Inserting M into N such that m starts at bit j and ends at bit i | Set-2

Given two 32-bit numbers, N and M, and two-bit positions, i and j. Write a method to insert M into N such that M starts at bit j and ends at bit i. You can assume that the bits j through i have enough space to fit all of M. Assuming index start from 0.
Examples:  

a)  N = 1024 (10000000000),
    M = 19 (10011),
    i = 2, j = 6 
    Output : 1100 (10001001100)
b)  N = 1201 (10010110001)
    M = 8 (1000)
    i = 3, j = 6
    Output: 1217 (10011000001)

This problem has been already discussed in the previous post. In this post, a different approach is discussed. 
Approach: The idea is very straightforward, following is the stepwise procedure –  

• Capture all the bits before i in N, that is from i-1 to 0.
• We don’t want to alter those bits so we will capture them and use later
• Clear all bits from j to 0
• Insert M into N at position j to i
• Insert captured bits at their respective position ie. from i-1 to 0

Let’s try to solve example b for a clear explanation of the procedure.

Capturing bits i-1 to 0 in N:
Create a mask whose i-1 to 0 bits are 1 and rest are 0. Shift 1 to i position in left and subtract 1 from this to get a bit mask having i-1 to 0 bits set.  

capture_mask = ( 1 << i ) - 1  

So for example b, mask will be –  

capture_mask = ( 1 << 3 ) - 1 
Now capture_mask = 1(001)

Now AND this mask with N, i-1 to 0 bits will remain same while rest bits become 0. Thus we are left with i-1 to 0 bits only.  

captured_bits = N & capture_mask

for example b, captured bits will be – 

captured_bits = N & capture_mask
Now capture_mask = 1 (001)

Clearing bits from j to 0 in N:
Since bits have been captured, clear the bits j to 0 without losing bits i-1 to 0. Create a mask whose j to 0 bits are 0 while the rest are 1. Create such mask by shifting -1 to j+1 position in left.  

clear_mask = -1 << ( j + 1 )

Now to clear bits j to 0, AND mask with N.  

N  &= clear_mask 

For example b will be: 

N &= clear_mask 
Now N = 1152 (10010000000)

Inserting M in N:

Now because N has bits from j to 0 cleared, just fit in M into N and shift M i position to left to align the MSB of M with position j in N.  

M <<= i

For example b-  

M <<= 3
Now M = 8 << 3 = 64 (1000000)

Do a OR of M with N to insert M at desired position.  

N |= M

For example b – 

N |= M 
Now N = 1152 | 64 = 1216 (10011000000) 

Inserting captured bits into N:

Till now M has been inserted into N. Now the only thing left is merging back the captured bits at position i-1 to 0. This can be simply done by OR of N and captured bits –  

N |= captured_bits

For example b –  

N |= captured_bits
N = 1216 | 1 = 1217 (10011000001)

So finally after insertion, the below bitset is obtained. 

N(before) = 1201 (10010110001)
N(after) = 1201 (10011000001)

Below is the implementation of the above approach:
 

N = 1201(10010110001)
M = 8(1000)
After inserting M into N from 3 to 6
N = 1217(10011000001)

Time complexity: O(log(M)+log(N))=O(log(M*N)
Auxiliary space: O(1)