fix: update

Author: yanglbme

solution/1800-1899/1893.Check if All the Integers in a Range Are Covered/README.md

@@ -71,7 +71,7 @@ tags:
 
 接着，我们遍历差分数组 $\textit{diff}$，维护一个前缀和 $s$，对于每个位置 $i$，我们令 $s$ 自增 $\textit{diff}[i]$，如果 $s \le 0$ 且 $left \le i \le right$，则说明区间 $[left, right]$ 中有一个整数 $i$ 没有被覆盖，返回 $\textit{false}$。
 
-如果遍历完差分数组 $\textit{diff}$ 后都没有返回 $\textit{False}$，则说明区间 $[left, right]$ 中的每个整数都被 $\textit{ranges}$ 中至少一个区间覆盖，返回 $\textit{true}$。
+如果遍历完差分数组 $\textit{diff}$ 后都没有返回 $\textit{false}$，则说明区间 $[left, right]$ 中的每个整数都被 $\textit{ranges}$ 中至少一个区间覆盖，返回 $\textit{true}$。
 
 时间复杂度 $O(n + M)$，空间复杂度 $O(M)$。其中 $n$ 是数组 $\textit{ranges}$ 的长度，而 $M$ 是区间的最大值，本题中 $M \le 50$。
 

solution/1800-1899/1893.Check if All the Integers in a Range Are Covered/README_EN.md

@@ -69,7 +69,7 @@ Next, we iterate through the array $\textit{ranges}$. For each interval $[l, r]$
 
 Then, we iterate through the difference array $\textit{diff}$, maintaining a prefix sum $s$. For each position $i$, we increment $s$ by $\textit{diff}[i]$. If $s \le 0$ and $left \le i \le right$, it indicates that an integer $i$ within the interval $[left, right]$ is not covered, and we return $\textit{false}$.
 
-If we finish iterating through the difference array $\textit{diff}$ without returning $\textit{False}$, it means that every integer within the interval $[left, right]$ is covered by at least one interval in $\textit{ranges}$, and we return $\textit{true}$.
+If we finish iterating through the difference array $\textit{diff}$ without returning $\textit{false}$, it means that every integer within the interval $[left, right]$ is covered by at least one interval in $\textit{ranges}$, and we return $\textit{true}$.
 
 The time complexity is $O(n + M)$, and the space complexity is $O(M)$. Here, $n$ is the length of the array $\textit{ranges}$, and $M$ is the maximum value of the interval, which in this case is $M \le 50$.