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if (pre == i) {
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++ans;
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pre = mx;
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}
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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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### **Go**
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```go
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func minTaps(n int, ranges []int) int {
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last := make([]int, n+1)
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for i, v := range ranges {
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l, r := max(0, i-v), min(n, i+v)
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last[l] = max(last[l], r)
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}
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ans, mx, pre := 0, 0, 0
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for i := 0; i < n; i++ {
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mx = max(mx, last[i])
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if mx <= i {
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return -1
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}
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if pre == i {
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ans++
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pre = mx
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/README_EN.md

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### **Python3**
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```python
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class Solution:
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def minTaps(self, n: int, ranges: List[int]) -> int:
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last = [0] * (n + 1)
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for i, v in enumerate(ranges):
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l, r = max(0, i - v), min(n, i + v)
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last[l] = max(last[l], r)
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ans = mx = pre = 0
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for i in range(n):
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mx = max(mx, last[i])
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if mx <= i:
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return -1
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if pre == i:
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ans += 1
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pre = mx
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int minTaps(int n, int[] ranges) {
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int[] last = new int[n + 1];
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for (int i = 0; i < n + 1; ++i) {
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int v = ranges[i];
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int l = Math.max(0, i - v), r = Math.min(n, i + v);
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last[l] = Math.max(last[l], r);
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}
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int ans = 0, mx = 0, pre = 0;
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for (int i = 0; i < n; ++i) {
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mx = Math.max(mx, last[i]);
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if (mx <= i) {
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return -1;
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}
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if (pre == i) {
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++ans;
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pre = mx;
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}
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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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### **C++**
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```cpp
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class Solution {
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public:
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int minTaps(int n, vector<int>& ranges) {
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vector<int> last(n + 1);
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for (int i = 0; i < n + 1; ++i) {
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int v = ranges[i];
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int l = max(0, i - v), r = min(n, i + v);
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last[l] = max(last[l], r);
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}
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int ans = 0, mx = 0, pre = 0;
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for (int i = 0; i < n; ++i) {
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mx = max(mx, last[i]);
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if (mx <= i) {
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return -1;
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}
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if (pre == i) {
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++ans;
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pre = mx;
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}
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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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### **Go**
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```go
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func minTaps(n int, ranges []int) int {
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last := make([]int, n+1)
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for i, v := range ranges {
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l, r := max(0, i-v), min(n, i+v)
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last[l] = max(last[l], r)
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}
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ans, mx, pre := 0, 0, 0
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for i := 0; i < n; i++ {
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mx = max(mx, last[i])
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if mx <= i {
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return -1
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}
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if pre == i {
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ans++
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pre = mx
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/Solution.cpp

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class Solution {
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public:
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int minTaps(int n, vector<int>& ranges) {
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vector<int> last(n + 1);
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for (int i = 0; i < n + 1; ++i) {
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int v = ranges[i];
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int l = max(0, i - v), r = min(n, i + v);
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last[l] = max(last[l], r);
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}
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int ans = 0, mx = 0, pre = 0;
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for (int i = 0; i < n; ++i) {
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mx = max(mx, last[i]);
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if (mx <= i) {
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return -1;
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}
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if (pre == i) {
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++ans;
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pre = mx;
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}
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}
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return ans;
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}
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};

‎solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/Solution.go

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func minTaps(n int, ranges []int) int {
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last := make([]int, n+1)
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for i, v := range ranges {
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l, r := max(0, i-v), min(n, i+v)
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last[l] = max(last[l], r)
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}
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ans, mx, pre := 0, 0, 0
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for i := 0; i < n; i++ {
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mx = max(mx, last[i])
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if mx <= i {
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return -1
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}
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if pre == i {
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ans++
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pre = mx
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}
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}

‎solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/Solution.java

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class Solution {
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public int minTaps(int n, int[] ranges) {
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int[] last = new int[n + 1];
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for (int i = 0; i < n + 1; ++i) {
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int v = ranges[i];
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int l = Math.max(0, i - v), r = Math.min(n, i + v);
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last[l] = Math.max(last[l], r);
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}
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int ans = 0, mx = 0, pre = 0;
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for (int i = 0; i < n; ++i) {
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mx = Math.max(mx, last[i]);
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if (mx <= i) {
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return -1;
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}
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if (pre == i) {
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++ans;
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pre = mx;
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}
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}
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return ans;
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}
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}

‎solution/1300-1399/1326.Minimum Number of Taps to Open to Water a Garden/Solution.py

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class Solution:
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def minTaps(self, n: int, ranges: List[int]) -> int:
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last = [0] * (n + 1)
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for i, v in enumerate(ranges):
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l, r = max(0, i - v), min(n, i + v)
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last[l] = max(last[l], r)
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ans = mx = pre = 0
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for i in range(n):
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mx = max(mx, last[i])
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if mx <= i:
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return -1
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if pre == i:
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ans += 1
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pre = mx
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return ans

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