## Monthly Archives: September 2023

## C# || How To Find The Minimum Speed To Arrive On Time Using C#

The following is a module with functions which demonstrates how to find the minimum speed to arrive on time using C#.

1. Min Speed On Time – Problem Statement

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 **i ^{th}** 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
**1**train ride takes^{st}**1.5**hours, you must wait for an additional**0.5**hours before you can depart on the**2**train ride at the 2 hour mark.^{nd}

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 **10 ^{7}** 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.

2. Min Speed On Time – Solution

The following is a solution which demonstrates how to find the minimum speed to arrive on time.

```
```
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// ============================================================================// Author: Kenneth Perkins// Date: Sep 1, 2023// Taken From: http://programmingnotes.org/// File: Solution.cs// Description: Demonstrates how to find the minimum speed to arrive on time// ============================================================================public class Solution { public int MinSpeedOnTime(int[] dist, double hour) { int left = 1; int right = 10000000; int minSpeed = -1; while (left <= right) { int mid = left + (right - left) / 2; // Move to the left half. if (TimeRequired(dist, mid) <= hour) { minSpeed = mid; right = mid - 1; } else { // Move to the right half. left = mid + 1; } } return minSpeed; } double TimeRequired(int[] dist, int speed) { double time = 0.0; for (int i = 0 ; i < dist.Length; i++) { double t = (double) dist[i] / (double) speed; // Round off to the next integer, if not the last ride. time += (i == dist.Length - 1 ? t : Math.Ceiling(t)); } return time; }}// http://programmingnotes.org/

**QUICK NOTES**:

The highlighted lines are sections of interest to look out for.

The code is heavily commented, so no further insight is necessary. If you have any questions, feel free to leave a comment below.

Once compiled, you should get this as your output for the example cases:

1

3

-1