Suppose we have an array nums. We can perform two types of operations on any element of the array any number of times
For even elements, divide it by 2
For odd elements, multiply it by 2.
Now the deviation of the array is the maximum difference between any two elements in the array. We have to find the minimum deviation the array can have after performing some number of operations. So, if the input is like nums = [6,3,7,22,5], then the output will be 5 because we can make our array in one operation [6,6,7,22,5] and in second operation [6,6,7,22,10], and in another operation [6,6,7,11,10], now the deviation is 11-6 = 5.
To solve this, we will follow these steps −
sort the list nums
max_v := maximum element of nums
min_v := minimum element of nums
heapify nums
res := max_v - min_v
while nums[0] is odd, do
v := poped element from heap queue nums
v := 2 * v
insert v into heap queue nums
min_v := nums[0]
max_v := maximum of v and max_v
res := minimum of res and (max_v - min_v)
nums := a list of all numbers in n and in negative order
heapify heap queue nums
while nums[0] is even, do
v := -(poped element from heap queue nums)
v := quotient of (v/2)
insert -v into the heap queue nums
max_v := -nums[0]
min_v := minimum of min_v and v
res := minimum of res and (max_v - min_v)
return res
Example
Let us see the following implementation to get better understanding
import heapq
def solve(nums):
nums.sort()
max_v,min_v = nums[-1],nums[0]
heapq.heapify(nums)
res = max_v-min_v
while nums[0]%2==1:
v = heapq.heappop(nums)
v = 2 * v
heapq.heappush(nums, v)
min_v = nums[0]
max_v = max(v, max_v)
res = min(res, max_v - min_v)
nums = [-n for n in nums]
heapq.heapify(nums)
while nums[0]%2==0:
v = -heapq.heappop(nums)
v = v // 2
heapq.heappush(nums, -v)
max_v = -nums[0]
min_v = min(min_v,v)
res = min(res, max_v - min_v)
return res
nums = [6,3,7,22,5]
print(solve(nums))Input
[6,3,7,22,5]
Output
5