feat: add solutions to lc problems: No.1604~1608 (#2478)

Author: yanglbme

solution/1600-1699/1604.Alert Using Same Key-Card Three or More Times in a One Hour Period/README.md

@@ -59,7 +59,7 @@
 
 最后，将答案数组按照字典序排序，即可得到答案。
 
-时间复杂度 $O(n \times \log n)$，其中 $n$ 是数组 $keyName$ 的长度。
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(n)$。其中 $n$ 是打卡记录的数量。
 
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@@ -171,6 +171,38 @@ func alertNames(keyName []string, keyTime []string) (ans []string) {
 }
 ```
 
+```ts
+function alertNames(keyName: string[], keyTime: string[]): string[] {
+    const d: { [name: string]: number[] } = {};
+    for (let i = 0; i < keyName.length; ++i) {
+        const name = keyName[i];
+        const t = keyTime[i];
+        const minutes = +t.slice(0, 2) * 60 + +t.slice(3);
+        if (d[name] === undefined) {
+            d[name] = [];
+        }
+        d[name].push(minutes);
+    }
+    const ans: string[] = [];
+    for (const name in d) {
+        if (d.hasOwnProperty(name)) {
+            const ts = d[name];
+            if (ts.length > 2) {
+                ts.sort((a, b) => a - b);
+                for (let i = 0; i < ts.length - 2; ++i) {
+                    if (ts[i + 2] - ts[i] <= 60) {
+                        ans.push(name);
+                        break;
+                    }
+                }
+            }
+        }
+    }
+    ans.sort();
+    return ans;
+}
+```
+
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 <!-- end -->

solution/1600-1699/1604.Alert Using Same Key-Card Three or More Times in a One Hour Period/README_EN.md

@@ -47,7 +47,15 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Hash Table + Sorting
+
+First, we use a hash table $d$ to record all the clock-in times of each employee.
+
+Then we traverse the hash table. For each employee, we first check whether the number of clock-in times is greater than or equal to 3. If not, we skip this employee. Otherwise, we sort all the clock-in times of this employee in chronological order, and then traverse the sorted clock-in times to check whether the two times at a distance of 2 indices are within the same hour. If so, we add this employee to the answer array.
+
+Finally, we sort the answer array in lexicographical order to get the answer.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the number of clock-in records.
 
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@@ -159,6 +167,38 @@ func alertNames(keyName []string, keyTime []string) (ans []string) {
 }
 ```
 
+```ts
+function alertNames(keyName: string[], keyTime: string[]): string[] {
+    const d: { [name: string]: number[] } = {};
+    for (let i = 0; i < keyName.length; ++i) {
+        const name = keyName[i];
+        const t = keyTime[i];
+        const minutes = +t.slice(0, 2) * 60 + +t.slice(3);
+        if (d[name] === undefined) {
+            d[name] = [];
+        }
+        d[name].push(minutes);
+    }
+    const ans: string[] = [];
+    for (const name in d) {
+        if (d.hasOwnProperty(name)) {
+            const ts = d[name];
+            if (ts.length > 2) {
+                ts.sort((a, b) => a - b);
+                for (let i = 0; i < ts.length - 2; ++i) {
+                    if (ts[i + 2] - ts[i] <= 60) {
+                        ans.push(name);
+                        break;
+                    }
+                }
+            }
+        }
+    }
+    ans.sort();
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1600-1699/1604.Alert Using Same Key-Card Three or More Times in a One Hour Period/Solution.ts

@@ -0,0 +1,29 @@
+function alertNames(keyName: string[], keyTime: string[]): string[] {
+    const d: { [name: string]: number[] } = {};
+    for (let i = 0; i < keyName.length; ++i) {
+        const name = keyName[i];
+        const t = keyTime[i];
+        const minutes = +t.slice(0, 2) * 60 + +t.slice(3);
+        if (d[name] === undefined) {
+            d[name] = [];
+        }
+        d[name].push(minutes);
+    }
+    const ans: string[] = [];
+    for (const name in d) {
+        if (d.hasOwnProperty(name)) {
+            const ts = d[name];
+            if (ts.length > 2) {
+                ts.sort((a, b) => a - b);
+                for (let i = 0; i < ts.length - 2; ++i) {
+                    if (ts[i + 2] - ts[i] <= 60) {
+                        ans.push(name);
+                        break;
+                    }
+                }
+            }
+        }
+    }
+    ans.sort();
+    return ans;
+}

solution/1600-1699/1605.Find Valid Matrix Given Row and Column Sums/README_EN.md

@@ -49,7 +49,19 @@ Another possible matrix is: [[1,2],
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Greedy + Construction
+
+We can first initialize an $m$ by $n$ answer matrix $ans$.
+
+Next, we traverse each position $(i, j)$ in the matrix, set the element at this position to $x = \min(rowSum[i], colSum[j])$, and subtract $x$ from $rowSum[i]$ and $colSum[j]$ respectively. After traversing all positions, we can get a matrix $ans$ that meets the requirements of the problem.
+
+The correctness of the above strategy is explained as follows:
+
+According to the requirements of the problem, we know that the sum of $rowSum$ and $colSum$ is equal, so $rowSum[0]$ must be less than or equal to $\sum_{j = 0}^{n - 1} colSum[j]$. Therefore, after $n$ operations, $rowSum[0]$ can definitely be made $0$, and for any $j \in [0, n - 1]$, $colSum[j] \geq 0$ is guaranteed.
+
+Therefore, we reduce the original problem to a subproblem with $m-1$ rows and $n$ columns, continue the above operations, until all elements in $rowSum$ and $colSum$ are $0$, we can get a matrix $ans$ that meets the requirements of the problem.
+
+The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Where $m$ and $n$ are the lengths of $rowSum$ and $colSum$ respectively.
 
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solution/1600-1699/1608.Special Array With X Elements Greater Than or Equal X/README.md

@@ -59,9 +59,9 @@ x 不能取更大的值，因为 nums 中只有两个元素。</pre>
 
 ### 方法一：暴力枚举
 
-在 $[1..n]$ 范围内枚举 $x$，然后统计数组中大于等于 $x$ 的元素个数，记为 $cnt$。若存在 $cnt$ 与 $x$ 相等，直接返回 $x$。
+我们在 $[1..n]$ 范围内枚举 $x$，然后统计数组中大于等于 $x$ 的元素个数，记为 $cnt$。若存在 $cnt$ 与 $x$ 相等，直接返回 $x$。
 
-时间复杂度 $O(n^2)$。
+时间复杂度 $O(n^2)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
 
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@@ -165,7 +165,7 @@ impl Solution {
 
 接下来同样枚举 $x$，利用二分查找，找到 `nums` 中第一个大于等于 $x$ 的元素，快速统计出 `nums` 中大于等于 $x$ 的元素个数。
 
-时间复杂度 $O(n\log n)$。
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组的长度。
 
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solution/1600-1699/1608.Special Array With X Elements Greater Than or Equal X/README_EN.md

@@ -51,7 +51,11 @@ x cannot be greater since there are only 2 numbers in nums.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Brute Force Enumeration
+
+We enumerate $x$ in the range of $[1..n]$, and then count the number of elements in the array that are greater than or equal to $x$, denoted as $cnt$. If there exists $cnt$ equal to $x$, return $x$ directly.
+
+The time complexity is $O(n^2)$, where $n$ is the length of the array. The space complexity is $O(1)$.
 
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@@ -149,7 +153,13 @@ impl Solution {
 
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-### Solution 2
+### Solution 2: Sorting + Binary Search
+
+We can also sort `nums` first.
+
+Next, we still enumerate $x$, and use binary search to find the first element in `nums` that is greater than or equal to $x$, quickly counting the number of elements in `nums` that are greater than or equal to $x$.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array.
 
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