Apple Interview Question for Backend Developers

Country: United States
Interview Type: Phone Interview

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of 3 vote

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For the first problem a hash set of {IPs} may simply solve it.

However in the follow-ups as the amount of data cumulates, if the IPs are stored as strings, 4GB is becomes absolutely insufficient, since there are over 4 billion IPs out there and each ip as a string takes at least 9 bytes.
Even if convert ip to integer(8B), it wouldn't be enough.

An approach is to go with bit set. Then each ip takes only 1 bit. Have a bit set to store whether an ip occurs and another set to store whether an ip repeats.

This takes 2 * 2^32 bit = ~1GB -> fits in RAM.

public class IPFilter {

    long[] map;          //mark all ip that showed up
    long[] repeatedIP;    //mark all ips that repeatedly showed up

    public IPFilter() {
        //there's 2^32 IP in total.
        // each long integer is identifies 64 IPs.
        // Need 2^32 / 2^6 long integers in the bit map
        int size = 1 << (32 - 6);
        map = new long[size];
        repeatedIP = new long[size];

    public void addToMap(List<String> IPs) {
        for(String ip: IPs) {
            long decimal = IPToLong(ip);
            int idx = (int)(decimal / 64);
            int res = (int)(decimal % 64);

            if((map[idx] >> res) == 1) {    //repeated ip
                repeatedIP[idx] |= (1 << res);
            } else {                        //first occurred ip
                map[idx] |= (1 << res);

    private long IPToLong(String ipAddress) { //convert ip (base 256) to decimal
        String[] ipAddressInArray = ipAddress.split("\\.");
        long result = 0;
        for (int i = 0; i < ipAddressInArray.length; i++) {
            int power = 3 - i;
            int ip = Integer.parseInt(ipAddressInArray[i]);
            result += ip * Math.pow(256, power);
        return result;

- aonecoding July 23, 2017 | Flag Reply
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of 0 vote

@aonecoding, I believe

if((map[idx] >> res) == 1

should be

if((map[idx] >> res) % 10 == 1

. This will get the bit value at location res from the bitset

- Arunava Saha November 08, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
of 1 vote

It is better to avoid writing a single line of code, when you can write it in 2 lines.
[ ]

./ | sort | uniq -u

and this, will run no matter how less memory you have.
Aws guys actually run it, so does azure folks. I am pretty sure Apple will do the same.

- NoOne July 23, 2017 | Flag Reply

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