Facebook Interview Question for Developer Program Engineers


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6
of 6 vote

We can find K largest elements, using a min heap, and find sum of the K largest elements. O(N * log K) time and O(K) space.

- Alex December 16, 2017 | Flag Reply
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1
of 1 vote

@dora. Yes, your solution works great for consecutive elements! Though, they usually call consecutive elements a subarray rather than a subset.

- Alex December 19, 2017 | Flag Reply
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1
of 1 vote

Questions to ask the interviewer
- does the array contain negative numbers.
- are we trying to find the sum of consecutive numbers in a subarray.
- if it's not consecutive, construct a min heap using priority queue.

The code below is to find the sum of consecutive numbers in a subarray using the sliding window method.

Time: O(N)
Space: O(1)

public int maxSumSubArray(int[] input, int k) {
		int n = input.length;
		if(k > n || k == n) return 0;
		
		int sum = input[0];
		for(int i = 1; i < k; i++) {
			sum += input[i];
		}
		
		int maxSum = sum;
		
		for(int i = k; i < n; i++) {
			sum = sum - input[i - k] + input[i];
			if(sum > maxSum) {
				maxSum = sum;
			}
		}
		
		return maxSum;
	}

- basilanathan January 03, 2018 | Flag Reply
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0
of 0 vote

Sort it first then sum last K elements

- Joe December 16, 2017 | Flag Reply
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0
of 0 vote

@dora. For list = {3, 1, 2}, k = 2, it returns 4.

- Alex December 18, 2017 | Flag Reply
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0
of 0 vote

Right. The question can be interpreted in many ways, so something to discuss with the interviewer.

I assumed the subset would be of consecutive elements. My solution is a O(n) way of finding K consecutive elements.

If the ask is "max of any k numbers", then using a min heap (klogn + n) is the most optimal way.

- dora December 18, 2017 | Flag Reply
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0
of 0 vote

Agree with alex

- kala kutta December 19, 2017 | Flag Reply
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0
of 0 vote

What about using the quick sort with random partition? It would find the biggest k elements but it would use quick sort only in half of the array and throw away the other part? with random partition we would get O(log n) on average.

- Anonymous December 19, 2017 | Flag Reply
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0
of 0 vote

Would this work??

int sum = 0;
		int maxSum = 0;
		for(int i = 0; i < k; i++) {
			for(int j = i; j < k; j++) {
				sum = a[i] + a[j];
				maxSum = sum;
				if(a[i + 1] + a[j + 1] > sum) {
					sum = a[i + 1] + a[j + 1];
					maxSum = sum;
				}
			}
		}
		System.out.print(maxSum);

- user1994 January 08, 2018 | Flag Reply
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0
of 0 vote

Taking into account fi they want:
- to find the max sum of consecutive numbers
- the array contains negative numbers
- sum array length may be less than or equal to a size K

int MaximumSum(int[] arr, int subLength){
var maxSum = 0;

var currentSum = 0;
for(int i = 0; i < arr.Length + subLength; i ++){
if(i < arr.Length){
currentSum += arr[i];
}

var startIndex = i - subLength;
if(startIndex >= 0){
currentSum -= arr[startIndex];
}

maxSum = Math.Max(maxSum, currentSum);
}

return maxSum;
}

- Egor January 28, 2018 | Flag Reply
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0
of 0 vote

int MaximumSum(int[] arr, int subLength){
        var maxSum = 0;
        
        var currentSum = 0;        
        for(int i = 0; i < arr.Length + subLength; i ++){
          if(i < arr.Length){
            currentSum += arr[i];
          }
            
          var startIndex = i - subLength;
          if(startIndex >= 0){
            currentSum -= arr[startIndex];
          }
            
          maxSum = Math.Max(maxSum, currentSum);
        }
        
        return maxSum;
    }

- egor4eg January 28, 2018 | Flag Reply
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0
of 0 vote

Expected O(n)

int maxSum(int[] a, int k) {
		quickPartition(a, k);
		
		int sum = 0;
		for(int i = 0; i < k; i++)
			sum += a[i];
		
		return sum;
	}
	
	void quickPartition(int[] a, int k) {
 		int st = 0;
		int en = a.length;
		
		while(st < en) {
			int p = partition(a, st, en);
			if(p == k)
				return;
			
			if(p < k)
				st = p;
			else
				en = p;
		}
	}

	int partition(int[] a, int st, int en) {
		swap(a, st, st + new Random().nextInt(en - st));
		
		int p = st;
		for(int i = st + 1; i < en; i++) 
			if(a[i] >= a[st])
				swap(a, i, ++p);
		
		swap(a, st, p);
		
		return p + 1;
	}

	void swap(int[] a, int i, int j) {
		int t = a[i];
		a[i] = a[j];
		a[j] = t;
	}

- Anonymous March 26, 2018 | Flag Reply
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0
of 0 vote

@O(N) approach

I like the solution, but this is O(N * log(N))

You essentially have to scan the entire array for the first partition, then continue to half it until p == k.

I like the solution, and studied it for a while to make sure I knew exactly what you were doing. But this method contains a certain amount of luck to find the right number which has K elements greater than it. In theory, this could loop forever.

- Verde March 28, 2018 | Flag Reply
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0
of 0 vote

I like the solution, but this is O(N * log(N))

You essentially have to scan the entire array for the first partition, then continue to half it until p == k.

I like the solution, and studied it for a while to make sure I knew exactly what you were doing. But this method contains a certain amount of luck to find the right number which has K elements greater than it. In theory, this could loop forever.

- Verde March 28, 2018 | Flag Reply
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0
of 0 vote

Modification of QuickSelect to partition by the kth largest element and sum the larger elements left from it. Runtime same as with QuickSelect, O(N)

In scala:

object MaxSubsetSum extends App {
  val list = Array(8, -5, 3, 7, 9 ,0)

  println(largestSubsetSum(3, 0, list.length - 1))

  def swap(from: Int, to: Int): Unit = {
    val temp = list(to)
    list(to) = list(from)
    list(from) = temp
  }

  def partition(left: Int, right: Int, pivot: Int): Int = {
    var index = left
    var i = left
    swap(pivot, right)
    while (i < right) {
      if (list(i) > list(right)) {
        swap(i, index)
        index += 1
      }
      i += 1
    }
    swap(right, index)
    index
  }

  def largestSubsetSum(k: Int, left: Int, right: Int): Int = {
    val v = partition(left, right, (right - left) / 2)
    if (v == k) list.take(k).sum
    else if (v < k) largestSubsetSum(k, v + 1, right)
    else largestSubsetSum(k, left, v - 1)
  }
}

- jvmakine April 08, 2018 | Flag Reply
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-1
of 1 vote

public int sumK(int[] list, int k) {
	if (k > list.length) return ERROR;

	int sum = 0, maxSum = 0, l = 0;
	for (int i = 0; i < k; i++) {
		sum += list[i];
	}

	maxSum = sum;
	for (int j = k; j < list.length; j++, l++) {
		sum = sum + list[j] - list[l];
		if (sum > maxSum) maxSum = sum;
	}

	return maxSum;
}

- dora December 18, 2017 | Flag Reply


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