Suppose we have a list of values called tasks where each different value represents a different task type, and we also have a non-negative integer k. Each task wants one minute to complete, but we must wait k minutes between doing two tasks of the same type. At any time, we can be doing a task or waiting. We have to find the smallest amount of time it takes to complete all the tasks.
So, if the input is like nums = [2, 2, 2, 3, 3, 2], k = 1, then the output will be 7, as the optimal ordering is [2, 3, 2, 3, 2, WAITING, 2].
To solve this, we will follow these steps −
c := count of all values in nums
ans := 0, lastsize := 0
while c is non−zero, do
lastsize := size of c
for each value x in most common (k + 1) values of c, do
c[x] := c[x] − 1
if c[x] is same as 0, then
ans := ans + k + 1
return ans + lastsize - (k + 1)
Let us see the following implementation to get better understanding −
class Solution: def solve(self, nums, k): from collections import Counter c = Counter(nums) ans = 0 lastsize = 0 while c: lastsize = len(c) for x, _ in c.most_common(k + 1): c[x] -= 1 if c[x] == 0: del c[x] ans += k + 1 return ans + lastsize - (k + 1) ob1 = Solution() nums = [2, 2, 2, 3, 3, 2] k = 1 print(ob1.solve(nums, k))
[2, 2, 2, 3, 3, 2], 1