SUBARRAYS WITH DISTINCT ELEMENTS 2testcases failed

#1

2 testcases failed…
logic— using hashmap and array when a distict element comes just add it in hashmap if a same element encountered then reform the hashmap by only having elemnts from last occurence of the elent to the new occurence.
and calculation length of contiguous subarrays by (l+1)(l)(l+2)/6…and summing all length and moduling at each go.

#2

can i just know that how are you dividing by 6?
if you use simple divide, then the answer might go wrong .
therefore use modular inverse to divide .
a/b=(a*modular_inverse(b))%mod.
where, modular_inverse(b)=power(b,mod-2);

#3

why mod-2??
and i am getting (l+1)(l)(l+2)/6 by doing summation of the series is it wrong??

#4

yes that is wrong too.
let say we have an array A and with indexes from i to j we have all distinct elements.
Here j is the maximum of all the indexes which satisfy the condition.
than total number of subarray with distinct elements starting from i are (j-i)(j-i+1)/2;
ie let the length of the subarray be n.
all subarrays with
1 element are n.
2 element are n-1.
3 element are n-2.
.
.
.
.
n element are 1.
adding we have n(n+1)/2

#5

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#6

please tell what if i encounter a duplicate element how to handle it?

#7

I am unable to understand the approach pls help?