Facebook Interview Question for SDE1s


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My algorithm uses the following idea:
1. keep a HashMap of <Char, Int> as each char and its occurence in string.
2. Use a Deque to keep the current used window of K last inserted chars.
3. Build result picking the best available candidate as:
. has the max. number of count and was not recently inserted in the deque of K last chars.
. if none available pick an already existing element from deque starting from head and keep track of its cost of insertion to result.

Using a custom Tuple<Char, Int> class since Java 8 doesn't support Tuples out-of-the-box as to keep track of selected candidate <Char> and its cost to add to result <Int>.

Implementation in Java:

public class TaskCountString {
	public static void main(String[] args) {
		System.out.println(rearrangeTasks("AAABBBCCC", 3));		
		System.out.println(rearrangeTasks("AAABC", 2));
		System.out.println(rearrangeTasks("AAADBBCC", 2));
	}
	
	public static int rearrangeTasks(String s, int k) {
		
		List<Character> listC = new ArrayList<Character>();
		for (char c : s.toCharArray()) {
		    listC.add(c);
		}

		// create a map of each char as key and its count num. of occurences in string 
		Map<Character, Long> mapCount = listC.stream()
	            .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

		int i = 0;
		int numChars = s.length();
		
		StringBuilder sb = new StringBuilder();	
		Deque<Character> deque = new LinkedList<>(); // use a deque to keep track of last used chars
		int totalCost = 0;
		
		while(i < numChars) {			
			Tuple<Character,Integer> t = getMaxCount(mapCount, deque, k);			
			sb.append(t.x);
			totalCost += t.y;
			deque.addLast(t.x);
			long countOfChar = mapCount.get(t.x); 			
			if (countOfChar == 1) {
				mapCount.remove(t.x);
			} else {
				mapCount.put(t.x, countOfChar - 1);
			}
			if (deque.size() > k) {
				deque.removeFirst();
			}			
			i++;
		}
				
		System.out.println(sb);		
		return totalCost;
	}

	private static Tuple<Character,Integer> getMaxCount(Map<Character, Long> mapCount, Deque<Character> deque, int k) { 	
		long max = 0;
		Character maxChar = null;		
		Map<Character, Long> mapCopy = new HashMap<>(mapCount);
		
		for(Character c : deque) {
			mapCopy.remove(c);
		}
		
		if (!mapCopy.isEmpty()) {
			for(Map.Entry<Character, Long> e : mapCopy.entrySet()) {
				if (e.getValue() > max) {
					max = e.getValue();
					maxChar = e.getKey();
				}
			}			
			return new Tuple<>(maxChar, 1);
		}				
		int cost = 2;
		for(Character c : deque) {
			if (mapCount.get(c) != null)
				return new Tuple<>(c, cost);
			cost++;
		}
		
		return null;
	}	
}
class Tuple<X, Y> { 
	  public final X x; 
	  public final Y y; 
	  public Tuple(X x, Y y) { 
	    this.x = x; 
	    this.y = y; 
	  } 
}

- guilhebl May 08, 2017 | Flag Reply
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1. Use a hashmap to store the Character Count.
2. Then iterate through the char array again, and keep track of the last character used.
3. Remove the entries from hashmap when the count becomes 0.

public static int getJob(char[] arr, int k){
		if(arr.length == 0){
			return 0;
		}
		HashMap<Character, Integer> map = new HashMap<>();
		for(int i=0; i<arr.length; i++){
			if(map.containsKey(arr[i]))
				map.put(arr[i], map.get(arr[i])+ 1);
			else
				map.put(arr[i], 1);
		}
		int res = 0;
		char l = ' ';
		int i = 0;
		while(!map.isEmpty()){
			if(i == arr.length){
				i = 0;
			}
			if(arr[i] == ' '){
				i++;
				continue;
			}
			else if(arr[i] != l){
				res +=1;
				map.put(arr[i], map.get(arr[i])-1);
				if(map.get(arr[i]) == 0){
					map.remove(arr[i]);
				}
				l = arr[i];
				arr[i] = ' ';
				i++;
			}
			else if(map.size() == 1){
				int val = map.get(arr[i]);
				res += val * k + val;
				map.remove(arr[i]);
				i++;
			}else
				i++;
		}
		return res;
	}
	public static void main(String[] args){
		char[] arr1 = {'A','A','A','B','B','B','C','C','C'};
		char[] arr2 = {'A','A','A','B','C'};
		System.out.println(getJob(arr1, 3));
		System.out.println(getJob(arr2, 2));
	}

- Anon May 09, 2017 | Flag Reply
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public static int reArrangeTasks(String tasks,int coolDown){
Map<Character,Integer> map=new HashMap<Character, Integer>();
//Storing string in may by character as key and number of repetions of it as value
for(int i=0;i<tasks.length();i++){
char c=tasks.charAt(i);
if(map.containsKey(c)){
map.put(c, map.get(c)+1);
}
else{
map.put(c, 1);
}
}
//looping through map and removing each character in every unique alphabet for cooltime period
int c=0;
boolean newLoop=false;
while(!map.isEmpty()){
newLoop=true;
Iterator<Character> it=map.keySet().iterator();
while(it.hasNext()){
char ch=it.next();
if(!newLoop && c!=0 && c%(coolDown+1)==0) break;
newLoop=false;
map.put(ch, map.get(ch)-1);
c++;
if(map.get(ch)==0) it.remove();
}
while(c%(coolDown+1)!=0 && !map.isEmpty()) c++;
}
return c;

}

- anvy123 May 09, 2017 | Flag Reply
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public static int reArrangeTasks(String tasks,int coolDown){
		Map<Character,Integer> map=new HashMap<Character, Integer>();
		//Storing string in may by character as key and number of repetions of it as value
		for(int i=0;i<tasks.length();i++){
			char c=tasks.charAt(i);
			if(map.containsKey(c)){
				map.put(c, map.get(c)+1);
			}
			else{
				map.put(c, 1);
			}
		}
		//looping through map and removing each character in every unique alphabet for cooltime period
		int c=0;
		boolean newLoop=false;
		while(!map.isEmpty()){
			newLoop=true;
			Iterator<Character> it=map.keySet().iterator();
			while(it.hasNext()){
				char ch=it.next();
				if(!newLoop && c!=0 && c%(coolDown+1)==0) break;
				newLoop=false;
				map.put(ch, map.get(ch)-1);
				c++;
				if(map.get(ch)==0) it.remove();
			}
			while(c%(coolDown+1)!=0 && !map.isEmpty()) c++;
		}
		return c;
		
	}

- anvy123 May 09, 2017 | Flag Reply
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string OptimizeTasks(string const &tasks)
{
	unordered_map<char, int> task_count;
	for (char c : tasks) {
		++task_count[c];
	}
	string optimized;
	while (!task_count.empty()) {
		for (auto it = task_count.begin(); it != task_count.end(); ++it) {
			optimized += it->first;
			if (--it->second == 0) {
				task_count.erase(it->first);
			}
		}
	}
	return optimized;
}

- Alex May 09, 2017 | Flag Reply
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//C# Code working for me
//optTaskSeqK("AAABBBCCC", 3);

private static Tuple<string, int> optTaskSeqK(string tasks, int coolTime)
        {
            var taskGp = tasks.GroupBy(s => s).Select(s => new { task = s.Key, count = s.Count() });
            var taskDic = taskGp.ToDictionary(g => g.task, g => g.count);
            var recTaskDic = taskGp.ToDictionary(g => g.task, g => 0);
            var coolTaskDic = taskGp.ToDictionary(g => g.task, g => 0);
            String taskOrder = string.Empty;
            int totalExeTime = 0;
            optTask cuTask = new optTask();
            //int cuTaskExeTime=0;
            for (int seq = 0; seq < tasks.Length; seq++)
            {
                foreach (var key in recTaskDic.Keys)
                {
                    if (coolTaskDic[key] != 0)
                        coolTaskDic[key] -= 1;
                }
                cuTask = getNextOptTask(seq);
                totalExeTime += 1 + coolTaskDic[cuTask.taskCh];
                coolTaskDic[cuTask.taskCh] += coolTime;
                taskOrder += cuTask.taskCh;
            }
            optTask getNextOptTask(int seq)
            {
                var taskGP = taskDic.Where(i => i.Value > 0).GroupBy(i => i.Value).OrderByDescending(i => i.Key).First().ToList();
                //if (taskGp.Count() == 1)
                //    return taskGP.First().Key;              

                var optTask = new optTask { taskCh = '\0', exeTime = int.MaxValue, seq = int.MaxValue };
                var tempTask = new optTask();

                foreach (var ti in taskDic)
                {
                    if (ti.Value > 0)
                    {
                        tempTask = new optTask();
                        tempTask.taskCh = ti.Key;
                        tempTask.count = ti.Value;
                        tempTask.seq = recTaskDic[ti.Key];
                        tempTask.exeTime = 1 + coolTaskDic[ti.Key];
                        optTask = compOptTask(tempTask, optTask);
                    }
                }
                taskDic[optTask.taskCh]--;
                recTaskDic[optTask.taskCh] = seq;
                return optTask;
            };

            optTask compOptTask(optTask t1, optTask t2)
            {
                if (t1.exeTime != t2.exeTime)
                    return t1.exeTime < t2.exeTime ? t1 : t2;
                if (t1.count != t2.count)
                    return t1.count > t2.count ? t1 : t2;
                if (t1.seq != t2.seq)
                    return t1.seq < t2.seq ? t1 : t2;
                return t2;
            }
            return Tuple.Create(taskOrder, totalExeTime);
        }

class optTask
        {
            public char taskCh;
            public int exeTime;
            public int count { get; set; }
            public int seq { get; set; }
        }

- syam.bollu May 09, 2017 | Flag Reply
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Example 1 (k=3, ie distance should be 3 in between same tasks) does not match with examples 2 & 3

test runs:

t.rearrangeTasks("AAABBBCCC", 3); => ABC_ABC_ABC 11
t.rearrangeTasks("AAABC", 2); => ABCA__A 7
t.rearrangeTasks("AAADBBCC", 2); => ABCADBAC 8
t.rearrangeTasks("AAADBBC", 2); => ABCABDA 7
t.rearrangeTasks("CCCBBB", 2); => BC_BC_BC 8
t.rearrangeTasks("CCCBBB", 3); => BC__BC__BC 10
t.rearrangeTasks("AAAAAABBCCCCCDDDD", 2); => ABCADCADCABCADCAD 17
t.rearrangeTasks("AAAAAAAAAABBCCCCCDDDD", 2); => ABCAD_AC_A_DA_CA_BAC_AD_ADCA 28

public int rearrangeTasks(String tasks, int cooldown){ 

		HashMap<Character, Integer> occurMap = new HashMap<>();

	 	//compute map
	    for (int i=0; i < tasks.length();i++) {
           char c = tasks.charAt(i);
           Integer occurence = occurMap.get(c);
           if (occurence == null) {
                occurMap.put(c, 1);
		   } else {
                occurMap.put(c, occurence+1);
           }
        }

		HashMap<Character, Integer> distMap = new HashMap<>();
	    final AtomicInteger startPos = new AtomicInteger(0);
	    AtomicInteger minDist = new AtomicInteger(tasks.length());
        occurMap.entrySet().stream().forEach( (e) -> {
        	char c = e.getKey();
        	int occurence = e.getValue();
		
			//placing same tasks far apart within minDist     
			int charDist = minDist.get() / occurence;
            if (charDist <= cooldown) {
            	 int newCharDist = (cooldown + 1);
            	 distMap.put(c, newCharDist);
            	 int newDist = (occurence-1) * newCharDist + 
                		 startPos.incrementAndGet();
            	 if (minDist.get() < newDist) {
            		 minDist.set(newDist);
            	 }
            } else {
            	distMap.put(c, charDist);
            }
        });
		
        //only to print output task order, not required otherwse
        AtomicInteger writePos = new AtomicInteger(-1);
        char[] out = new char[minDist.get()];
        Arrays.fill(out, '_');
        occurMap.entrySet().stream().forEach( (e) -> {
        	char c = e.getKey();
        	int occurence = e.getValue();
        	
        	int pos = writePos.incrementAndGet();
        	int charDist = distMap.get(c);
        	while(occurence > 0) {
        		while (out[pos] != '_' || 
        				anyInRange(out, pos-cooldown, pos+cooldown, c)) {
        			pos += (charDist + 1); //conflicts, moveover
        			pos %= minDist.get();
        		}
        		out[pos] = c;
        		occurence--;
        		pos += (charDist);
        		pos %= minDist.get();
        	}
        });
        System.out.println(new String(out) + " " + out.length);
        
	    return minDist.get();
	}

	private boolean anyInRange(char[] out, int i, int j, char c) {
		if (i < 0) {
			i = 0;
		}
		
		if (j >= out.length) {
			j = out.length - 1;
		}
		
		for (int k = i; k <= j; k++) {
			if (out[k] == c) {
				return true;
			}
		}
		
		return false;
	}

- ramyaack July 14, 2017 | Flag Reply
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Output:

ABC_ABC_ABC 11
ABCDEABCDEABCDEABCDEABCDE 25
ABCA__A 7
ABCADBAC 8
ABCABDA 7
BC_BC_BC 8
BC__BC__BC 10
ABCADCADCABCADCAD 17
ABCAD_AC_A_DA_CA_BAC_AD_ADCA 28

public class TaskDistance {

	public static void main(String[] args) {
		TaskDistance t = new TaskDistance();
		t.rearrangeTasks("AAABBBCCC", 3);
		t.rearrangeTasks("AAAAABBBBBCCCCCDDDDDEEEEE", 3);
		t.rearrangeTasks("AAABC", 2);
		t.rearrangeTasks("AAADBBCC", 2);
		t.rearrangeTasks("AAADBBC", 2);
		t.rearrangeTasks("CCCBBB", 2);
		t.rearrangeTasks("CCCBBB", 3);
		t.rearrangeTasks("AAAAAABBCCCCCDDDD", 2);
		t.rearrangeTasks("AAAAAAAAAABBCCCCCDDDD", 2);
	}
	
	public int rearrangeTasks(String tasks, int cooldown){ 

		HashMap<Character, Integer> occurMap = new HashMap<>();

	 	//compute map
	    for (int i=0; i < tasks.length();i++) {
           char c = tasks.charAt(i);
           Integer occurence = occurMap.get(c);
           if (occurence == null) {
                occurMap.put(c, 1);
		   } else {
                occurMap.put(c, occurence+1);
           }
        }

		HashMap<Character, Integer> distMap = new HashMap<>();
	    final AtomicInteger startPos = new AtomicInteger(0);
	    AtomicInteger minDist = new AtomicInteger(tasks.length());
        occurMap.entrySet().stream().forEach( (e) -> {
        	char c = e.getKey();
        	int occurence = e.getValue();
		
			//placing same tasks far apart within minDist     
			int charDist = minDist.get() / occurence;
            if (charDist <= cooldown) {
            	 int newCharDist = (cooldown + 1);
            	 distMap.put(c, newCharDist);
            	 int newDist = (occurence-1) * newCharDist + 
                		 startPos.incrementAndGet();
            	 if (minDist.get() < newDist) {
            		 minDist.set(newDist);
            	 }
            } else {
            	distMap.put(c, charDist);
            }
        });
		
        //only to print output task order, not required otherwse
        AtomicInteger writePos = new AtomicInteger(-1);
        char[] out = new char[minDist.get()];
        Arrays.fill(out, '_');
        occurMap.entrySet().stream().forEach( (e) -> {
        	char c = e.getKey();
        	int occurence = e.getValue();
        	
        	int pos = writePos.incrementAndGet();
        	int charDist = distMap.get(c);
        	while(occurence > 0) {
        		while (out[pos] != '_' || 
        				anyInRange(out, pos-cooldown, pos+cooldown, c)) {
        			pos += (charDist + 1); //conflicts, moveover
        			pos %= minDist.get();
        		}
        		out[pos] = c;
        		occurence--;
        		pos += (charDist);
        		pos %= minDist.get();
        	}
        });
        System.out.println(new String(out) + " " + out.length);
        
	    return minDist.get();
	}

	private boolean anyInRange(char[] out, int i, int j, char c) {
		if (i < 0) {
			i = 0;
		}
		
		if (j >= out.length) {
			j = out.length - 1;
		}
		
		for (int k = i; k <= j; k++) {
			if (out[k] == c) {
				return true;
			}
		}
		
		return false;
	}

}

- ramyaack July 14, 2017 | Flag Reply
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public class TaskDistance {

	public static void main(String[] args) {
		TaskDistance t = new TaskDistance();
		t.rearrangeTasks("AAABBBCCC", 3);
		t.rearrangeTasks("AAAAABBBBBCCCCCDDDDDEEEEE", 3);
		t.rearrangeTasks("AAABC", 2);
		t.rearrangeTasks("AAADBBCC", 2);
		t.rearrangeTasks("AAADBBC", 2);
		t.rearrangeTasks("CCCBBB", 2);
		t.rearrangeTasks("CCCBBB", 3);
		t.rearrangeTasks("AAAAAABBCCCCCDDDD", 2);
		t.rearrangeTasks("AAAAAAAAAABBCCCCCDDDD", 2);
	}
	
	public int rearrangeTasks(String tasks, int cooldown){ 

		HashMap<Character, Integer> occurMap = new HashMap<>();

	 	//compute map
	    for (int i=0; i < tasks.length();i++) {
           char c = tasks.charAt(i);
           Integer occurence = occurMap.get(c);
           if (occurence == null) {
                occurMap.put(c, 1);
		   } else {
                occurMap.put(c, occurence+1);
           }
        }

		HashMap<Character, Integer> distMap = new HashMap<>();
	    final AtomicInteger startPos = new AtomicInteger(0);
	    AtomicInteger minDist = new AtomicInteger(tasks.length());
        occurMap.entrySet().stream().forEach( (e) -> {
        	char c = e.getKey();
        	int occurence = e.getValue();
		
			//placing same tasks far apart within minDist     
			int charDist = minDist.get() / occurence;
            if (charDist <= cooldown) {
            	 int newCharDist = (cooldown + 1);
            	 distMap.put(c, newCharDist);
            	 int newDist = (occurence-1) * newCharDist + 
                		 startPos.incrementAndGet();
            	 if (minDist.get() < newDist) {
            		 minDist.set(newDist);
            	 }
            } else {
            	distMap.put(c, charDist);
            }
        });
		
        //only to print output task order, not required otherwse
        AtomicInteger writePos = new AtomicInteger(-1);
        char[] out = new char[minDist.get()];
        Arrays.fill(out, '_');
        occurMap.entrySet().stream().forEach( (e) -> {
        	char c = e.getKey();
        	int occurence = e.getValue();
        	
        	int pos = writePos.incrementAndGet();
        	int charDist = distMap.get(c);
        	while(occurence > 0) {
        		while (out[pos] != '_' || 
        				anyInRange(out, pos-cooldown, pos+cooldown, c)) {
        			pos += (charDist + 1); //conflicts, moveover
        			pos %= minDist.get();
        		}
        		out[pos] = c;
        		occurence--;
        		pos += (charDist);
        		pos %= minDist.get();
        	}
        });
        System.out.println(new String(out) + " " + out.length);
        
	    return minDist.get();
	}

	private boolean anyInRange(char[] out, int i, int j, char c) {
		if (i < 0) {
			i = 0;
		}
		
		if (j >= out.length) {
			j = out.length - 1;
		}
		
		for (int k = i; k <= j; k++) {
			if (out[k] == c) {
				return true;
			}
		}
		
		return false;
	}

}

Output:

ABC_ABC_ABC 11
ABCDEABCDEABCDEABCDEABCDE 25
ABCA__A 7
ABCADBAC 8
ABCABDA 7
BC_BC_BC 8
BC__BC__BC 10
ABCADCADCABCADCAD 17
ABCAD_AC_A_DA_CA_BAC_AD_ADCA 28

- rck July 14, 2017 | Flag Reply
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of 0 vote

public int rearrange(List<Character> tasks, int k) {
	if (tasks.size() == 1) return 1;

	HashMap<Character, Integer> hm = new HashMap<>();

	for (Character c: tasks) {
		if (hm.contains(c)) hm.put(c, hm.get(c) + 1);
		else (hm.add (c, 1));
	}

	// Sort tasks by frequency
	ArrayList<Map.Entry> sortedTasks = Collections.sort(hm.entrySet().toArray(), (p1, p2) -> p1.value > p2.value);

	// Add all tasks except the first, S
	int sum = 0;
	for (int i = 1; i < sortedTasks.size(); i++) {
		sum += sortedTasks.get(i).value;
	}

	// If X is #of tasks of highest frequency, the formula for total duration is
	// X + (X-1)k - S
	return ((sortedTasks.get(0).value - 1) * k) - sum + sortedTasks.get(0).value;
}

- dora January 24, 2018 | Flag Reply


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