Amazon Interview Question for SDE1s


Country: United States
Interview Type: Phone Interview




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0
of 0 vote

# assumes sorted and unique
def get_union(a,b):
	if not b:
		return a
	if not a:
		return b
	p = q = 0
	union = []
	while p < len(a) and q < len(b):
		if a[p] < b[q]:
			union.append(a[p])
			p += 1
		elif a[p] > b[q]:
			union.append(b[q])
			q += 1
		else:
			union.append(a[p])
			p += 1
			q += 1
	if p < len(a):
		union.extend(a[p:])
	if q < len(b):
		union.extend(b[q:])
	return union

a = [2, 10, 14, 19, 51, 71]
b = [2, 9, 19, 40, 51]
u = get_union(a,b)
print u
u_from_stdlib = set(a) | set(b)
assert len(u) == len(set(u))
assert len(set(u) ^ u_from_stdlib) == 0, "symmetric difference exists"

- marcopolo May 16, 2018 | Flag Reply
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0
of 0 vote

I would rather put them both in a set. Set would filter out distinct elements from each of them. Question doesn't say output should be sorted. So I would get the union that is distinct element from both the lists into the set.

- Archana Kumari May 17, 2018 | Flag Reply
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0
of 0 vote

{

public static void main(String[] args){
int a[] = {2, 10, 14, 19, 51, 71};
int b[] = {2, 9, 19, 40, 51} ;
int union[]= getUnion(a,b);
for(int i=0;i<union.length-1;i++){ System.out.println(union[i]);
}
}
private static int[] getUnion(int[] a, int[] b) {
int u[] = new int[a.length+b.length];
int i=0,j=0,k=0;
boolean flag =true;
while(flag){
if(a[i] == b[j]){
u[k] =b[j];
k+=1;
j+=1;
i+=1;
}
else if(a[i]<b[j]){
u[k]=a[i];
k+=1;
i+=1;
}
else{
u[k]=b[j];
k+=1;
j+=1;
}
if(j==b.length){
flag=false;
}
}
return u;
}

}

- Harshad Deo May 17, 2018 | Flag Reply
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0
of 0 vote

{public static void main(String[] args){
int a[] = {2, 10, 14, 19, 51, 71};
int b[] = {2, 9, 19, 40, 51} ;
int union[]= getUnion(a,b);
for(int i=0;i<union.length-1;i++){ System.out.println(union[i]);
}
}
private static int[] getUnion(int[] a, int[] b) {
int u[] = new int[a.length+b.length];
int i=0,j=0,k=0;
boolean flag =true;
while(flag){
if(a[i] == b[j]){
u[k] =b[j];
k+=1;
j+=1;
i+=1;
}
else if(a[i]<b[j]){
u[k]=a[i];
k+=1;
i+=1;
}
else{
u[k]=b[j];
k+=1;
j+=1;
}
if(j==b.length){
flag=false;
}
}
return u;
}
}

- Harshad Deo May 17, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

{

public static void main(String[] args){
int a[] = {2, 10, 14, 19, 51, 71};
int b[] = {2, 9, 19, 40, 51} ;
int union[]= getUnion(a,b);
for(int i=0;i<union.length-1;i++){
System.out.println(union[i]);
}
}
private static int[] getUnion(int[] a, int[] b) {
int u[] = new int[a.length+b.length];
int i=0,j=0,k=0;
boolean flag =true;
while(flag){
if(a[i] == b[j]){
u[k] =b[j];
k+=1;
j+=1;
i+=1;
}
else if(a[i]<b[j]){
u[k]=a[i];
k+=1;
i+=1;
}
else{
u[k]=b[j];
k+=1;
j+=1;
}
if(j==b.length){
flag=false;
}
}
return u;
}

}

- Harshad Deo May 17, 2018 | Flag Reply
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0
of 0 vote

def get_union(a,b):
  a_len = len(a)
  b_len = len(b)
  print(a_len, b_len)

  result = []
  hashmap = {}
  for i in range(len(a)-1):
    if (i > a_len or i>b_len):
        break

    # print(mn, mx)
    if hashmap.get(a[i]) == None:
      result.append(a[i])
      hashmap[a[i]] = True

    if hashmap.get(b[i]) == None:
      result.append(b[i])
      hashmap[b[i]] = True

  return result

- Anonymous May 19, 2018 | Flag Reply
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0
of 0 vote

def get_union(a,b):
  a_len = len(a)
  b_len = len(b)
  print(a_len, b_len)

  result = []
  hashmap = {}
  for i in range(len(a)-1):
    if (i > a_len or i>b_len):
        break

    # print(mn, mx)
    if hashmap.get(a[i]) == None:
      result.append(a[i])
      hashmap[a[i]] = True

    if hashmap.get(b[i]) == None:
      result.append(b[i])
      hashmap[b[i]] = True

  return result

- sk May 19, 2018 | Flag Reply
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0
of 0 vote

static void union(int arr1[], int[] arr2) {
		int i = 0, j = 0, k = 0;
		List<Integer> list = new ArrayList<>();
		boolean flag = true;
		while (flag) {
			if (arr1[i] == arr2[j]) {
				list.add(arr1[i]);
				i++;
				j++;
			} else if (arr1[i] < arr2[j]) {
				list.add(arr1[i]);
				i++;
			} else {
				list.add(arr2[j]);
				j++;
			}

			if (j == arr2.length) {
				flag = false;
			}
		}
		if (arr1.length > arr2.length) {
			int min = arr2.length;
			for (int x = min; x < arr1.length; x++) {
				list.add(arr1[x]);
			}
		} else {
			int min = arr1.length;
			for (int x = min; x < arr2.length; x++) {
				list.add(arr2[x]);
			}
		}
		System.out.println();
		for (Integer x : list) {
			System.out.print(x + " ");
		}
	}

- Somendra Raj May 19, 2018 | Flag Reply
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0
of 0 vote

In Swift

let setA:Set = [2, 10, 14, 19, 51, 71]
let setB:Set = [2, 9, 19, 40, 51]

let unionOfAandB = setA.union(setB)

for item in unionOfAandB
{
print("items is \(item)")
}

- Rahul Jain May 20, 2018 | Flag Reply
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0
of 0 vote

Use 'sorted' built-in function if you want the union to be sorted

a = {2, 10, 14, 19, 51, 71}
b = {2, 9, 19, 40, 51}
c = sorted(a.union(b))
print (c)

- Anonymous May 22, 2018 | Flag Reply
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0
of 0 vote

Use python's built-in function 'sorted' if you want the union to be sorted

a = {2, 10, 14, 19, 51, 71}
b = {2, 9, 19, 40, 51}
c = sorted(a.union(b)

- Goutham May 22, 2018 | Flag Reply
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0
of 0 vote

public List<Integer> union(int[] nums1, int[] nums2) {
	List<Integer> result = new LinkedList<Integer>();
	helper(result, nums1, nums2, 0, 0);
	return result;
}

void helper(List<Integer> result, int[] nums1, int[] nums2, int nums1Index, int nums2Index) {
	if(nums1Index >= nums1.length && nums2Index >= nums2.length) return;
	if(nums1Index >= nums1.length) {
		result.add(nums2[nums2Index]);
		helper(result, nums1, nums2, nums1Index, nums2Index + 1);
		return;
	}

	if(nums2Index >= nums2.length) {
		result.add(nums1[nums1Index]);
		helper(result, nums1, nums2, nums1Index + 1, nums2Index);
		return;
	}

	if(nums1[nums1Index] == nums2[nums2Index]) {
		result.add(nums1[nums1Index]);
		helper(result, nums1, nums2, nums1Index + 1, nums2Index + 1);
		return;
	}

	if(nums1[nums1Index] < nums2[nums2Index]) {
		result.add(nums1[nums1Index]);
		helper(result, nums1, nums2, nums1Index + 1, nums2Index);
		return;
	}

	result.add(nums2[nums2Index]);
	helper(result, nums1, nums2, nums1Index, nums2Index + 1);
}

- Anonymous May 23, 2018 | Flag Reply


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