Find N Distinct Numbers Whose Bitwise OR is Equal to K in Python

Suppose we have two integers N and K; we have to find N unique values whose bit-wise OR is same as K. If there is no such result, then return -1

So, if the input is like N = 4 and K = 6, then the output will be [6,0,1,2].

To solve this, we will follow these steps −

• MAX := 32

• visited := a list of size MAX and fill with False

• res := a new list

• Define a function add() . This will take num

• point := 0

• value := 0

• for i in range 0 to MAX, do

  • if visited[i] is non-zero, then

    • go for next iteration

  • otherwise,

    • if num AND 1 is non-zero, then

      • value := value +(2^i)

    • num := num/2 (take only integer part)

• insert value at the end of res

• From the main method, do the following −

• pow2 := an array of power of 2 from 2^0 to 2^31

• insert k at the end of res

• cnt_k := number of bits in k

• if pow2[cnt_k] < n, then

  • return -1

• count := 0

• for i in range 0 to pow2[cnt_k] - 1, do

  • add(i)

  • count := count + 1

  • if count is same as n, then

    • come out from the loop

• return res

Example

Let us see the following implementation to get better understanding −

 Live Demo

MAX = 32
visited = [False for i in range(MAX)]
res = []
def set_bit_count(n):
   if (n == 0):
      return 0
   else:
      return (n & 1) + set_bit_count(n >> 1)
def add(num):
   point = 0
   value = 0
   for i in range(MAX):
      if (visited[i]):
         continue
      else:
         if (num & 1):
            value += (1 << i)
         num = num//2
   res.append(value)
def solve(n, k):
   pow2 = [2**i for i in range(MAX)]
   res.append(k)
   cnt_k = set_bit_count(k)
   if (pow2[cnt_k] < n):
      return -1
   count = 0
   for i in range(pow2[cnt_k] - 1):
      add(i)
      count += 1
      if (count == n):
         break
   return res

n = 4
k = 6
print(solve(n, k))

Input

4, 6

Output

[6, 0, 1, 2]