feat: add solutions to lc problems: No.2208,2209 (#2482)

Author: yanglbme

solution/2200-2299/2208.Minimum Operations to Halve Array Sum/README_EN.md

@@ -53,7 +53,13 @@ It can be shown that we cannot reduce the sum by at least half in less than 3 op
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Greedy + Priority Queue (Max Heap)
+
+According to the problem description, each operation will halve a number in the array. To minimize the number of operations that reduce the array sum by at least half, each operation should halve the current maximum value in the array.
+
+Therefore, we first calculate the total sum $s$ that the array needs to reduce, and then use a priority queue (max heap) to maintain all the numbers in the array. Each time we take the maximum value $t$ from the priority queue, halve it, and then put the halved number back into the priority queue, while updating $s$, until $s \le 0$. The number of operations at this time is the answer.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array.
 
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solution/2200-2299/2209.Minimum White Tiles After Covering With Carpets/README_EN.md

@@ -49,7 +49,20 @@ Note that the carpets are able to overlap one another.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Memoization Search
+
+Design a function $dfs(i, j)$ to represent the minimum number of white bricks that are not covered starting from index $i$ using $j$ carpets. The answer is $dfs(0, numCarpets)$.
+
+For index $i$, we discuss different cases:
+
+-   If $i \ge n$, it means that all bricks have been covered, return $0$;
+-   If $floor[i] = 0$, there is no need to use a carpet, just skip it, that is, $dfs(i, j) = dfs(i + 1, j)$;
+-   If $j = 0$, we can directly calculate the number of remaining white bricks that have not been covered using the prefix sum array $s$, that is, $dfs(i, j) = s[n] - s[i]$;
+-   If $floor[i] = 1$, we can choose to use a carpet to cover it, or choose not to use a carpet to cover it, and take the minimum of the two, that is, $dfs(i, j) = min(dfs(i + 1, j), dfs(i + carpetLen, j - 1))$.
+
+Use memoization search.
+
+The time complexity is $O(n\times m)$, and the space complexity is $O(n\times m)$. Where $n$ and $m$ are the lengths of the string $floor$ and the value of $numCarpets$ respectively.
 
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solution/2200-2299/2214.Minimum Health to Beat Game/README.md

@@ -72,7 +72,7 @@
 
 我们可以贪心地选择在伤害值最大的回合中使用一次护甲技能，假设伤害值最大为 $mx$，那么我们可以免受 $min(mx, armor)$ 的伤害，因此我们需要的最小生命值为 $sum(damage) - min(mx, armor) + 1$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `damage` 的长度。
+时间复杂度 $O(n)$，其中 $n$ 为数组 `damage` 的长度。空间复杂度 $O(1)$。
 
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solution/2200-2299/2214.Minimum Health to Beat Game/README_EN.md

@@ -65,7 +65,11 @@ Note that you did not use your armor ability.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Greedy
+
+We can greedily choose to use the armor skill in the round with the maximum damage. Suppose the maximum damage is $mx$, then we can avoid $min(mx, armor)$ damage, so the minimum life value we need is $sum(damage) - min(mx, armor) + 1$.
+
+The time complexity is $O(n)$, where $n$ is the length of the `damage` array. The space complexity is $O(1)$.
 
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