 #1

# Problem Link

Practice

Contest

Author: Noszály Áron

Tester: Misha Chorniy

Editorialist: Bhuvnesh Jain

EASY

# Prerequisites

BFS/Floyd Warshall, Observations

# Problem

You are to convert number A to number B using the following operation in the minimum number of steps:

• Write A as a binary number with arbitrary number of leading zeros (possibly none)
• Shuffle the binary digits of the A in an arbitrary order. Call it A'
• The output is (A' + 1). Let us call this number as X.

Remember the notation of A' and X as it is used everywhere in the editorial.

# Explanation

## Subtask 1: A, B ≤ 128

The problem asks to minimise the number of steps to go from A to B. Generally, these type of problems can be easily modelled as a graph and perform BFS to find the answer. Here is a general idea:

Let us model each step as an edge from initial number to final number with an edge of some cost. So, the path with the smallest weight from source to destination is the required answer and also gives us a way to achieve the transition.

In this problem, we can just do what the operations ask us to do. Just all possible shuffling of the digits of A in binary representation and consider only those ones which give end result within the desired range ([0, 128]). Let us call this number as X. Add an edge of weight 1 from A to X. Build the complete graph based on all possible transitions. Make sure that this graph will be directed one as transforming from A to X might not be the same as transforming from X to A due to the last operation which adds 1 to the result. Once, we have the directed graph, we can simply perform any all pair shortest path algorithm like BFS, Floyd Warshall (or Dynamic Programming as a graph is directed) and precalculate the answer for all possible cases. Once, the answers are precalculated, each test case can be answered in constant complexity. For more details, one can refer to the editorialist solution below.

The maximum number of edges from a number A emerging out will be 7! = 5040 in the worst case, but in practice, it will be quite less. The number of vertices in the graph will be 128. Using Floyd Warshall algorithm the precomputation can be achieved in O(128^3), i.e. O(A^3). This is enough to solve this subtask.

## Subtask 2: A, B ≤ {10}^{18}

The constraints in this subtask are quite large and the number of test cases also suggest we need a logarithmic approach. This leads us to write A and B in binary representation and finding some observations to proceed with the solution.

Let us first simplify the approach by trying to convert A to (B-1) as in the end 1 would be added as part of the last operation.

We can see that in one step we can shuffle the digits of A in such a manner that an extra 1 is introduced in the binary representation of the newly formed number. This can be done by simply shuffling the digits to contain binary digit 1 from the {2}^{nd} position. For example: Let A = 3. In binary representation A = 011. We can shuffle the digits as A' = 110. On adding 1, we get X = 111, i.e. X = 7. Thus, we can introduce a binary digit 1 in one step.

Also, we can decrease any number of binary digit 1 from A. This can be done by easily placing the required in the number of 1 in towards the end, followed by a 0 and then placing the digits in any order we like. For example: Let A = 13, i.e. A = 1101 in binary representation. Say we want to decrease one binary digit 1, we can arrange the digits as A' = 1011. On adding 1, we get X = 1100. If we wanted to decrease two binary digit 1, we can arrange the digits as A' = 0111. On adding 1, we get X = 1000. Thus, we decrease any number of binary digit 1 in one step.

With the above 2 scenarios, the following is the algorithm-

1. Find the number of ones in the binary representation of A and (B - 1). Let us denote then by OA and OB
2. If OA > OB, then we can achieve the operation in 2 steps. First decreasing the number of ones in the first step and then rearranging the digits in another step.
3. If OA <= OB, then we need (OB - OA) steps to first make the number of ones equal (see decrease operation takes place at one step each). Finally, we can arrange the digits to achieve the desired number.

The only corner case is as follows:

1. If B is 0 then, we can’t achieve the desired state whatever the value of A is. Since we are adding 1 in the last step we are guaranteed to have atleast one binary digit 1 in binary representation. Thus the answer for this case is -1.
2. If B is 1 then, we can achieve the desired state only if A = 0, in one step. In another case, it is impossible as even though decreasing the number of binary digit 1, the end result would be a number greater than 1. So the answer is 1 if A = 0, else -1.

The number of ones in binary representation can be calculated using the below pseudo-code:

``````
def count_ones(integer x):
ones = 0
while x > 0:
if x % 2 == 1:
ones += 1
x /= 2
return ones
```
```

The time complexity of the above pseudo-code will be O(\log{x}) as each iteration of the while loop decreases the value of x by 2.

Feel free to share your approach, if it was somewhat different.

# Time Complexity

O(\log{A} + \log{B}) per test case.

# Space Complexity

O(1)

### AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here. (It will pass subtask 1 only and uses dynammic programming approach)

Tester’s solution can be found here.

Editorialist’s solution for subtask 1 can be found here.

Editorialist’s solution for full problem can be found here.

#2

### Here : https://www.codechef.com/viewsolution/18719196

##Is a commented solution of this problem till soln gets linked…
I am in div1 though had written one for discuss Feel free to ask queries…
PS: The code is simple enough to understand and it ll help…

#3

You can use __builtin_popcountll(num) to find the number of 1s in binary rep of num.

https://www.codechef.com/viewsolution/18766756

Let me know if anyone has any query regarding the solution.

#4

Hello!
I understand this editorial but can someone help me through this example.
Let us say I got A =9 (1001)
and B =16 (10000)
So B-1=1111 in binary
and the steps are.

1001
Becomes 1100 (reshuffled) ,add one to get 1101,this was Step 1.
Next reshuffle this to get 1110,and then add 1,you get 1111,and this was step 2.

but this is B-1 but not B and I have completed OB-OA steps!?

#5

can someone tell me in the first testcase A=2 and B=4 , why we can not shuffle 010 to 100 in one step??

#6

Pretty straightforward solution with a slight modification. You can use the Brian Kernighan’s Algorithm for calculating set bits. It has a runtime of O(A) + O(B). counting set bits

#7

can you tell the mistake in my solution, its giving TLE.
https://www.codechef.com/viewsolution/18854022. Thanks in advance!

#8

yeah true… you need OB-OA+1 steps
to get B from A when
Cnt(A)<=Cnt(B-1)

#9

look at my solution it has AC… I printed OB-OA+1

#10

and that is what editorialist meant to say…

#11

Thanks Man! Appreciated!

#12

thanks and welcome #13

You have to add 1 as well in last step.

#14

You can but at the end you will have to add 1, and thus the final number will be 4 + 1 = 5. Read the problem again carefully. You can do arbitrary shuffles(including 0) and arbitrary prefixing of 0s(including 0) but for each operation you must add 1.

The alternate way to think of how to solve this problem is that you must get B - 1 from A(because in the last operation you will add 1 to it, getting B - 1 + 1 = B). Now to get B - 1 from A, they must have the same number of set bits. Now how to make the set bits same? This has been explained in the editorial well. Read the editorial carefully.

#15

Great one !!!

#16

Thanks .

#17

Thanks for upvoting guys… specially @iamjoker I got 10 from u at last and now I can edit wiki questions… #18

Check for infinite loops

#19

@shivan111, the logic mentioned is exactly the same as one used in fenwick trees. But your complexity analysis is wrong. it should be O(\log{A} + \log{B}). It is just that constant factor of the approach you mentioned is smaller.

#20

@poseidon_22, also there is an issue with the data types used by you in the solution. There will be overflow issues which can lead to unexpected behaviour in your code.