Print number of 1s in a sequence up to a number, without actually counting 1s [closed]

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An interview question:

Make a program which takes input 'N'(unsigned long) and prints two columns, 1st column prints numbers from 1 to N (in hexadecimal format) and second column prints the number of 1s in the binary representation of the number in the left column. Condition is that this program should not count 1s (so no computations 'per number' to get 1s/ no division operators).

I tried to implement this by leveraging fact that No of 1s in 0x0 to 0xF can be re-used to generate 1s for any number. I am pasting code ( basic one without error checking.) Its giving correct results but I am not happy with space usage. How can I improve on this? ( Also I am not sure if its what interviewer was looking for).

void printRangeFasterWay(){

    uint64_t num = ~0x0 ;
    cout << " Enter upper number " ;
    cin >> num ;

    uint8_t arrayCount[] = { 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4} ;
    // This array will store information needed to print 
    uint8_t * newCount = new uint8_t[num] ;
    uint64_t mask = 0x0 ;
    memcpy(newCount, &arrayCount[0], 0x10) ; 

    uint64_t lower = 0;
    uint64_t upper = 0xF;
    uint64_t count = 0 ;
    uint32_t zcount= 0 ; 

    do{
      upper = std::min(upper, num) ;

      for(count = lower ; count <= upper ; count++){
         newCount[count] = (uint32_t)( newCount[count & mask] + newCount[(count & ~mask)>>(4*zcount)]) ;
      }
      lower += count  ; 
      upper |= (upper<<4) ;
      mask =   ((mask<<4) | 0xF ) ;
      zcount++ ;
    }while(count<=num) ;

    for(uint64_t xcount=0 ; xcount <= num ; xcount++){
       cout << std::hex << " num = " << xcount << std::dec << "   number of 1s = " << (uint32_t)newCount[xcount] << endl;
    }

}

Edited to add sample run

Enter upper number 18
 num = 0   number of 1s = 0
 num = 1   number of 1s = 1
 num = 2   number of 1s = 1
 num = 3   number of 1s = 2
 num = 4   number of 1s = 1
 num = 5   number of 1s = 2
 num = 6   number of 1s = 2
 num = 7   number of 1s = 3
 num = 8   number of 1s = 1
 num = 9   number of 1s = 2
 num = a   number of 1s = 2
 num = b   n开发者_如何学Pythonumber of 1s = 3
 num = c   number of 1s = 2
 num = d   number of 1s = 3
 num = e   number of 1s = 3
 num = f   number of 1s = 4
 num = 10   number of 1s = 1
 num = 11   number of 1s = 2
 num = 12   number of 1s = 2

---

I have a slightly different approach which should solve your memory problem. Its based on the fact that the bitwise operation i & -i gives you the smallest power of two in the number i. For example, for i = 5, i & -i = 1, for i = 6, i & -i = 2. Now, for code:

void countBits(unsigned N) {
   for (int i = 0;i < N; i ++)
   {
       int bits = 0;
       for (int j = i; j > 0; j= j - (j&-j))
           bits++;
       cout <<"Num: "<<i <<" Bits:"<<bits<<endl;
   }
}

I hope I understood your question correctly. Hope that helps

Edit: Ok, try this - this is dynamic programming without using every bit in every number:

void countBits(unsigned N) {
   unsigned *arr = new unsigned[N + 1];
   arr[0]=0;
   for (int i = 1;i <=N; i ++)
   {
       arr[i] = arr[i - (i&-i)] + 1;
   }
   for(int i = 0; i <=N; i++)
    cout<<"Num: "<<i<<" Bits:"<<arr[i]<<endl;
}

Hopefully, this works better

---

Several of the answers posted so far make use of bit shifting (just another word for division by 2) or bit masking. This stikes me as a bit of a cheat. Same goes for using the '1' bit count in a 4 bit pattern then matching by chunks of 4 bits.

How about a simple recursive solution using an imaginary binary tree of bits. each left branch contains a '0', each right branch contains a '1'. Then do a depth first traversal counting the number of 1 bits on the way down. Once the bottom of the tree is reached add one to the counter, print out the number of 1 bits found so far, back out one level and recurse again.

Stop the recursion when the counter reaches the desired number.

I am not a C/C++ programmer, but here is a REXX solution that should translate without much imagination. Note the magic number 32 is just the number of bits in an Unsigned long. Set it to anything

/* REXX */

SAY 'Stopping number:'
pull StopNum

Counter = 0
CALL CountOneBits 0, 0
return

CountOneBits: PROCEDURE EXPOSE Counter StopNum
ARG Depth, OneBits

   If Depth = 32 then Return              /* Number of bits in ULong */
   if Counter = StopNum then return       /* Counted as high as requested */
   call BitCounter Depth + 1, OneBits     /* Left branch is a 0 bit */
   call BitCounter Depth + 1, OneBits + 1 /* Right branch is a 1 bit */
   Return

BitCounter: PROCEDURE EXPOSE Counter StopNum
ARG Depth, OneBits

   if Depth = 32 then do            /* Bottom of binary bit tree */
      say D2X(Counter) 'contains' OneBits 'one bits'
      Counter = Counter + 1
      end
   call CountOneBits Depth, OneBits
  return    

Results:

Stopping number:
18
0 contains 0 one bits
1 contains 1 one bits
2 contains 1 one bits
3 contains 2 one bits
4 contains 1 one bits
5 contains 2 one bits
6 contains 2 one bits
7 contains 3 one bits
8 contains 1 one bits
9 contains 2 one bits
A contains 2 one bits
B contains 3 one bits
C contains 2 one bits
D contains 3 one bits
E contains 3 one bits
F contains 4 one bits
10 contains 1 one bits
11 contains 2 one bits

This answer is resonably efficient in time and space.

---

Can be done relatively trivially in constant time with the appropriate bit switching. No counting of 1s and no divisions. I think you were on the right track with keeping the array of known bit values:

int bits(int x)
{
   // known bit values for 0-15
   static int bc[16] = {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};

   // bit "counter"
   int b = 0;
   // loop iterator
   int c = 0;

   do
   {
      // get the last 4 bits in the number
      char lowc = static_cast<char>(x & 0x0000000f);

      // find the count
      b += bc[lowc];

      // lose the last four bits
      x >>= 4;

      ++c;

      // loop for each possible 4 bit combination,
      // or until x is 0 (all significant bits lost)
   }
   while(c < 8 && x > 0);

   return b;
}

---

Explanation

The following algorithm is like yours, but expands on the idea (if I understood your approach correctly.) It does not do any computation 'per number' as directed by the question, but instead uses a recursion that exists between sequences of lengths that are powers of 2. Basically, the observation is that for the sequence 0, 1,..,2^n-1 , we can use the sequence 0, 1, ...,2^(n-1)-1 in the following way.

Let f(i) be the number of ones in number i then f(2^(n-1)+i)=f(i)+1 for all 0<=i<2^(n-1). (Verify this for yourself)

Algorithm in C++

#include <stdio.h>
#include <stdlib.h>

int main( int argc, char  *argv[] )
{
   const int N = 32;

   int* arr = new int[N];
   arr[0]=0;
   arr[1]=1;
   for ( int i = 1; i < 15; i++ )
   {
      int pow2 = 1 << i;
      int offset = pow2;
      for ( int k = 0; k < pow2; k++ )
      {
         if ( offset+k >= N )
            goto leave;
         arr[offset+k]=arr[k]+1;
      }
   }

leave:

   for ( int i = 0; i < N; i++ )
   {
      printf( "0x%8x %16d", i, arr[i] );
   }

   delete[] arr;
   return EXIT_SUCCESS;
}

Note that in the for loop

   for ( int i = 0; i < 15; i++ )

there may be overflow into negative numbers if you go higher than 15, otherwise use unsigned int's if you want to go higher than that.

Efficiency

This algorithm runs in O(N) and uses O(N) space.

---

Here is an approach that has O(nlogn) time complexity and O(1) memory usage. The idea is to get the Hex equivalent of the number and iterate over it to get number of ones per Hex digit.

int oneCount[] = { 0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};

int getOneCount(int n)
{
  char inStr[70];
  sprintf(inStr,"%X",n);
 int i;
int sum=0;
for(i=0; inStr[i];i++)
{
 if ( inStr[i] > '9' )
  sum += oneCount[inStr[i]-'A' + 10];
else
 sum+= oneCount[inStr[i] -'0'];
}
return sum;

}

int i,upperLimit;
cin>>upperLimit;

for(i=0;i<=upperLimit;i++)
{
 cout << std::hex << " num = " << i << std::dec << "   number of 1s = " << getOneCount(i) << endl;

}

---

enum bit_count_masks32
{
    one_bits= 0x55555555, // 01...
    two_bits= 0x33333333, // 0011...
    four_bits= 0x0f0f0f0f, // 00001111....
    eight_bits= 0x00ff00ff, // 0000000011111111...
    sixteen_bits= 0x0000ffff, // 00000000000000001111111111111111
};

unsigned int popcount32(unsigned int x)
{
    unsigned int result= x;
    result= (result & one_bits) + (result & (one_bits << 1)) >> 1;
    result= (result & two_bits) + (result & (two_bits << 2)) >> 2;
    result= (result & four_bits) + (result & (four_bits << 4)) >> 4;
    result= (result & eight_bits) + (result & (eight_bits << 8)) >> 8;
    result= (result & sixteen_bits) + (result & (sixteen_bits << 16)) >> 16;

    return result;
}

void print_range(unsigned int low, unsigned int high)
{
    for (unsigned int n= low; unsigned int n<=high; ++n)
    {
        cout << std::hex << " num = " << xcount << std::dec << "   number of 1s = " << popcount32(n) << endl;
    }
}