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Suppose we have a number start and another number end (start < end), we have to find the minimum number of operations required to convert start to end using these operations −

- Increment by 1
- Multiply by 2

So, if the input is like start = 5, end = 11, then the output will be 2, as we can multiply 2 to get 10, then add 1 to get 11.

To solve this, we will follow these steps −

- ct:= 0
- while end/2 >= start, do
- if end mod 2 is same as 1, then
- end := end - 1
- end := end/2
- ct := ct + 2

- otherwise,
- end:= end/2
- ct := ct + 1

- if end mod 2 is same as 1, then
- ct := ct +(end-start)
- return ct

Let us see the following implementation to get better understanding −

class Solution: def solve(self, start, end): ct=0 while(end/2>=start): if end%2==1: end-=1 end=end/2 ct+=2 else: end=end/2 ct+=1 ct+=(end-start) return ct ob = Solution() print(ob.solve(5,11))

5,11

2

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