| comments | difficulty | edit_url | rating | source | tags | ||
|---|---|---|---|---|---|---|---|
true |
Medium |
1675 |
Weekly Contest 242 Q2 |
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You are given a floating-point number hour, representing the amount of time you have to reach the office. To commute to the office, you must take n trains in sequential order. You are also given an integer array dist of length n, where dist[i] describes the distance (in kilometers) of the ith train ride.
Each train can only depart at an integer hour, so you may need to wait in between each train ride.
- For example, if the
1sttrain ride takes1.5hours, you must wait for an additional0.5hours before you can depart on the2ndtrain ride at the 2 hour mark.
Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1 if it is impossible to be on time.
Tests are generated such that the answer will not exceed 107 and hour will have at most two digits after the decimal point.
Example 1:
Input: dist = [1,3,2], hour = 6 Output: 1 Explanation: At speed 1: - The first train ride takes 1/1 = 1 hour. - Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second train takes 3/1 = 3 hours. - Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third train takes 2/1 = 2 hours. - You will arrive at exactly the 6 hour mark.
Example 2:
Input: dist = [1,3,2], hour = 2.7 Output: 3 Explanation: At speed 3: - The first train ride takes 1/3 = 0.33333 hours. - Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train ride takes 3/3 = 1 hour. - Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third train takes 2/3 = 0.66667 hours. - You will arrive at the 2.66667 hour mark.
Example 3:
Input: dist = [1,3,2], hour = 1.9 Output: -1 Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.
Constraints:
n == dist.length1 <= n <= 1051 <= dist[i] <= 1051 <= hour <= 109- There will be at most two digits after the decimal point in
hour.
We notice that if a speed value
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
Next, we define the left and right boundaries for the binary search as
The problem is transformed into determining whether a speed value
After the binary search ends, if the left boundary exceeds
The time complexity is
class Solution:
def minSpeedOnTime(self, dist: List[int], hour: float) -> int:
def check(v: int) -> bool:
s = 0
for i, d in enumerate(dist):
t = d / v
s += t if i == len(dist) - 1 else ceil(t)
return s <= hour
if len(dist) > ceil(hour):
return -1
r = 10**7 + 1
ans = bisect_left(range(1, r), True, key=check) + 1
return -1 if ans == r else ansclass Solution {
public int minSpeedOnTime(int[] dist, double hour) {
if (dist.length > Math.ceil(hour)) {
return -1;
}
final int m = (int) 1e7;
int l = 1, r = m + 1;
while (l < r) {
int mid = (l + r) >> 1;
if (check(dist, mid, hour)) {
r = mid;
} else {
l = mid + 1;
}
}
return l > m ? -1 : l;
}
private boolean check(int[] dist, int v, double hour) {
double s = 0;
int n = dist.length;
for (int i = 0; i < n; ++i) {
double t = dist[i] * 1.0 / v;
s += i == n - 1 ? t : Math.ceil(t);
}
return s <= hour;
}
}class Solution {
public:
int minSpeedOnTime(vector<int>& dist, double hour) {
if (dist.size() > ceil(hour)) {
return -1;
}
const int m = 1e7;
int l = 1, r = m + 1;
int n = dist.size();
auto check = [&](int v) {
double s = 0;
for (int i = 0; i < n; ++i) {
double t = dist[i] * 1.0 / v;
s += i == n - 1 ? t : ceil(t);
}
return s <= hour;
};
while (l < r) {
int mid = (l + r) >> 1;
if (check(mid)) {
r = mid;
} else {
l = mid + 1;
}
}
return l > m ? -1 : l;
}
};func minSpeedOnTime(dist []int, hour float64) int {
if float64(len(dist)) > math.Ceil(hour) {
return -1
}
const m int = 1e7
n := len(dist)
ans := sort.Search(m+1, func(v int) bool {
v++
s := 0.0
for i, d := range dist {
t := float64(d) / float64(v)
if i == n-1 {
s += t
} else {
s += math.Ceil(t)
}
}
return s <= hour
}) + 1
if ans > m {
return -1
}
return ans
}function minSpeedOnTime(dist: number[], hour: number): number {
if (dist.length > Math.ceil(hour)) {
return -1;
}
const n = dist.length;
const m = 10 ** 7;
const check = (v: number): boolean => {
let s = 0;
for (let i = 0; i < n; ++i) {
const t = dist[i] / v;
s += i === n - 1 ? t : Math.ceil(t);
}
return s <= hour;
};
let [l, r] = [1, m + 1];
while (l < r) {
const mid = (l + r) >> 1;
if (check(mid)) {
r = mid;
} else {
l = mid + 1;
}
}
return l > m ? -1 : l;
}impl Solution {
pub fn min_speed_on_time(dist: Vec<i32>, hour: f64) -> i32 {
if dist.len() as f64 > hour.ceil() {
return -1;
}
const M: i32 = 10_000_000;
let (mut l, mut r) = (1, M + 1);
let n = dist.len();
let check = |v: i32| -> bool {
let mut s = 0.0;
for i in 0..n {
let t = dist[i] as f64 / v as f64;
s += if i == n - 1 { t } else { t.ceil() };
}
s <= hour
};
while l < r {
let mid = (l + r) / 2;
if check(mid) {
r = mid;
} else {
l = mid + 1;
}
}
if l > M {
-1
} else {
l
}
}
}/**
* @param {number[]} dist
* @param {number} hour
* @return {number}
*/
var minSpeedOnTime = function (dist, hour) {
if (dist.length > Math.ceil(hour)) {
return -1;
}
const n = dist.length;
const m = 10 ** 7;
const check = v => {
let s = 0;
for (let i = 0; i < n; ++i) {
const t = dist[i] / v;
s += i === n - 1 ? t : Math.ceil(t);
}
return s <= hour;
};
let [l, r] = [1, m + 1];
while (l < r) {
const mid = (l + r) >> 1;
if (check(mid)) {
r = mid;
} else {
l = mid + 1;
}
}
return l > m ? -1 : l;
};