Programming interview question

print_ipv4.js

'use strict'

// Take a valid octet from a string. Octet can be 1-3 characters
function takeOctet(octets, string) {
    // If we have 4 octets, and nothing is left on the string, we're done
    if (octets.length === 4) {
        if (string.length === 0) {
            console.log(`${octets[0]}.${octets[1]}.${octets[2]}.${octets[3]}`)
        }
        // Return anyway
        return
    }

    // Check for no possible solutions: the longest octet is 3 chars
    // String too long:
    if (string.length > (4 - octets.length) * 3) return
    // String too short:
    if (string.length < 4 - octets.length) return

    // Take an octet, 1-3 characters
    for (let n = 1; n < 4; ++n) {
        // Not enough characters, continue
        if (n > string.length) continue

        const octet = string.substring(0, n)

        // Recursively call takeOctet() if this octet is valid
        if (+octet < 256) {
            takeOctet(octets.concat(octet), string.substring(n))
        }
    }
}

function printIPs(string) {
    takeOctet([], string)
}

// Test case
console.log('\n\n999666\n------')
printIPs('999666')

task.md

Write a function that receives a string, which is an IP address without dots. Like this: "127001"

The function should print all valid IPs that you can get from that string by adding dots. For example, for "123123" there are many valid IPs:

• 123.1.2.3
• 1.23.12.3
• 1.2.31.23

...and so on. We need to print them all, in any order. For simplicity, let's say IP always has 3 dots.