Merge pull request #89 from remy727/1870-minimum-speed-to-arrive-on-time

1870. Minimum Speed to Arrive on Time

Author: remy727

README.md

@@ -331,6 +331,7 @@
 | 1802 | Maximum Value at a Given Index in a Bounded Array | [Ruby](./algorithms/ruby/1802-maximum-value-at-a-given-index-in-a-bounded-array.rb) | Medium |
 | 1822 | Sign of the Product of an Array | [Ruby](./algorithms/ruby/1822-sign-of-the-product-of-an-array.rb) | Easy |
 | 1857 | Largest Color Value in a Directed Graph | [Ruby](./algorithms/ruby/1857-largest-color-value-in-a-directed-graph.rb) | Hard |
+| 1870 | Minimum Speed to Arrive on Time | [Ruby](./algorithms/ruby/1870-minimum-speed-to-arrive-on-time.rb) | Medium |
 | 1926 | Nearest Exit from Entrance in Maze | [Ruby](./algorithms/ruby/1926-nearest-exit-from-entrance-in-maze.rb) | Medium |
 | 1964 | Find the Longest Valid Obstacle Course at Each Position | [Ruby](./algorithms/ruby/1964-find-the-longest-valid-obstacle-course-at-each-position.rb) | Hard |
 | 1970 | Last Day Where You Can Still Cross | [Ruby](./algorithms/ruby/1970-last-day-where-you-can-still-cross.rb) | Hard |

algorithms/ruby/1870-minimum-speed-to-arrive-on-time.rb

@@ -0,0 +1,78 @@
+# frozen_string_literal: true
+
+# 1870. Minimum Speed to Arrive on Time
+# Medium
+# https://leetcode.com/problems/minimum-speed-to-arrive-on-time
+
+=begin
+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 1st train ride takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2nd train 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.length
+* 1 <= n <= 105
+* 1 <= dist[i] <= 105
+* 1 <= hour <= 109
+* There will be at most two digits after the decimal point in hour.
+=end
+
+# @param {Integer[]} dist
+# @param {Float} hour
+# @return {Integer}
+def min_speed_on_time(dist, hour)
+  return -1 if hour.ceil <= dist.size - 1
+
+  i, j = 1, 10**7 + 1
+
+  while i < j
+    mid = (i + j) / 2
+    time = dist[0..-2].sum { |d| (d + mid - 1) / mid } + dist[-1] / mid.to_f
+    if time > hour
+      i = mid + 1
+    else
+      j = mid
+    end
+  end
+
+  i == 10**7 + 1 ? -1 : i
+end
+
+# **************** #
+#       TEST       #
+# **************** #
+
+require "test/unit"
+class Test_min_speed_on_time < Test::Unit::TestCase
+  def test_
+    assert_equal 1, min_speed_on_time([1, 3, 2], 6)
+    assert_equal 3, min_speed_on_time([1, 3, 2], 2.7)
+    assert_equal(-1, min_speed_on_time([1, 3, 2], 1.9))
+    assert_equal(-1, min_speed_on_time([1, 1], 1.0))
+  end
+end