Update README_EN.md

Author: yanglbme

solution/1800-1899/1870.Minimum Speed to Arrive on Time/README_EN.md

@@ -85,7 +85,9 @@ tags:
 
 We notice that if a speed value $v$ allows us to arrive within the stipulated time, then for any $v' > v$, we can also definitely arrive within the stipulated time. This exhibits monotonicity, hence we can use binary search to find the smallest speed value that meets the condition.
 
-We can enumerate the speed value using binary search. First, we define the left and right boundaries of the binary search as $l = 1$, $r = 10^7 + 1$, and then we take the middle value $\text{mid} = \frac{l + r}{2}$ each time to check if it meets the condition. If it does, we move the right boundary to $\text{mid}$; otherwise, we move the left boundary to $\text{mid} + 1$.
+Before conducting the binary search, we need to first determine if it is possible to arrive within the stipulated time. If the number of trains is greater than the ceiling of the stipulated time, then it is definitely impossible to arrive within the stipulated time, and we should directly return $-1$.
+
+Next, we define the left and right boundaries for the binary search as $l = 1$, $r = 10^7 + 1$, and then we take the middle value $\text{mid} = \frac{l + r}{2}$ each time to check if it meets the condition. If it does, we move the right boundary to $\text{mid}$; otherwise, we move the left boundary to $\text{mid} + 1$.
 
 The problem is transformed into determining whether a speed value $v$ can allow us to arrive within the stipulated time. We can traverse each train trip, calculate the running time of each trip $t = \frac{d}{v}$, if it is the last trip, we directly add $t$; otherwise, we round up and add $t$. Finally, we check if the total time is less than or equal to the stipulated time, if so, it means the condition is met.