90 lines
3.2 KiB
C++
90 lines
3.2 KiB
C++
// Source : https://leetcode.com/problems/minimum-speed-to-arrive-on-time/
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// Author : Hao Chen
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// Date : 2021-05-30
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/*****************************************************************************************************
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*
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* You are given a floating-point number hour, representing the amount of time you have to reach the
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* office. To commute to the office, you must take n trains in sequential order. You are also given an
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* integer array dist of length n, where dist[i] describes the distance (in kilometers) of the i^th
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* train ride.
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*
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* Each train can only depart at an integer hour, so you may need to wait in between each train ride.
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*
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* For example, if the 1^st train ride takes 1.5 hours, you must wait for an additional 0.5
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* hours before you can depart on the 2^nd train ride at the 2 hour mark.
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*
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* Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel
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* at for you to reach the office on time, or -1 if it is impossible to be on time.
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*
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* Tests are generated such that the answer will not exceed 10^7 and hour will have at most two digits
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* after the decimal point.
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*
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* Example 1:
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*
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* Input: dist = [1,3,2], hour = 6
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* Output: 1
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* Explanation: At speed 1:
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* - The first train ride takes 1/1 = 1 hour.
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* - Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second
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* train takes 3/1 = 3 hours.
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* - Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third
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* train takes 2/1 = 2 hours.
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* - You will arrive at exactly the 6 hour mark.
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*
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* Example 2:
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*
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* Input: dist = [1,3,2], hour = 2.7
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* Output: 3
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* Explanation: At speed 3:
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* - The first train ride takes 1/3 = 0.33333 hours.
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* - Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train
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* ride takes 3/3 = 1 hour.
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* - Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third
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* train takes 2/3 = 0.66667 hours.
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* - You will arrive at the 2.66667 hour mark.
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*
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* Example 3:
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*
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* Input: dist = [1,3,2], hour = 1.9
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* Output: -1
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* Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.
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*
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* Constraints:
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*
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* n == dist.length
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* 1 <= n <= 10^5
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* 1 <= dist[i] <= 10^5
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* 1 <= hour <= 10^9
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* There will be at most two digits after the decimal point in hour.
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******************************************************************************************************/
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class Solution {
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public:
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bool verify(vector<int>& dist, double hour, int speed) {
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double t = 0;
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int n = dist.size();
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for (int i=0; i<n-1; i++){
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t += (dist[i] + speed -1) / speed;
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}
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t += (double)dist[n-1]/speed;
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return t <= hour;
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}
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int minSpeedOnTime(vector<int>& dist, double hour) {
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int n = dist.size();
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if (hour <= n-1) return -1;
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int low = 1, high = 1e7;
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while (low < high) {
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int mid = low + (high - low) / 2;
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if (verify(dist, hour, mid)) {
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high = mid;
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} else {
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low = mid + 1;
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}
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}
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return high;
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}
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};
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