2462. Total Cost to Hire K Workers

Author: remy727

README.md

@@ -231,6 +231,7 @@
 | 2413 | Smallest Even Multiple | [Ruby](./algorithms/ruby/2413-smallest-even-multiple.rb) | Easy |
 | 2439 | Minimize Maximum of Array | [Ruby](./algorithms/ruby/2439-minimize-maximum-of-array.rb) | Medium |
 | 2444 | Count Subarrays With Fixed Bounds | [Ruby](./algorithms/ruby/2444-count-subarrays-with-fixed-bounds.rb) | Hard |
+| 2462 | Total Cost to Hire K Workers | [Python3](./algorithms/python3/2462-total-cost-to-hire-k-workers.py) | Medium |
 | 2466 | Count Ways To Build Good Strings | [Ruby](./algorithms/ruby/2466-count-ways-to-build-good-strings.rb) | Medium |
 | 2477 | Minimum Fuel Cost to Report to the Capital | [Ruby](./algorithms/ruby/2477-minimum-fuel-cost-to-report-to-the-capital.rb) | Medium |
 | 2492 | Minimum Score of a Path Between Two Cities | [Ruby](./algorithms/ruby/2492-minimum-score-of-a-path-between-two-cities.rb) | Medium |

algorithms/python3/2462-total-cost-to-hire-k-workers.py

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+# 2462. Total Cost to Hire K Workers
+# https://leetcode.com/problems/total-cost-to-hire-k-workers
+# Medium
+
+from heapq import heapify, heappush, heappop
+
+class Solution:
+  def totalCost(self, costs: list[int], k: int, candidates: int) -> int:
+    q = costs[:candidates]
+    qq = costs[max(candidates, len(costs)-candidates):]
+    heapify(q)
+    heapify(qq)
+    ans = 0
+    i, ii = candidates, len(costs)-candidates-1
+    for _ in range(k):
+      if not qq or q and q[0] <= qq[0]:
+        ans += heappop(q)
+        if i <= ii:
+          heappush(q, costs[i])
+          i += 1
+      else:
+        ans += heappop(qq)
+        if i <= ii:
+          heappush(qq, costs[ii])
+          ii -= 1
+    return ans
+
+# ********************#
+#       TEST         #
+# ********************#
+
+import unittest
+
+class TestMinimumTime(unittest.TestCase):
+  def test_minimumTime(self):
+    self.assertEqual(Solution.totalCost(self, [17,12,10,2,7,2,11,20,8], 3, 4), 11)
+    self.assertEqual(Solution.totalCost(self, [1,2,4,1], 3, 3), 4)
+
+if __name__ == '__main__':
+  unittest.main()

algorithms/ruby/2462-total-cost-to-hire-k-workers.rb

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+# frozen_string_literal: true
+
+# 2462. Total Cost to Hire K Workers
+# https://leetcode.com/problems/total-cost-to-hire-k-workers
+# Medium
+# TODO: Implement
+
+=begin
+You are given a 0-indexed integer array costs where costs[i] is the cost of hiring the ith worker.
+
+You are also given two integers k and candidates. We want to hire exactly k workers according to the following rules:
+
+You will run k sessions and hire exactly one worker in each session.
+In each hiring session, choose the worker with the lowest cost from either the first candidates workers or the last candidates workers. Break the tie by the smallest index.
+For example, if costs = [3,2,7,7,1,2] and candidates = 2, then in the first hiring session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2].
+In the second hiring session, we will choose 1st worker because they have the same lowest cost as 4th worker but they have the smallest index [3,2,7,7,2]. Please note that the indexing may be changed in the process.
+If there are fewer than candidates workers remaining, choose the worker with the lowest cost among them. Break the tie by the smallest index.
+A worker can only be chosen once.
+Return the total cost to hire exactly k workers.
+
+Example 1:
+Input: costs = [17,12,10,2,7,2,11,20,8], k = 3, candidates = 4
+Output: 11
+Explanation: We hire 3 workers in total. The total cost is initially 0.
+- In the first hiring round we choose the worker from [17,12,10,2,7,2,11,20,8]. The lowest cost is 2, and we break the tie by the smallest index, which is 3. The total cost = 0 + 2 = 2.
+- In the second hiring round we choose the worker from [17,12,10,7,2,11,20,8]. The lowest cost is 2 (index 4). The total cost = 2 + 2 = 4.
+- In the third hiring round we choose the worker from [17,12,10,7,11,20,8]. The lowest cost is 7 (index 3). The total cost = 4 + 7 = 11. Notice that the worker with index 3 was common in the first and last four workers.
+The total hiring cost is 11.
+
+Example 2:
+Input: costs = [1,2,4,1], k = 3, candidates = 3
+Output: 4
+Explanation: We hire 3 workers in total. The total cost is initially 0.
+- In the first hiring round we choose the worker from [1,2,4,1]. The lowest cost is 1, and we break the tie by the smallest index, which is 0. The total cost = 0 + 1 = 1. Notice that workers with index 1 and 2 are common in the first and last 3 workers.
+- In the second hiring round we choose the worker from [2,4,1]. The lowest cost is 1 (index 2). The total cost = 1 + 1 = 2.
+- In the third hiring round there are less than three candidates. We choose the worker from the remaining workers [2,4]. The lowest cost is 2 (index 0). The total cost = 2 + 2 = 4.
+The total hiring cost is 4.
+
+Constraints:
+1 <= costs.length <= 105
+1 <= costs[i] <= 105
+1 <= k, candidates <= costs.length
+=end
+
+# @param {Integer[]} costs
+# @param {Integer} k
+# @param {Integer} candidates
+# @return {Integer}
+def total_cost(costs, k, candidates)
+end
+
+# ********************#
+#       TEST         #
+# ********************#
+
+require "test/unit"
+class Test_total_cost < Test::Unit::TestCase
+  def test_
+    assert_equal 11, total_cost([17,12,10,2,7,2,11,20,8], 3, 4)
+    assert_equal 4, total_cost([1,2,4,1], 3, 3)
+  end
+end