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Interview Question for Software Engineer at Google:
Feb 5, 2012
Helpful Question?
Yes |
No
Inappropriate?
Answers & Comments (3)
0 of 1 people found this helpful
Feb 14, 2012
I tried to implement it using HashTable technique
public ArrayList<Integer> intersection(int list1[], int list2[]) {
int i, max = 0, min = 0;
Hashtable<Integer, Integer> hash;
ArrayList<Integer> list = new ArrayList<Integer>();
if (list1.length > list2.length) {
hash = new Hashtable<Integer, Integer>(list1.length, list1.length);
max = list1.length;
min = list2.length;
} else {
hash = new Hashtable<Integer, Integer>(list2.length, list2.length);
max = list2.length;
min = list1.length;
}
for (i = 0; i < max; i++) {
if (max == list1.length)
hash.put(list1[i], list1[i]);
else
hash.put(list2[i], list2[i]);
}
for (i = 0; i < min; i++) {
if (min == list1.length) {
if (hash.put(list1[i], list1[i]) != null) {
list.add(list1[i]);
}
} else {
if (hash.put(list2[i], list2[i]) != null) {
list.add(list2[i]);
}
}
}
return list;
}
public ArrayList<Integer> intersection(int list1[], int list2[]) {
int i, max = 0, min = 0;
Hashtable<Integer, Integer> hash;
ArrayList<Integer> list = new ArrayList<Integer>();
if (list1.length > list2.length) {
hash = new Hashtable<Integer, Integer>(list1.length, list1.length);
max = list1.length;
min = list2.length;
} else {
hash = new Hashtable<Integer, Integer>(list2.length, list2.length);
max = list2.length;
min = list1.length;
}
for (i = 0; i < max; i++) {
if (max == list1.length)
hash.put(list1[i], list1[i]);
else
hash.put(list2[i], list2[i]);
}
for (i = 0; i < min; i++) {
if (min == list1.length) {
if (hash.put(list1[i], list1[i]) != null) {
list.add(list1[i]);
}
} else {
if (hash.put(list2[i], list2[i]) != null) {
list.add(list2[i]);
}
}
}
return list;
}
Helpful Answer?
Yes |
No
Inappropriate?
0 of 1 people found this helpful
Apr 7, 2012
why do you need this big code?
Try to be simple while explaining simple ideas..
1) add list1 elements to hash.
2) iterate through list2. check if the item is present in hash and add to output list.
3) return output list.
Best way is to leave it here and ask politely if he wants to code this simple naive idea?
Over explaining simple things will cast impression of small knowledge.
Try to be simple while explaining simple ideas..
1) add list1 elements to hash.
2) iterate through list2. check if the item is present in hash and add to output list.
3) return output list.
Best way is to leave it here and ask politely if he wants to code this simple naive idea?
Over explaining simple things will cast impression of small knowledge.
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No
Inappropriate?
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1 of 1 people found this helpful
by Harry:
http://codercareer.blogspot.com/2011/11/no-24-intersection-of-sorted-arrays.html