feat: update solutions to lc problem: No.3362

Author: yanglbme

solution/3300-3399/3362.Zero Array Transformation III/README.md

@@ -97,7 +97,34 @@ tags:
 
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-### 方法一：贪心 + 差分数组 + 优先队列（大根堆）
+### 方法一：贪心 + 差分数组 + 优先队列
+
+我们希望尽可能多地「移除」区间查询，但又要保证对每个位置 $i$ 来说，被选中、覆盖到它的查询次数 $s(i)$ 至少达到原数组值 $\textit{nums}[i]$，这样才能把该位置的值“减”到 0 或以下。如果对于某个位置 $i$ 无法满足 $s(i)\ge\textit{nums}[i]$，就说明再多选任何查询都不可能让该位置归零，返回 $-1$。
+
+为了做到这一点，我们按查询区间的左端点从小到大遍历，并维护：
+
+1. **差分数组** `d`：用于记录当前已应用的查询在何处分界——当我们在区间 $[l,r]$ 上“应用”一次查询时，立刻在差分数组位置 `d[l] += 1`，并在 `d[r+1] -= 1`，这样在遍历到下标 $i$ 时累加前缀和就能知道有多少次查询覆盖了 $i$。
+2. **最大堆** `pq`：存放当前「候选」区间查询的右端点（取负数以便 Python 最小堆模拟最大堆）。为什么选「最晚结束」的区间？因为它能覆盖更远的位置，我们的贪心策略是：**在每个 $i$ 处，只在必要时才摘取堆顶最长的区间来增加一次覆盖**，这样能为后续位置保留更多可用的区间。
+
+具体步骤如下：
+
+1. 先对 `queries` 按左端点 `l` 升序排序；
+2. 初始化差分数组 `d` 长度为 `n+1`（用来处理 `r+1` 处的减操作），以及当前覆盖次数 `s=0`、堆指针 `j=0`；
+3. 从 $i=0$ 遍历到 $n-1$：
+
+    - 先把 `d[i]` 累加到 `s`，即时更新已有的覆盖次数；
+    - 将所有左端点 $\le i$ 的查询 $[l,r]$ 压入最大堆 `pq`（存 `-r`），并推进 `j`；
+    - 当当前覆盖次数 `s` 小于所需值 `nums[i]`，且堆不空且堆顶区间仍能覆盖 $i$（即 $-pq[0]\ge i$）时：
+
+        1. 弹出堆顶（最长的区间），等价于“应用”这次查询；
+        2. 令 `s += 1` 并在 `d[r+1] -= 1`（使得在跨过 `r` 后覆盖次数自动回退）；
+
+    - 重复上述步骤，直到 `s>=nums[i]` 或无法再选区间；
+    - 若此时 `s<nums[i]`，说明无法将位置 $i$ 归零，直接返回 $-1$。
+
+4. 全部遍历完毕后，剩在堆里的区间都是「未被弹出」的，也就是真正被**保留**（即未被用来完成“归零”任务）的查询。堆大小即为答案。
+
+时间复杂度 $O(n + m \times \log m)$，空间复杂度 $O(n + m)$，其中 $n$ 是数组的长度，而 $m$ 是查询的个数。
 
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solution/3300-3399/3362.Zero Array Transformation III/README_EN.md

@@ -94,7 +94,34 @@ tags:
 
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-### Solution 1
+### Solution 1: Greedy + Difference Array + Priority Queue
+
+We want to "remove" as many interval queries as possible, while ensuring that for each position $i$, the number of selected queries covering it, $s(i)$, is at least the original array value $\textit{nums}[i]$, so that the value at that position can be reduced to 0 or below. If for some position $i$ we cannot satisfy $s(i) \ge \textit{nums}[i]$, it means that no matter how many more queries we select, it is impossible to make that position zero, so we return $-1$.
+
+To achieve this, we traverse the queries in order of their left endpoints and maintain:
+
+1. **Difference array** `d`: Used to record where the currently applied queries take effect—when we "apply" a query on the interval $[l, r]$, we immediately do `d[l] += 1` and `d[r+1] -= 1`. This way, when traversing to index $i$, the prefix sum tells us how many queries cover $i$.
+2. **Max heap** `pq`: Stores the right endpoints of the current "candidate" interval queries (store as negative numbers to simulate a max heap in Python's min heap). Why choose the "latest ending" interval? Because it can cover farther positions. Our greedy strategy is: **at each $i$, only pick the longest interval from the heap when necessary to increase coverage**, so that more intervals are available for subsequent positions.
+
+The specific steps are as follows:
+
+1. Sort `queries` by the left endpoint `l` in ascending order;
+2. Initialize the difference array `d` with length `n+1` (to handle the decrement at `r+1`), and set the current coverage count `s=0`, heap pointer `j=0`;
+3. For $i=0$ to $n-1$:
+
+    - First, add `d[i]` to `s` to update the current coverage count;
+    - Push all queries $[l, r]$ with left endpoint $\le i$ into the max heap `pq` (store `-r`), and advance `j`;
+    - While the current coverage `s` is less than the required value `nums[i]`, and the heap is not empty, and the top interval in the heap still covers $i$ (i.e., $-pq[0] \ge i$):
+
+        1. Pop the top of the heap (the longest interval), which is equivalent to "applying" this query;
+        2. Increment `s` by 1 and do `d[r+1] -= 1` (so that after passing $r$, the coverage count automatically decreases);
+
+    - Repeat the above steps until `s \ge nums[i]` or no more intervals can be selected;
+    - If at this point `s < nums[i]`, it means it is impossible to make position $i$ zero, so return $-1$.
+
+4. After traversing all positions, the intervals remaining in the heap are those that were **not popped**, i.e., the queries that are truly **retained** (not used for the "zeroing" task). The heap size is the answer.
+
+The time complexity is $O(n + m \times \log m)$, and the space complexity is $O(n + m)$, where $n$ is the length of the array and $m$ is the number of queries.
 
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