Sum of distinct element in a range

Question

Consider a array (0 based index) i have to find the sum of distinct element of all possible range[i,n] where 0< i < n

Example:

arr={1,2,1,3}

sum_range[0,3]={1,2,3}=6
sum_range[1,3]={1,2,3}=6
sum_range[2,3]={1,3}=4
sum_range[3,3]={3}=3

O(n^2) solution is one possible solution and i have read persistent segment tree can also do this though i can't find good tutorial.

Can it be solved in less than O(N^2) time?

If someone points persistent segment tree please explain or provide some good link?

dasblinkenlight's algorithm is better, but another strategy would be to sort the part of the array that is within the range (`O(n log n)`), then traverse the sorted array while summing the elements that are different than their predecessors (`O(n)`), giving an overall time complexity of `O(n log n)`. — Jeff Irwin, Jan 01 '16 at 19:25

Sergey Kalinichenko · Answer 1 · 2016-01-01T17:02:54.220

5

This can be done in O(n) with a simple dynamic programming algorithm.

Start from the back of the array, and use a hash-based container to store the numbers that you have already encountered. The value of the last element, i.e. sum_range[n-1] is set to arr[n-1]. After that, values of sum_range[i] should be computed as follows:

If arr[i] is not in the set of seen numbers, sum_range[i] = sum_range[i+1]+arr[i]
Otherwise, sum_range[i] = sum_range[i+1]

Since the cost of checking a hash container for a value is O(1) and the cost of adding an item to hash container is amortized O(1) for n items, the overall cost of this algorithm is O(n).

Compared to O(n²) algorithm that uses O(1) space, this algorithm uses additional O(n) space for the hash container.

edited Jan 01 '16 at 17:02

answered Jan 01 '16 at 16:50

Sergey Kalinichenko

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Do you have a source for hash insert and lookup complexity ? I always heard this was O(log(n)) at best, because of collisions. I fail to find any source arguing for or against this. – Diane M Jan 01 '16 at 17:07
1

@ArthurHavlicek [Hash table : performance analysis](https://en.wikipedia.org/wiki/Hash_table#Performance_analysis) – Sergey Kalinichenko Jan 01 '16 at 17:09

Diane M · Answer 2 · 2016-01-01T20:29:29.757

0

Slice the array (O(n)), then use a set and add values to the set.

In python3.x code, edited after a comment:

array = [1, 2, 1, 3, 5]
range_sum = 0
total_sum = 0
valueset = set ()
for val in reversed(array):
    if val not in valueset :
        valueset.add (val)
        range_sum += val
    total_sum += range_sum

print (total_sum)

edited Jan 01 '16 at 20:29

answered Jan 01 '16 at 16:56

Diane M

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Where do you get a log n with set? – Untitled123 Jan 01 '16 at 17:01
OP asks for "sum of distinct element of all possible range[i,n]". This doesn't answer that. – Juan Lopes Jan 01 '16 at 19:09

score 0 · Answer 3 · edited Feb 14 '22 at 09:20

There are multiple ways to solve this.

sort the array by Arrays.sort();

Using set

long sumOfDistinct(long arr[], int N)
{
     int size = arr.length;
     HashSet<Long> unique = new HashSet<Long>(); 
     long sum = 0;
     for(int i=0;i<size;i++){
             if(!unique.contains(arr[i])){
                 sum = sum + arr[i];
                 unique.add(arr[i]);
             }
     }

     return sum;
 }

Sum of distinct element in a range

Example:

3 Answers3