Amazon Interview Question for Backend Developers


Country: United States




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

Similar to Binary Search

public class KMEFinder {
	
	public int find(int[] sa, int k) {
		return find(sa, 0, sa.length-1, k, 0);
	}
	
	private int find(int[] sa, int stIdx, int endIdx, int k, int leftmisses) {
		if (endIdx - stIdx == 1) {
			return sa[stIdx]+(k - leftmisses);
		}
		
		int midIdx = stIdx + (endIdx-stIdx)/2;
		int lms = (sa[midIdx] - sa[stIdx]) - (midIdx - stIdx);
		if (leftmisses + lms >= k) {
			return find(sa, stIdx, midIdx, k, leftmisses);
		} else {
			leftmisses = leftmisses + lms;
			return find(sa, midIdx, endIdx, k, leftmisses);
		}
	}
}

- Misc1 April 18, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def findMissingElement(arr,k):
mid = int (len(arr)-1)/2
mid = int(mid)+k
if arr[mid+1] - arr[mid] > 1:
return arr[int(mid)]+1
else:
return 0

- SachinK April 15, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int findK(int [] list, int num){
		 int k = 0, val = 0;
		 if(list.length < 2)return -1;
		 val = list[0];
		 for(int i = 0; i < list.length - 1; val = list[++i])
			 while((list[i + 1] - val++) > 1) 
				 if(k++ == num)return val;
		 return -1;
	 }

- Punjabi_Jatt April 17, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

import java.util.Arrays;

public class MissingElement
{
  public static void main(String[] args)
  {
    int[] intArray = {2,3,5,7,8,9,11,12,13};
    int k = 3;
    System.out.println(missingK(intArray,k));
    
  }
    
  static int missingK(int[] arr, int k){
    int m = arr[1]-arr[0];
    int x = 0;
    int val = 0;
    do{
      for(int i=0;i<arr.length;i++){ 
        val = arr[i]+m;
        if(!contains(arr,arr[i]+m)){ 
          x++;
          if(k==x)break;
        }
      }
    }
    while(k!=x);
    return val;
  }
  
     public static boolean contains(final int[] array, final int v) {
        boolean result = false;
        for(int i : array){
            if(i == v){
                result = true;
                break;
            }
        }
        return result;
    }
}

- Udy April 20, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public int findKthMissing(int[] arr, int k){
		return kthMissingHelper(arr, 0, arr.length-1, k);
	}
	
	public int kthMissingHelper(int[] arr, int low, int high, int k){
		int numRange = arr[high] - arr[low];
		int numPresent = high - low;
		int totalMissing = numRange - numPresent;
		if(totalMissing == 0)
			return arr[low] - 1;
		if(k > totalMissing)
			return -1;
		int middle = (low + high)/2;
		int missingCountInLeft = (arr[middle] - arr[low]) - (middle - low);
		if(missingCountInLeft >= k)
			return kthMissingHelper(arr, low, middle, k);
		else
			return kthMissingHelper(arr, middle+1, high, k - (missingCountInLeft));
	}

- Amish kumar May 03, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Ok, I missed the log(n) part last time, my bad. Following code uses binary search style solution and has log(n) average time. worst case is still O(n/2). For example when K = 0 and no K exists in the list.

static int FindK(int [] list, int K){
	 	if(list.length < 2)return -1;
	 	int d = list.length - 1;
		int m = d/2, dif = 0;
		
		while(d >= 0 && d < list.length){
			d = m;
			dif = (list[d] - list[0]) - d;
			if(K == dif){
				for(m = d; m < list.length - 1; m++)
					if(list[m + 1] - list[m] > 1)
						return ++list[d];
				break;
			}
			if(K < dif){
				m = d - (m + 1)/2;
			}else{
				m = d + (m + 1)/2;
			}
		}
		return -1;
	}

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

import org.junit.Test
import org.junit.Assert._

class Problem {
  def findMissingElement(array: Array[Int], target: Int, current: Int = 0): Option[Int] =
    if (current == (array.length - 1)) {
      None
    } else {
      val missingCount = array(current + 1) - array(current) - 1

      if (missingCount >= (target + 1)) {
        Some(array(current) + target + 1)
      } else {
        findMissingElement(array, target - missingCount, current + 1)
      }
    }

  @Test
  def test(): Unit = {
    val numbers = Array(2, 3, 5, 7)

    assertEquals(Some(4), findMissingElement(numbers, 0))
    assertEquals(Some(6), findMissingElement(numbers, 1))
    assertEquals(None, findMissingElement(numbers, 2))
    assertEquals(None, findMissingElement(numbers, 3))
  }
}

- emir10xrec June 01, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

public class App7 {

    public static void main(String... arg) {

        System.out.println(countingSort(new int[]{1, 3, 4, 6, 7, 8,10,11,13}, 0, 0, 8));
        System.out.println(countingSort(new int[]{1, 3, 4, 6, 7, 8,10,11,13}, 1, 0, 8));
        System.out.println(countingSort(new int[]{1, 3, 4, 6, 7, 8,10,11,13}, 2, 0, 8));
        System.out.println(countingSort(new int[]{1, 3, 4, 6, 7, 8,10,11,13}, 3, 0, 8));

    }

    public static int countingSort(int[] arr, int k, int s, int e) {
        int mid = s + ((e - s) / 2);
        if (mid == s) {
            return arr[mid] + 1;
        }
        int i1 =   arr[mid]-((mid - s) + arr[s]);
        if (i1 >= (k + 1)) {
            return countingSort(arr, k, s, mid);
        } else {
            return countingSort(arr, k - i1, mid, e);
        }
    }
}

- abrahamimohiosen April 13, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
-2
of 2 vote

A simple python solution use list comprehensive.

def findMissing(alist, k):
	return [a for a in range(alist[0], alist[-1]+1) if a not in alist][k]

- mazhuo79 April 12, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.


Add a Comment
Name:

Writing Code? Surround your code with {{{ and }}} to preserve whitespace.

Books

is a comprehensive book on getting a job at a top tech company, while focuses on dev interviews and does this for PMs.

Learn More

Videos

CareerCup's interview videos give you a real-life look at technical interviews. In these unscripted videos, watch how other candidates handle tough questions and how the interviewer thinks about their performance.

Learn More

Resume Review

Most engineers make critical mistakes on their resumes -- we can fix your resume with our custom resume review service. And, we use fellow engineers as our resume reviewers, so you can be sure that we "get" what you're saying.

Learn More

Mock Interviews

Our Mock Interviews will be conducted "in character" just like a real interview, and can focus on whatever topics you want. All our interviewers have worked for Microsoft, Google or Amazon, you know you'll get a true-to-life experience.

Learn More