Please help to solve this problem

general
2

I am pretty sure this problem is from one of the contests - Hackwithinfy and Infytq. It was one of the problems my friend had to solve. We later thought of upsolving it. I tried finding a pattern in the output - Wrote a brute-force solution and generated answers for smaller values of N. Luckily, I found the pattern on OEIS.

Here’s how it worked.

The sequence available on OEIS:

http://oeis.org/A064194

What you would like to stress more is this part: (The recurrence relation)

a(2n) = 3\times a(n),\ a(2n+1) = 2\times a(n+1)+a(n), with\ a(1) = 1.

The implementation.

Python Code 
import sys

sys.setrecursionlimit(10**6)

map = dict()
mod = 10**9 + 7

def solve(N):
    if N in map:
        return map[N]
    elif N == 0:
        return 0
    elif N == 1:
        return 1
    elif N % 2 == 0:
        ans = solve(N // 2)
        map[N] = (ans * 3) % mod
        return map[N]
    else:
        ans = 2 * solve(N // 2) + solve(N // 2 + 1)
        map[N] = ans % mod
        return map[N]

def main():
    N = int(input(), 2) + 1
    print(solve(N) % mod)


main()

This is a DP approach. Unfortunately, this approach was giving Segmentation fault in local machine for strings of length 10^6. Though it was still working fine with strings of length 10^5. The worst-case time it had taken to run against inputs of length 10^5 was around 1-2 seconds though.

Maybe, @cubefreak777 can help us with more optimal approach.

3

I hope this thing isn’t ongoing?

4

Yup, these are recruiting contests (I guess) and are over ( I am sure).

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5

I didn’t go through your solution can you tell me your solution’s complexity? And some samples would be great if you have any.
Also, do we have a report number of ordered pairs or unordered pairs?

6

I guess I have the problem statement.

You are given a positive Integer N in base two. Your task is to print the total number of pairs of non-negative integers (a, b) that satisfy the following conditions.

  • a + b \le N
  • a + b = a \oplus b

Here \oplus operation represents the bitwise XOR Operation.

Since the answer could be extremely large print it modulo 10^9 + 7

Note: The Input does not contain leading zeroes.

Input Format
The first line contains a string, N, denoting the Integer N in base two.

Constraints
1 \le |N| \le 10^5

Sample Input(s)

Input 1

10

Output 1

5

Explanation 1

Pairs satisfying the conditions are: (0, 0), (0, 1), (0, 2), (1, 0), (2, 0)

Input 2

100

Output 2

11
Explanation 2

There are many, but I guess you can understand

Input 3

1111111111

Output 3

59049

Explanation 3

Sample case for debugging purpose

My Approach (In between Naive and Optimal)

I read the given String into an Integer. In Python, int(input(), 2) converts the given binary string into an integer. Then, I applied the Recurrence relation using a Dictionary to store previously calculated values. But still, it didn’t work for large values of N. :slightly_frowning_face:

Coming to the time complexity, I guess it’s O(N\log N), where N is the length of Binary String.

7

For the first sample shouldn’t the answer be 7 ? or may be I got the question wrong.
[i,j] = [0, 0]
[i,j] = [0, 1]
[i,j] = [0, 2]
[i,j] = [1, 0]
[i,j] = [1, 2]
[i,j] = [2, 0]
[i,j] = [2, 1]
these pairs ?

8

The following is mentioned bro :sweat_smile:

  • a + b \le N
1 Like
9

Ah I see, my bad.

10

Sample test cases:
1)input : 10
Output: 2
Explanation : As decimal equivalent 10 is 2.Only pairs are (1,2) and (2,1)

2)input: 101
Output: 10
Explanation : As decimal equivalent 101 is 5.Few pairs are (1,4),(1,2) and (2,1)

12

I think you’re mistaken, a and b can be zero as well, according to the constraints given by Suman, also it makes more sense to include 0.

13

Sorry, I got a little caught up with something important so couldn’t reply. I think we can solve it with digit dp. I’m attaching the code it’s pretty neat to understand, though feel free to ping me if you need an explanation of my approach.

CODE 
#include "bits/stdc++.h"
#define ll long long   
using namespace std ;
const int M = 1e9+7,mxN=1e5;
int dp[mxN+1] ;
ll pM(ll b,int p){
  ll r=1 ;
  for(;p;b=(b*b)%M,p/=2)
    if(p&1)
      r=(r*b)%M ;
  return r ;
}
int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);     
  string s ; cin >> s ;
  int n = s.size() ;  
  dp[n]=1;
  for(int i=n-1,j=0;~i;--i,j++)
    dp[i]=(s[i]=='0'?dp[i+1]:(2*dp[i+1]+pM(3,j))%M)  ;
  cout << dp[0] ;
}

@anon41318104

1 Like
14

Please explain the approach .

15

hey @cubefreak777 how do you put your code in a drop-down way(▼)? using some latex code?

16
Summary

image

image
image850×301 16 KB

1 Like
17
Thanks

That’s a really cool thing. thanks a lot @cubefreak777 . btw got to know about your final exams from a thread. All the very best! :slightly_smiling_face:

1 Like
18

Thanks! :smile:

1 Like
19

If the i’th bit is set in the string then if you keep that bit unset you have 3 combinations for all remaining bits , those are {{0,1},{1,0},{0,0}}.
You cannot have {1,1} as its xor will be 0 whereas sum will be more than 0.
Where as to keep that bit set you have 2 combination available {{1,0},{0,1}}

2 Likes
20

What is the use of j in the function

21

Same question

22

NOTATION:

  1. S is the binary string given.
  2. N is length of S.
  3. (A,B) is any valid pair such that conditions given hold true.
  4. X = A+B=A \oplus B
  5. D(S) is the decimal representation of binary string S.
  6. C(i) is the number of set bits in i.
  7. S[i \dots N] is the suffix of S after index i.

We can reduce the problem to find the following summation, since it is given in the problem that A+B \le D \implies X \le D. So we just have sum up the number of pairs for all possible values of X. i.e. the following summation.

R= \sum_{X=0}^{D(S)} 2^{C(X)}

But how to find this? obviously, we can’t naively iterate overall all values from 0 to 2^N where N\le 10^5.
So we’ll use digit dp to find R.

DIGIT DP STATES AND TRANSITIONS:

DP[i] → \displaystyle \sum_{k=0}^{D(S[i \dots N])}2^{C(k)} i.e. Sum of 2^{C(k)} for all 0 \le k \le D(S[i \dots N]) in the suffix of S after index i.
Transitions are pretty straightforword.

IF S[i]=0:

Then it’d be the same as the answer we obtained at index i+1 (Since we have no option but to put a zero in here otherwise it’ll be greater than D(S[i \dots N])

\boxed{DP[i]=DP[i+1]}

IF S[i]=1:

Now we can put either a zero or a 1 at index i.
If we happen to put a 0 then for the entire suffix from i+1 to N we can have any number of set bits we want. So the number of numbers with k set bits would just be \displaystyle \binom{N-1-i}{k} but since the contribution of each number with k set bits is 2^k so total contribution would be \displaystyle \sum_{k=0}^{N-1-i}\binom{N-1-i}{k}2^k
But what if we put a 1 at index i then it is clearly evident that it is the sub subproblem for suffix from i+1 that we solved already i.e. DP[i+1] but with one extra set bit so the contribution from this subpart will be 2 \times DP[i+1]. So the transition will look like.

DP[i]=2 \times DP[i+1]+\sum_{k=0}^{N-1-i}\binom{N-1-i}{k}2^k \\ \sum_{k=0}^{N-1-i}\binom{N-1-i}{k}2^k=(1+2)^{N-1-i}=3^{N-1-i} \\ \boxed{DP[i]=2 \times DP[i+1]+3^{N-1-i}}

Base case is DP[N]=1 and final answer R would just be DP[0].

ASYMPOTICS.

Time & Space complexity for this solution is \mathcal{O}(N)

CODE FOR REFERENCE 
#include "bits/stdc++.h"
#define ll long long   
using namespace std ;
const int M = 1e9+7,mxN=1e7;
int dp[mxN+1] ;
ll pM(ll b,int p){
  ll r=1 ;
  for(;p;b=(b*b)%M,p/=2)
    if(p&1)
      r=(r*b)%M ;
  return r ;
}
int main(){
  string s ; cin >> s ;
  int n = s.size() ;  
  dp[n]=1;
  for(int i=n-1;~i;--i)
    dp[i]=(s[i]=='0'?dp[i+1]:(2*dp[i+1]+pM(3,n-1-i))%M)  ;
  cout << dp[0] ;
}

Solutions that involve converting the entire string to integer as presented by @suman_18733097, won’t work for strings greater than length 10^4(thanks to python, C++ would’ve failed to handle strings of length 60 ) it is because you’re essentially storing 2^{10^5} in int data type which is way way too much!!.

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