Subarrays with distinct elements

Given an array, the task is to calculate the sum of lengths of contiguous subarrays having all elements distinct.


Input :  arr[] = {1, 2, 3}
Output : 10
{1, 2, 3} is a subarray of length 3 with 
distinct elements. Total length of length
three = 3.
{1, 2}, {2, 3} are 2 subarray of length 2 
with distinct elements. Total length of 
lengths two = 2 + 2 = 4
{1}, {2}, {3} are 3 subarrays of length 1
with distinct element. Total lengths of 
length one = 1 + 1 + 1 = 3
Sum of lengths = 3 + 4 + 3 = 10

Input :  arr[] = {1, 2, 1}
Output : 7

Input :  arr[] = {1, 2, 3, 4}
Output : 20

A simple solution is to consider all subarrays and for every subarray check if it has distinct elements or not using hashing. And add lengths of all subarrays having distinct elements. If we use hashing to find distinct elements, then this approach takes O(n2) time under the assumption that hashing search and insert operations take O(1) time.

An efficient solution is based on the fact that if we know all elements in a subarray arr[i..j] are distinct, sum of all lengths of distinct element subarrays in this sub array is ((j-i+1)*(j-i+2))/2. How? the possible lengths of subarrays are 1, 2, 3,……, j – i +1. So, the sum will be ((j – i +1)*(j – i +2))/2.

We first find largest subarray (with distinct elements) starting from first element. We count sum of lengths in this subarray using above formula. For finding next subarray of the distinct element, we increment starting point, i and ending point, j unless (i+1, j) are distinct. If not possible, then we increment i again and move forward the same way.

Below is C++ implementation of this approach:

[sourcecode language=”CPP” highlight=”6-37″]
// C++ program to calculate sum of lengths of subarrays
// of distinct elements.
using namespace std;

// Returns sum of lengths of all subarrays with distinct
// elements.
int sumoflength(int arr[], int n)
// For maintaining distinct elements.
unordered_set<int> s;

// Initialize ending point and result
int j = 0, ans = 0;

// Fix starting point
for (int i=0; i<n; i++)
// Find ending point for current subarray with
// distinct elements.
while (j < n && s.find(arr[j]) == s.end())

// Calculating and adding all possible length
// subarrays in arr[i..j]
ans += ((j – i) * (j – i + 1))/2;

// Remove arr[i] as we pick new stating point
// from next

return ans;

// Driven Code
int main()
int arr[] = {1, 2, 3, 4};
int n = sizeof(arr)/sizeof(arr[0]);
cout << sumoflength(arr, n) << endl;
return 0;


Time Complexity of this solution is O(n). Note that the inner loop runs n times in total as j goes from 0 to n across all outer loops. So we do O(2n) operations which is same as O(n).

This article is contributed by Anuj Chauhan(anuj0503). If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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