751. IP to CIDR - Detailed Explanation

Problem Statement

You are given a starting IP address (as a string) and a number n that represents how many consecutive IP addresses need to be covered. Your task is to return a list of CIDR (Classless Inter-Domain Routing) blocks that exactly cover the range of IP addresses starting from the given IP address.

A CIDR block is represented as "A.B.C.D/E", where E is the prefix length that indicates the number of fixed bits (from the left) in the address. For example, "255.0.0.7/32" covers exactly one IP address, while "255.0.0.8/29" covers 8 addresses.

Example 1

• Input:
  ip = "255.0.0.7", n = 10
• Output:
  ["255.0.0.7/32", "255.0.0.8/29", "255.0.0.16/32"]
• Explanation:
  1. The first CIDR block "255.0.0.7/32" covers 1 IP (255.0.0.7).
  2. The second block "255.0.0.8/29" covers 8 IPs (255.0.0.8 to 255.0.0.15).
  3. The final block "255.0.0.16/32" covers the last IP (255.0.0.16).
     In total, these blocks cover exactly 10 IP addresses.

Example 2

• Input:
  ip = "117.145.102.62", n = 5
• Output:
  ["117.145.102.62/32", "117.145.102.63/32", "117.145.102.64/31", "117.145.102.66/32"]
• Explanation:
  A valid answer covers exactly 5 addresses starting from 117.145.102.62. (Note: There can be more than one valid answer.)

Constraints

• The starting IP address is a valid IPv4 address.
• 1 ≤ n ≤ 10⁹

Hints and Intuition

• Hint 1:
  Convert the IP address from its dotted-decimal string format to a 32-bit integer. This makes it easier to perform arithmetic on the IP addresses.

• Hint 2:
  At any step, determine the largest CIDR block you can form from the current IP that does not exceed the remaining count n and is also aligned correctly (i.e., the block must start at an address that is a multiple of its size).

Detailed Approach

A. Converting IP to an Integer

Every IPv4 address has 4 parts (A.B.C.D). You can convert it to a 32-bit integer by:

• Shifting the first part (A) left by 24 bits,
• The second part (B) left by 16 bits,
• The third part (C) left by 8 bits,
• And then adding the fourth part (D).

For example, for "255.0.0.7":

ip_int = (255 << 24) | (0 << 16) | (0 << 8) | 7

B. Determining the Maximum Block Size at Each Step

When you have a starting integer start and a remaining count n:

1. Alignment Constraint:
   The largest block you can take is determined by the lowest set bit in start. For example, if start ends with 3 zeros (i.e., divisible by 8), the maximum block size that is aligned is 8.

   You can calculate this using:

   max_block = start & -start
   

2. Count Constraint:
   The block size must not exceed n. If max_block is greater than n, you need to choose the largest power of 2 less than or equal to n.

3. CIDR Prefix Calculation:
   If you choose a block size of size (which is a power of 2), the corresponding CIDR prefix length is:

   prefix_length = 32 - log2(size)
   

C. Forming the CIDR Block and Moving Forward

1. Convert the Integer Back to IP String:
   Convert start back to its dotted-decimal string by extracting each byte:

   • (start >> 24) & 255
   • (start >> 16) & 255
   • (start >> 8) & 255
   • start & 255

2. Build the CIDR Notation:
   Concatenate the IP string with "/" and the calculated prefix length.

3. Update for Next Block:

   • Subtract the chosen block size from n.
   • Increase start by the block size.

4. Repeat until n becomes zero.

Step-by-Step Walkthrough (Example 1: "255.0.0.7", n = 10)

1. Initialization:

   • Convert "255.0.0.7" to an integer (let’s call it start).
   • n is 10.

2. First Iteration:

   • Determine Alignment:
     Calculate start & -start to get the maximum aligned block. Suppose the lowest set bit gives 1 (meaning start is odd). Then the maximum block size by alignment is 1.
   • Count Check:
     The largest block you can take is min(1, 10) = 1.
   • CIDR Block:
     Block size 1 corresponds to prefix length 32 (because 2^(32-32) = 1).
     Create block: "255.0.0.7/32".
   • Update:
     Subtract 1 from n (now n = 9) and increase start by 1.

3. Second Iteration:

   • Now start is updated (which is now even) so that it is aligned to a larger power of 2.
   • Determine Alignment:
     Compute start & -start; assume it yields 8 (if start is divisible by 8).
   • Count Check:
     Compare 8 with n (9). The maximum block size that does not exceed n is 8.
   • CIDR Block:
     For block size 8, the prefix length is 32 - log2(8) = 29.
     Create block: for example, "255.0.0.8/29".
   • Update:
     Subtract 8 from n (now n = 1) and increase start by 8.

4. Third Iteration:

   • With n = 1, no matter the alignment, you can only take 1 IP.
   • CIDR Block:
     This block is "255.0.0.16/32".
   • Update:
     Now n becomes 0 and the process stops.

The final CIDR blocks are:
["255.0.0.7/32", "255.0.0.8/29", "255.0.0.16/32"].

Code Implementation

Python Implementation