given 3 arrays check if there is any common number - arrays

Question

**I have 3 arrays a[1...n] b[1...n] c[1....n] which contain integers.
It is not mentioned if the arrays are sorted or if each array has or has not duplicates.
The task is to check if there is any common number in the given arrays and return true or false.
For example : these arrays a=[3,1,5,10] b=[4,2,6,1] c=[5,3,1,7] have one common number : 1
I need to write an algorithm with time complexity O(n^2).
I let the current element traversed in a[] be x, in b[] be y and in c[] be z and have following cases inside the loop : If x, y and z are same, I can simply return true and stop the program,something like:
for(x=1;x<=n;x++)
for(y=1;y<=n;y++)
for(z=1;z<=n;z++)
if(a[x]==b[y]==c[z])
return true
But this algorithm has time complexity O(n^3) and I need O(n^2).Any suggestions?

Answer 1

There is a pretty simple and efficient solution for this.
Sort a and b. Complexity = O(NlogN)
For each element in c, use binary search to check if it exists in both a and b. Complexity = O(NlogN).
That'll give you a total complexity of O(NlogN), better than O(N^2).

Answer 2

Create a new array, and save common elements in a and b arrays. Then find common elements in this array with c array.
python solution
def find_d(a, b, c):
for i in a:
for j in b:
if i==j:
d.append(i)
def findAllCommon(c, d):
for i in c:
for j in d:
if i==j:
e.append(i)
break
a = [3,1,5,10]
b = [4,2,6,1]
c = [5,3,1,7]
d = []
e = []
find_d(a, b, c)
findAllCommon(c, d)
if len(e)>0:
print("True")
else:
print("False")

Answer 3

Since I haven't seen a solution based on sets, so I suggest looking for how sets are implemented in your language of choice and do the equivalent of this:
set(a).intersection(b).intersection(c) != set([])
This evaluates to True if there is a common element, False otherwise. It runs in O(n) time.

Answer 4

All solutions so far either require O(n) additional space (creating a new array/set) or change the order of the arrays (sorting).
If you want to solve the problem in O(1) additional space and without changing the original arrays, you indeed can't do better than O(n^2) time:
foreach(var x in a) { // n iterations
if (b.Contains(x) && c.Contains(x)) return true; // max. 2n
} // O(n^2)
return false;

Answer 5

A suggestion:
Combine into a single array(z) where z = sum of the entries in each array. Keep track of how many entries there were in Array 1, Array 2, Array 3.
For each entry Z traverse the array to see how many duplicates there are within the combined array and where they are. For those which have 2 or more (ie there are 3 or more of the same number), check that the location of those duplicates correspond to having been in different arrays to begin with (ruling our duplicates within the original arrays). If your number Z has 2 or more duplicates and they are all in different arrays (checked through their position in the array) then store that number Z in result array.
Report result array.
You will traverse the entire combined array once and then almost (no need to check if Z is a duplicate of itself) traverse it again for each Z, so n^2 complexity.
Also worth noting that the time complexity will now be a function of total number of entries and not of number of arrays (your nested loops would become n^4 with 4 arrays - this will stay as n^2)
You could make it more efficient by always checking the already found duplicates before checking for a new Z - if the new Z is already found as a duplicate to an earlier Z you need not traverse to check for that number again. This will make it more efficient the more duplicates there are - with few duplicates the reduction in number of traverses is probably not worth the extra complexity.
Of course you could also do this without actually combining the values into a single array - you would just need to make sure that your traversing routine looks through the arrays and keeps track of what it finds the in the right order.
Edit
Thinking about it, the above is doing way more than you want. It would allow you to report on doubles, quads etc. as well.
If you just want triples, then it is much easier/quicker. Since a triple needs to be in all 3 arrays, you can start by finding those numbers which are in any of the 2 arrays (if they are different lengths, compare the 2 shortest arrays first) and then to check any doublets found against the third array. Not sure what that brings the complexity down to but it will be less than n^2...

Answer 6

there are many ways to solve this here few selected ones sorted by complexity (descending) assuming n is average size of your individual arrays:
Brute force O(n^3)
its basicaly the same as you do so test any triplet combination by 3 nested for loops
for(x=1;x<=n;x++)
for(y=1;y<=n;y++)
for(z=1;z<=n;z++)
if(a[x]==b[y]==c[z])
return true;
return false;
slightly optimized brute force O(n^2)
simply check if each element from a is in b and if yes then check if it is also in c which is O(n*(n+n)) = O(n^2) as the b and c loops are not nested anymore:
for(x=1;x<=n;x++)
{
for(ret=false,y=1;y<=n;y++)
if(a[x]==b[y])
{ ret=true; break; }
if (!ret) continue;
for(ret=false,z=1;z<=n;z++)
if(a[x]==c[z])
{ ret=true; break; }
if (ret) return true;
}
return false;
exploit sorting O(n.log(n))
simply sort all arrays O(n.log(n)) and then just traverse all 3 arrays together to test if each element is present in all arrays (single for loop, incrementing the smallest element array). This can be done also with binary search like one of the other answers suggest but that is slower still not exceeding n.log(n). Here the O(n) traversal:
for(x=1,y=1,z=1;(x<=n)&&(y<=n)&&(z<=n);)
{
if(a[x]==b[y]==c[z]) return true;
if ((a[x]<b[y])&&(a[x]<c[z])) x++;
else if ((b[y]<a[x])&&(b[y]<c[z])) y++;
else z++;
}
return false;
however this needs to change the contents of arrays or need additional arrays for index sorting instead (so O(n) space).
histogram based O(n+m)
this can be used only if the range of elements in your array is not too big. Let say the arrays can hold numbers 1 .. m then you add (modified) histogram holding set bit for each array where value is presen and simply check if value is present in all 3:
int h[m]; // histogram
for(i=1;i<=m;i++) h[i]=0; // clear histogram
for(x=1;x<=n;x++) h[a[x]]|=1;
for(y=1;y<=n;y++) h[b[y]]|=2;
for(z=1;z<=n;z++) h[c[z]]|=4;
for(i=1;i<=m;i++) if (h[i]==7) return true;
return false;
This needs O(m) space ...
So you clearly want option #2
Beware all the code is just copy pasted yours and modified directly in answer editor so there might be typos or syntax error I do not see right now...