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Amazon Interview Question for SDE1s


Country: United States
More Questions from This Interview



Comment hidden because of low score. Click to expand.
1
of 1 vote

C++ Solution.
Assumption: int m (level) is also the limit length for the key. If this assumption is not agreeable, it's easy to increase the level and filter strings that have more than m of the same character....

#include <iostream>
#include <string>
#include <set>

using namespace std;

// Recursive Function to generate IDs for one level.
set<string> makeIdsForLevel(set<string> dict, int level){
    if(level == 1){
        return dict;
    }

    set<string> mergedSet;
    for(auto its1 : dict)
        for(auto its2 : makeIdsForLevel(dict,level-1))
            mergedSet.insert(its1+its2);

    return mergedSet;
}

//Wrapper function.
set<string> makeUniqueIDs(set<char> charDict, int maxRep){

    set<string> strDict;
    for(auto c : charDict)
        strDict.insert(string(1,c));

    set<string> ids;
    for(int i=1; i<=maxRep; ++i){
        auto idLevel = makeIdsForLevel(strDict,i);
        ids.insert(idLevel.begin(), idLevel.end());
    }
    return ids;
}

//Comparator to order set by length of ID
struct CompareLenght
{
    bool operator()(const string& s1, const string& s2) {
        return (s1.length() <= s2.length());
    }
};

// Wrapper Class around ID generator.
// The set is sorter in increasing length order.
// next() always returns the next available ID then deletes it.
class IdGenerator{
    set<string,CompareLenght> ids;
public:
    IdGenerator(set<char> charDict, int level){
        auto idsUnordered = makeUniqueIDs(charDict,level);
        for(auto id : idsUnordered)
            ids.insert(id);

        cout << "Constructed Ids: [" << endl;
        for(auto id : ids)
            cout << id << " ";
        cout << "]" << endl;
    }
    string next(){
        if(ids.begin()==ids.end()) return "";

        string first = *ids.begin();
        ids.erase(ids.begin());
        return first;
    }
};

int main()
{
    set<char> charDict = {'1','2', '3'};

    IdGenerator idGen(charDict,3);

    string id = idGen.next();
    while(!id.empty()){
        cout << "next id: " << id << endl;
        id = idGen.next();
    }

    cout << "All IDs have been used.";

    return 0;
}

OUTPUT:

Constructed Ids: [
3 2 1 33 32 31 23 22 21 13 12 11 333 332 331 323 322 321 313 312 311 233 232 231 223 222 221 213 212 211 133 132 131 123 122 121 113 112 111 ]
next id: 3
next id: 2
next id: 1
next id: 33
next id: 32
next id: 31
next id: 23
next id: 22
next id: 21
next id: 13
next id: 12
next id: 11
next id: 333
next id: 332
next id: 331
next id: 323
next id: 322
next id: 321
next id: 313
next id: 312
next id: 311
next id: 233
next id: 232
next id: 231
next id: 223
next id: 222
next id: 221
next id: 213
next id: 212
next id: 211
next id: 133
next id: 132
next id: 131
next id: 123
next id: 122
next id: 121
next id: 113
next id: 112
next id: 111
All IDs have been used.
Process returned 0 (0x0) execution time : 0.047 s
Press any key to continue.

- Gabriel Feyer May 23, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Java Version:

public class IdsGenerator {
	static Set<String> idsSet = new TreeSet<>(new Comparator<String>(){

		@Override
		public int compare(String o1, String o2) {
			if(o1.length() <= o2.length())
				return -1;
			else
				return 1;
		}
		
	});
	
	static Queue<String> q = new LinkedList<String>();
	
	public static void idGenerator(Set<Character> set, int m){
		Set<String> dict = new HashSet<String>();
		
		for(Character ch : set){
			dict.add(String.valueOf(ch));
		}
		
		idsSet.addAll(makeUniqueIds(dict, m));
		
		q.addAll(idsSet);
		//System.out.println(idsSet);
	}
	
	public static Set<String> makeUniqueIds(Set<String> dict, int level){
		Set<String> ids = new HashSet<String>();
		
		for(int i = 1; i <= level; i++){
			ids.addAll(makeIdsForLevel(dict, i));
		}
		return ids;
	}
	
	public static Set<String> makeIdsForLevel(Set<String> dict, int level){
		if(level == 1)
			return dict;
		Set<String> mergedSet = new HashSet<String>();
		
		for(String str : dict){
			for(String ids : makeIdsForLevel(dict, level - 1)){
				mergedSet.add(str + ids);
			}
		}
		
		return mergedSet;
	}
	
	public static boolean hasNext(){
		return q.size() > 0;
	}
	
	public static String next(){
		return q.poll();
	}
	// To Do: DP
	public static void main(String[] args) {
		Set<Character> set = new HashSet<>();
		set.add('1');
		set.add('2');
		set.add('3');
		idGenerator(set,3);
		
		while(hasNext()){
			System.out.print(next() + " ");
		}
	}

}

- AJ June 24, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Java DP Version:

public class IdsGenerator {
	static Set<String> idsSet = new TreeSet<>(new Comparator<String>(){

		@Override
		public int compare(String o1, String o2) {
			if(o1.length() <= o2.length())
				return -1;
			else
				return 1;
		}
		
	});
	
	static Queue<String> q = new LinkedList<String>();
	
	public static void idGenerator(Set<Character> set, int m){
		Set<String> dict = new HashSet<String>();
		Map<Integer, Set<String>> cache = new HashMap<Integer, Set<String>>();
		
		for(Character ch : set){
			dict.add(String.valueOf(ch));
		}
		
		cache.put(1, dict);
		idsSet.addAll(makeUniqueIds(dict, m, cache));
		
		q.addAll(idsSet);
	}
	
	public static Set<String> makeUniqueIds(Set<String> dict, int level, Map<Integer, Set<String>> cache){
		Set<String> ids = new HashSet<String>();
		
		for(int i = 1; i <= level; i++){
			ids.addAll(makeIdsForLevel(dict, i, cache));
		}
		return ids;
	}
	
	public static Set<String> makeIdsForLevel(Set<String> dict, int level, Map<Integer, Set<String>> cache){
		if(cache.containsKey(level)){
			return cache.get(level);
		}
		
		Set<String> mergedSet = new HashSet<String>();
		
		for(String str : dict){
			for(String ids : makeIdsForLevel(dict, level - 1, cache)){
				mergedSet.add(str + ids);
			}
		}
		
		cache.put(level, mergedSet);
		return mergedSet;
	}
	
	public static boolean hasNext(){
		return q.size() > 0;
	}
	
	public static String next(){
		return q.poll();
	}
	// To Do: DP
	public static void main(String[] args) {
		Set<Character> set = new HashSet<>();
		set.add('1');
		set.add('2');
		set.add('3');
		idGenerator(set,3);
		
		while(hasNext()){
			System.out.print(next() + " ");
		}
	}

}

- AJ June 24, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

# expandable counter - a generator of a number represented as an array of integers in given base
# Yields an array with incremented number on each next() call.
# Each position is an integer in [0..base-1] where the first item is the number LSB
# The number of digits is incremented by one when the counter rolls-over.
# For example for base 10 it yields: [0], [1],...,[9], [0,0], [1,0],...,[9,9], [0,0,0], ...
# (1 digits counter 0..9, 2 digits counter 00..99, 3 digits counter 000..999, etc.)
#
def xcounter(base):
    counter = [0]
    yield counter

    while True:
        for i in xrange(len(counter)):
            counter[i] += 1
            if counter[i] >= base:
                counter[i] = 0
            else:
                break

        if i == len(counter) - 1 and counter[-1] == 0: 
            counter.append(0)

        yield counter

# A generator for id strings consisting fiven characters set were no character is repeated more than m times
# chars - a list of characters (e.g. ['0', '1', ... ,'9']
# m - maximum repetition of any character in yielded id
#
# Uses xcounter for generating expanded counter values and skip values which contain
# at least one consecutive appearance longest than m of any digit
#
def idgen(chars, m):
    counter = xcounter(len(chars))
    
    while True:
        a = counter.next()
        # skip counter with longer than m consecutive run of any digit
        if longSubarray(a, m):
            continue
        # converts counter value into a string, reversing the value array for lexicograpical order
        # (of same length ids)
        id = ''.join([chars[x] for x in reversed(a)])
        yield id

# Predicate for checking whether given list 'a' contains
# a consecutive repetition longer than given threshold of any value
#
def longSubarray(a, threshold):
    if len(a) <= threshold:
        return False
    
    currLen = 1
    for i in xrange(1, len(a)):
        if a[i] == a[i-1]:
            currLen += 1
            if currLen > threshold:
                return True
        else:
            currLen = 1

    return False

- Anonymous July 02, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

# expandable counter - a generator of a number represented as an array of integers in given base
# Yields an array with incremented number on each next() call.
# Each position is an integer in [0..base-1] where the first item is the number LSB
# The number of digits is incremented by one when the counter rolls-over.
# For example for base 10 it yields: [0], [1],...,[9], [0,0], [1,0],...,[9,9], [0,0,0], ...
# (1 digits counter 0..9, 2 digits counter 00..99, 3 digits counter 000..999, etc.)
#
def xcounter(base):
    counter = [0]
    yield counter

    while True:
        for i in xrange(len(counter)):
            counter[i] += 1
            if counter[i] >= base:
                counter[i] = 0
            else:
                break

        if i == len(counter) - 1 and counter[-1] == 0: 
            counter.append(0)

        yield counter

# A generator for id strings consisting fiven characters set were no character is repeated more than m times
# chars - a list of characters (e.g. ['0', '1', ... ,'9']
# m - maximum repetition of any character in yielded id
#
# Uses xcounter for generating expanded counter values and skip values which contain
# at least one consecutive appearance longest than m of any digit
#
def idgen(chars, m):
    counter = xcounter(len(chars))
    
    while True:
        a = counter.next()
        # skip counter with longer than m consecutive run of any digit
        if longSubarray(a, m):
            continue
        # converts counter value into a string, reversing the value array for lexicograpical order
        # (of same length ids)
        id = ''.join([chars[x] for x in reversed(a)])
        yield id

# Predicate for checking whether given list 'a' contains
# a consecutive repetition longer than given threshold of any value
#
def longSubarray(a, threshold):
    if len(a) <= threshold:
        return False
    
    currLen = 1
    for i in xrange(1, len(a)):
        if a[i] == a[i-1]:
            currLen += 1
            if currLen > threshold:
                return True
        else:
            currLen = 1

    return False

Example:

>>> idg = idgen(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'], 3)
idg = idgen(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'], 3)
>>> ids = [idg.next() for _ in xrange(20)]
ids = [idg.next() for _ in xrange(20)]
>>> ids
ids
['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '00', '01', '02', '03', '04', '05', '06', '07', '08', '09']
>>> idg = idgen(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'], 3)
idg = idgen(['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'], 3)
>>> ids = [idg.next() for _ in xrange(2250)]
ids = [idg.next() for _ in xrange(2250)]
>>> ids[2218:2222]
ids[2218:2222]
['1109', '1110', '1112', '1113']

- Python solution: July 02, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

hoffman encoding!

- Anonymous February 01, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package com.harsha.cs;


import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

/**
 * Harsha.H
 */
public class Subsets {

    private static Set uniqueIds = new HashSet();

    public void generateUniqueIds(int[] givenArray){
        int[] subset = new int[givenArray.length];
        helper(givenArray, subset, 0);
    }

    /**
     * Get all the subsets for the given array.
     * The algorithm goes like, if you pick or not pick the current item, how does it evolve.
     * @param givenArray
     * @param subset
     * @param i
     */
    private void helper(final int[] givenArray, int[] subset, int i) {
        if(i == givenArray.length) {
            addToSet(subset);
        } else {
            subset[i] = 0;
            helper(givenArray, subset, i+1);
            subset[i] = givenArray[i];
            helper(givenArray, subset, i+1);
        }
    }

    /**
     * Remove dups from the string.
     * @param subset
     */
    private void addToSet(int[] subset) {
        String str = Arrays.stream(subset).filter(x -> x != 0).mapToObj(a -> ((Integer)a).toString()).collect(Collectors.joining(","));
        uniqueIds.add(str);
    }


    public static void main(String[] args){
        Subsets subsets = new Subsets();
        int[] arr = {1,2,3};
        int m = 3;
        int[] newArray = Arrays.stream(arr).flatMap(a -> IntStream.iterate(a, j -> j).limit(m)).toArray();
        System.out.println(Arrays.toString(newArray)); // Create exploded array with elements of m times.

        subsets.generateUniqueIds(newArray);

        uniqueIds.stream().sorted(new Comparator() {
            @Override
            public int compare(Object o1, Object o2) {
                if(((String)o1).length() < ((String)o2).length()) return -1;
                else if(((String)o1).length() > ((String)o2).length()) return 1;
                else return 0;
            }
        }).forEach(System.out::println);
    }

}

- Harsha.Halgaswatta February 16, 2019 | Flag Reply


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