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I != j != k Condition Not Met

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#1

The problem states i != j != k. Wouldn’t we need to de-duplicate the array before counting?

I don’t see how the current solution guarantees not counting duplicated numbers. Thank you.

#2

We don’t need to find three different numbers. ‘i’, ‘j’, and ‘k’ are indices here, so we just need to have three numbers from the array, these numbers can all be same:

count all triplets in it such that arr[i] + arr[j] + arr[k] < target where i , j , and k are three different indices.

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#3

But your question contradicts the solution.

in the solution, you’ve mentioned:
“make sure that each number is not used more than once.”

and in the above comment, you’ve mentioned:
“we just need to have three numbers from the array, these numbers can all be same”

#4

@qqq: There are two things discussed here: 1) Indices, and 2) numbers pointed by these indices. The problem states that the indices need to be different (i.e., i != j != k) but the numbers pointed out by these indices could all be same, as there could be repeating numbers in the array.

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