Google Interview Question for Software Developers


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
3
of 3 vote

You can rotate all strings in the input list such that they all begin with the character "A". Afterwards, it's a simple matter of printing all pairs of duplicate strings (mapped to their original string).

This has a runtime of O(n).

def rotatate_to_A(str):
    if len(str) == 0:
        return ""
    delta = ord('A') - ord(str[0])
    output = ""
    for char in str:
        new_char = ord(char) + delta
        if new_char < ord('A'):
            new_char += 26
        output += chr(new_char)

    return output

def rot_finder(lst):
    rotated_lst = [rotatate_to_A(str) for str in lst]

    dict = {}
    output = []

    for i in range(len(rotated_lst)):
        rotated_word = rotated_lst[i]
        if rotated_word in dict:
            value = dict[rotated_word]
            for index in value:
                original_word_1 = lst[index]
                original_word_2 = lst[i]
                output.append([original_word_1, original_word_2])
            value.append(i)
            dict[rotated_word] = value
        else:
            dict[rotated_word] = [i]

    return output

- Anonymous May 04, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

I think that the output should also contain [bc, za] as it corresponds to ROT_2(ZA) = BC

public class RotFinder {
    String alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    Set<Pair<String>> findRotationPairs(List<String> input) {
        Set<Pair<String>> pairs = new HashSet<>();
        for (int i = 0 ; i < input.size() ; i++) {
            for (int j = i + 1 ; j < input.size() ; j++) {
                String source = input.get(i);
                String target = input.get(j);
                if (source.length() < 1) {
                    break;
                }
                if (source.length() == target.length()) {
                    char sourceLetter= source.charAt(0);
                    char targetLetter = target.charAt(0);
                    int rot = findRot(sourceLetter, targetLetter);
                    boolean recordPair = false;
                    for (int k = 1 ; k < source.length() ; k++) {
                        sourceLetter = source.charAt(k);
                        targetLetter = target.charAt(k);
                        int sourceIndex = alpha.indexOf(sourceLetter);
                        int targetIndex = (sourceIndex + rot) % 26;
                        if (targetLetter == alpha.charAt(targetIndex)) {
                            if (k == source.length() - 1) {
                                recordPair = true;
                            }
                        } else {
                            break;
                        }
                    }
                    if (recordPair) {
                        pairs.add(new Pair<>(source, target));
                    }
                }
            }
        }
        return pairs;
    }

    int findRot(char source, char target) {
        for (int i = 0 ; i < 26 ; i++) {
            char checked = alpha.charAt((alpha.indexOf(source) + i) % 26);
            if (checked == target) {
                return i;
            }
        }
        return -1;
    }

    class Pair<T> {
        T first;
        T second;

        Pair(T first, T second) {
            this.first = first;
            this.second = second;
        }

        public String toString() {
            return "[" + first + ", " + second + "] ";
        }
    }

    public static void main (String[] args) {
        List<String> strings = new LinkedList<>();
        strings.add("AB");
        strings.add("BC");
        strings.add("FOO");
        strings.add("ZA");
        strings.add("BAZ");
        strings.add("IRR");
        System.out.println(Arrays.toString(new RotFinder().findRotationPairs(strings).toArray()));
    }

}

- vroom April 22, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

You don't need to find rotation n all. Just find index of both character
Say cik and cjk obtained at index i and j of kth item.
Index of the is i1 and i2.
If i2 < i1 then
Return 26-(i1-i2)
Otherwise
Return i1-i2

- nitinguptaiit April 16, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Complexity O(L* n^2)
L is max length of word.

- nitinguptaiit April 16, 2019 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

Maybe something like this?

def source_and_rotation(words: List[str]) -> List[List[str]]:

    def rot_char_by(c, i):
        shift = ord('A')
        return chr((((ord(c) - shift) + i) % N) + shift)

    N = len(string.ascii_lowercase)

    transformed_to_origins = defaultdict(set)

    for w in words:
        for i in range(1, N):
            transformed_to_origins["".join([rot_char_by(c, i) for c in w])].add(w)

    res = []
    in_result = set()

    for w in words:
        if w in transformed_to_origins:
            for t in transformed_to_origins[w]:
                if t != w and (w, t) not in in_result:
                    res.append([w, t])
                    in_result.add((w, t)), in_result.add((t, w))

    return res

- M April 22, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def source_and_rotation(words: List[str]) -> List[List[str]]:

    def rot_char_by(c, i):
        shift = ord('A')
        return chr((((ord(c) - shift) + i) % N) + shift)
    
    N = len(string.ascii_lowercase)

    transformed_to_origins = defaultdict(set)

    for w in words:
        for i in range(1, N):
            transformed_to_origins["".join([rot_char_by(c, i) for c in w])].add(w)

    res = []
    in_result = set()

    for w in words:
        if w in transformed_to_origins:
            for t in transformed_to_origins[w]:
                if t != w and (w, t) not in in_result:
                    res.append([w, t])
                    in_result.add((w, t)), in_result.add((t, w))

    return res

- Anonymous April 22, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I think you can transform each string into an array of differences (FOO -> [F-O, O-O]), and then you will just have to use a data structure to find matching arrays ( a trie maybe ) or some smart hashes.

- Anonymous April 27, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is an idea - construct a prefix tree, where each node is character, child of a node exists if and only if there is a word in original collection that has these characters going one after another. E.g. for collection ['bar', 'baz', 'cat'], the tree would look like:

Tree:

        [z]
        /   
[b] -> [a] 
        \
        [r]

[c] -> [a] -> [t]

Now for each word in the collection, and for every rotation you can descend in the tree. If path exists - the rotation exists too. Note - it is also important to mark last symbol of a string as "terminal" in the tree, so that algorithm would not match "AB", "BCZ" as ["AB", BC'] -> C is not terminal.

Here is the full implementation in JS:

var alphabet = Array.from('ABCDEFGHIJKLMNOPQRSTUVWXYZ');
var charIndex = makeCharIndex(alphabet);

console.log(getRotationPairs(['AB', 'BC', 'BCD', 'FOO', 'ZA', 'BAZ']));

function getRotationPairs(wordCollection) {
  var prefixTree = makePrefixTree(wordCollection);
  var rotationPairs = [];
  
  wordCollection.forEach(word => {
    var foundPairs = getRotationPairsForWord(word, prefixTree);
    rotationPairs = rotationPairs.concat(foundPairs)
  });
  
  return rotationPairs;
}

function getRotationPairsForWord(word, prefixTree) {
  var foundPairs = [];
  
  for (var rotationNumber = 1; rotationNumber < alphabet.length; ++rotationNumber) {
    var foundRotation = findRotationInTree(rotationNumber, word, prefixTree);
    if (foundRotation) {
      foundPairs.push([word, foundRotation]);
    }
  }
  
  return foundPairs;
}

function findRotationInTree(rotationNumber, word, prefixTree) {
  var currentNode = prefixTree;
  var rotatedWord = '';
  for (var i = 0; i < word.length; ++i) {
    var char = rotate(word[i], rotationNumber);
    if (!currentNode[char]) return;
    rotatedWord += char;
    currentNode = currentNode[char];
  }

  if (currentNode.isTerminal) return rotatedWord;
}

function rotate(char, offset) {
  return alphabet[(charIndex[char] + offset) % alphabet.length];
}

function makePrefixTree(wordCollection) {
  var tree = Object.create(null);
  wordCollection.forEach(addWordToTree);
  return tree;
  
  function addWordToTree(word) {
    var currentNode = tree;
    for (var i = 0; i < word.length; ++i) {
      var char = word[i];
      var nextNode = currentNode[char];
      if (!nextNode) {
        nextNode = currentNode[char] = Object.create(null);
      }
      
      currentNode = nextNode;
    }
    
    if (word.length > 0) {
      // So that we can check if word truly ends here.
      currentNode.isTerminal = true;
    }
  }
}

function makeCharIndex(alphabet) {
  var charIndex = {};
  alphabet.forEach((char, index) => charIndex[char] = index);
  
  return charIndex;
}

Tree is constructed in O(m) time, where `m` is total number of characters in your collection. The lookup for rotation is also linear function of characters count, so the overall performance is `O(m)`, `m` - total number of characters, and a constant factor of rotation counts (length of your alphabet).

- anvaka May 01, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void rotTrans(String [] list){
		
		HashMap<String,ArrayList<String>> map = new HashMap<String,ArrayList<String>>();
		String key;
		int len = 0,i = 0, j = 0;
		
		for(String s: list){
			key = keyGen(s);
			if(map.containsKey(key)){
				map.get(key).add(s);
			}else{
				map.put(key, new ArrayList<String>());
				map.get(key).add(s);
			}
		}
		for(String k : map.keySet())
			len += map.get(k).size() - 1;
		
		String [][] result = new String [len][2];
		
		for(Map.Entry<String,ArrayList<String>> entry : map.entrySet()){
			if(entry.getValue().size() > 1){
				for(j = 1; j < entry.getValue().size(); j++){
					result[i][0] = entry.getValue().get(0);
					result[i][1] = entry.getValue().get(j);
					i++;
				}
			}
		}
		System.out.println(Arrays.deepToString(result));
	}
	public static String keyGen(String str){
		if(str.length() <= 1) return "1";
		
		StringBuilder sb = new StringBuilder();
		char ch1,ch2;
		ch1 = str.charAt(0);
		sb.append(1);
		for(int i = 1; i < str.length(); i++){
			ch2 = str.charAt(i);
			if(ch2 >= ch1){
				sb.append((ch2 - ch1) + 1);
			}else{
				sb.append((ch1 - 'Z' + 1) + (('A' - ch2) + 1));
			}
			ch1 = ch2;
		}
		return sb.toString();
	}

- Raminder May 28, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

const normalize = string => string.split('').map(x => x.charCodeAt(0));
const transform = arr => arr.map(val => ((val -  arr[0]) + 26) % 26);
const normalizeString = string => transform( normalize(string)).join('');
const generateBuckets = strings => {
	const buckets = {  };
	const values = strings.forEach(string => {
		const normal = normalizeString(string);
		if (buckets[normal] === undefined){ buckets[normal] = [ ] };
		buckets[normal].push(string);
	});
	return buckets;
};

- Anonymous May 29, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I assume, to each word, we can apply any number of ROT_1 or any number of ROT_25.

#include <unordered_map>
#include <vector>
#include <string>
#include <iostream>

using namespace std;

pair<string, string> GetPatterns(const string& s)
{
    string p1, p2;
    for (int i = 0; i + 1 < s.size(); ++i)
    {
        int delta1 = (static_cast<int>(s[i + 1]) - s[i]);
        if (delta1 < 0)
        {
            delta1 += 26;
        }
        int delta2 = (static_cast<int>(s[i]) - s[i + 1]);
        if (delta2 < 0)
        {
            delta2 += 26;
        }
        p1 += to_string( delta1) + ",";
        p2 += to_string(-delta2) + ",";
    }
    return pair<string, string>(p1, p2);
};

vector<string> RotationDups(const vector<string>& words)
{
    vector<string> out;
    unordered_map<string, int> patterns;
    for (auto w : words)
    {
        pair<string, string> p = GetPatterns(w);
        ++patterns[p.first];
        ++patterns[p.second];
    }
    for (auto w : words)
    {
        pair<string, string> p = GetPatterns(w);
        if (patterns[p.first] > 1 ||
            patterns[p.second] > 1)
        {
            out.push_back(w);
        }
    }
    return out;
}


int main()
{
    vector<string> out = RotationDups({"AB", "BC", "FOO", "ZA", "BAZ"});
    for (auto w : out)
    {
        cout << w << ", ";
    }
    cout << "\n";
    return 0;
}

- Alex October 25, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is some code;
Bruteforce and optimised;

package Java;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

/**
 * Author: Nitin Gupta(nitin.gupta@walmart.com)
 * Date: 16/04/19
 * Description:
 */
public class StringRotationMatch {

    static class Pair {
        String s1, s2;

        @Override
        public String toString() {
            return "{" + s1 + "," + s2 + '}';
        }

        public Pair(String s1, String s2) {
            this.s1 = s1;
            this.s2 = s2;
        }
    }

    public static void main(String args[]) {
        String[] arr1 = {"AB", "BC", "FOO", "ZA", "BAZ"};
        System.out.println(rotationPairsBruteForce(arr1));
        System.out.println(rotationPairs(arr1));

        String[] arr2 = {"AB", "BC", "FOO", "ZA", "BAZ", "CBA"};
        System.out.println(rotationPairsBruteForce(arr2));
        System.out.println(rotationPairs(arr2));
    }


    //O(Ln^2)
    private static List<Pair> rotationPairsBruteForce(String arr[]) {
        String str = " ABCDEFGHIJKLMNOPQRSTUVWXYZ";
        List<Pair> out = new ArrayList<>();
        for (int i = 0; i < arr.length; i++) { //O(n)
            for (int j = i + 1; j < arr.length; j++) { //O(n)
                if (i == j)
                    continue;

                String s1 = arr[i];
                String s2 = arr[j];

                if (s1.length() != s2.length())
                    continue;

                char c1 = s1.charAt(0);
                char c2 = s2.charAt(0);
                int diff = getDiff(c1, c2, str);
                int k;
                for (k = 1; k < s1.length(); k++) { //O(L)

                    if (diff != getDiff(s1.charAt(k), s2.charAt(k), str))
                        break;
                }

                if (k == s1.length()) {
                    out.add(new Pair(s1, s2));
                }

            }
        }
        return out;
    }

    private static int getDiff(char c1, char c2, String str) { //O(1)

        int index1 = str.indexOf(c1);
        int index2 = str.indexOf(c2);

        if (index2 < index1)
            return 26 - (index1 - index2);
        return index2 - index1;
    }

    //Optimized-> O(nL)
    private static List<Pair> rotationPairs(String[] arr) {

        List<Pair> out = new ArrayList<>();
        List<String> transformed = transform(arr); //O(nL)

        Map<String, List<Integer>> duplicates = new HashMap<>();

        int i = 0;
        //This loop will run at most O(n) time since in inner loop even all of them map to one entry only, then each element will touch at most 2 times
        for (String s : transformed) { //O(n)

            List<Integer> entries = duplicates.getOrDefault(s, new ArrayList<>());


            for (Integer entry : entries) {

                out.add(new Pair(arr[entry], arr[i]));
            }
            entries.add(i);
            duplicates.put(s, entries);
            i++;
        }


        return out;
    }

    //O(nL)
    private static List<String> transform(String ar[]) {
        List<String> transformed = new ArrayList<>();

        for (int i = 0; i < ar.length; i++) { //O(n)

            String x = ar[i];

            int delta = 'A' - x.charAt(0);

            StringBuilder str = new StringBuilder();
            for (int j = 0; j < x.length(); j++) { //O(L)
                char current = x.charAt(j);

                current = (char) ((int) current + delta);

                if (current < 'A')
                    current = (char) ((int) current + 26);

                str.append(current);
            }
            transformed.add(str.toString());
        }

        return transformed;

    }
}

- nitinguptaiit April 16, 2019 | Flag Reply


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