Software Engineer · Phone Screen · Coding

OpenAI — Software Engineer

Level: Senior-Level

Round: Phone Screen · Type: Coding · Difficulty: 5/10 · Duration: 60 min · Interviewer: Neutral

Topics: Iterator, IPv4 Address

Location: San Francisco Bay Area

Interview date: 2026-01-20

Summary

I had to implement an IPv4 iterator with hasNext() and next() methods.

Details

The coding question I got was:

` class IPv4Iterator: def init(self, startIp): # Initializes the iterator with the startIp. # The first address yielded by the iterator must be exactly startIp. pass

def hasNext(self):
    # Returns true if there is at least one address left to be produced.
    # Otherwise, returns false.
    pass

def next(self):
    # Returns the current address as a string in dotted-decimal form,
    # then advances the internal cursor to the next sequential address.
    # It is guaranteed that next() is called only when hasNext() returns true.
    # The iterator visits addresses in strict ascending order (e.g., after "x.y.z.255" comes "x.y.(z+1).0").
    pass

`

Constraints:

• startIp is a valid IPv4 address string.
• 0 ≤ startIp ≤ 255.255.255.255.
• The number of next() calls will not exceed the remaining addresses in the range.

Example:

` Input: ["IPv4Iterator", "next", "next", "next", "hasNext"] [["255.255.255.253"], [], [], [], []]

Output: [null, "255.255.255.253", "255.255.255.254", "255.255.255.255", false]

Explanation:

IPv4Iterator iterator = new IPv4Iterator("255.255.255.253"); iterator.next(); // Returns "255.255.255.253". iterator.next(); // Returns "255.255.255.254". iterator.next(); // Returns "255.255.255.255". iterator.hasNext(); // Returns false. `