Farthest equal elements in an array - algorithm

Question

Suppose you have an unsorted array, how will you find 2 equal elements such that they are the farthest in the array. For eg 8 7 3 4 7 5 3 9 3 7 9 0 the ans will be 7(9) - 7(1) = 8.
I have thought of the following
initialise max = 0
using hashing, store the elements along with its index
whenever there is a collision, compare the current index with the hashed one
if it is greater, set max = current index - hashed one and copy element along with index to some other location.
runs in time - O(n) and space - O(n). Is this correct? Is there a better solution.

Answer 1

O(n) running time and O(n) space seems to be optimal AFAIK.
Here's python implementation:
#!/usr/bin/python
hashindex = {}
l = [8,7,3,4,7,5,3,9,3,7,9,0]
max_diff = 0
value = 0
for i in range(len(l)):
indices = hashindex.get(l[i], None)
if indices:
hashindex[l[i]] = (indices[0], i)
else:
hashindex[l[i]] = (i, i)
diff = hashindex[l[i]][1] - hashindex[l[i]][0]
if diff > max_diff:
max_diff = diff
value = l[i]
print max_diff, value

Answer 2

I very much doubt this is the "correct" answer but if I were given this problem at work, I would do a quick sort along the lines of sort by value then by index. Then I would walk through my ordered array and for each value, grab the first and last index. That would become a candidate max distance. Each time I encountered a new value, I would compare it against my current max distance and then store only the max. In this way, run time would be
sort + walk = total
N Log N + N = N Log N
Ex.:
Values in unordered array:
1,2,3,1,5,3,3,3
1,1,2,3,3,3,3,5 = Value
0,3,1,2,5,6,7,4 = Index
1's = max of 3-0 = 3
2's = x
3's = max of 7-2 = 5
Max of 5
Keep in mind this is probably NOT the textbook answer, but it still runs in N Log N so I'd use it at work with no qualms for anything under a million elements.
Have you thought of any other potential answers to the problem? How about ways to improve this one?

Answer 3

Here is my idea of the best solution:
Since the two elements is going to be the farthest apart, I would make a loop that looks at the start of the array, and at the end of the array at the same time, and look inwards, the first corresponding elements that they find, will be the farthest apart.
For example:
8 7 3 4 7 5 3 9 3 7 9 0
First look at 8, 0, and store them in set A, and in set B respectively.
Look at 7, 9, and store them in the same sets fashion.
Look at 3, 7, and same as above. Now this 7 is already in the opposite set, A.
Therefore they must be the farthest apart. Just calculate the distance between the the 7 elements.
Now that I think about it, you just need to make some kind of check to make sure that the farthest elements are now within a set itself.
So, if either of sets have 1/3 of all the elements, you have to also start checking if the largest distance is not within the set itself.

Answer 4

My solution in Java:
int[] testArr = {1,2,3,5,1,7,5,4,3,1,2,3,2,3};
int maxLen = 0;
int minidx = 0;
int maxidx = 0;
int numVal = testArr[0];
int counter = 0; //Just to check how many times its entering inside the below if condition
for (int ii=0;ii<testArr.length-maxidx;ii++) {
for (int jj=testArr.length-1;jj>=ii+maxidx;jj--){
if(testArr[ii]==testArr[jj] /*&& maxLen<(jj-ii)*/){
System.out.println(counter++);
minidx = ii;
maxidx = jj;
maxLen = jj-ii;
numVal = testArr[ii];
break;
}
}
}
System.out.println(maxLen +" "+minidx+" "+maxidx+" "+numVal);

Answer 5

runs in time - O(n) and space - O(n). Is this correct?
Yes.
Is there a better solution.
I highly doubt it.
JavaScript implementation:
function furthest(a) {
const first= {};
return a.reduce((longest, elt, idx) => {
if (!(elt in first)) first[elt] = idx;
return Math.max(longest, idx - first[elt]);
}, Infinity);
}

Answer 6

Solution in obj-c. Should be O(N). Just one pass over the array.
//NSMutableArray *numbers = [[NSMutableArray alloc] initWithArray:#[#3,#2,#1,#7,#1,#2,#3,#6,#5,#4,#4,#5,#6,#7]];
//int solution = [self maxDistanceBetweenPairs:numbers];
//NSLog(#"%d", solution);
- (int)maxDistanceBetweenPairs:(NSMutableArray *)A {
NSMutableDictionary *dictionary = [NSMutableDictionary dictionary];
__block int max = 0;
int i = 0;
for (NSNumber *num in A) {
if ([dictionary objectForKey:num]) {
int first = [[dictionary objectForKey:num] intValue];
if (i - first > max) {
max = i - first;
}
} else {
[dictionary setObject:[NSNumber numberWithInt:i] forKey:num];
}
i++;
}
return max;
}