We shall consider each $$$A_i$$$ and count number of $$$j > i$$$ such that $$$A_i$$$ and $$$A_j$$$ satisfy the condition. Then the requirement is $$$j - i = A_i + A_j$$$, which can be rewritten as $$$j - A_j = i + A_i$$$. So basically, we need to count the number of $$$j$$$'s for which $$$j-A_j$$$ matches a given value. We can precalculate this value for all indices, store in a map, and query it every time to add to the final answer. Be careful to reduce the count of $$$i - A_i$$$ while moving to the next one. My submission

1. The question is asking us to find all pairs of indices i,j where j> i such that j-i = a[i]+a[j]
2. interchange the above formula into getting j terms on one side and i terms on one side => j-a[j] = a[i]+i.
3. This changes the problem statement to find all i,j j > i such that j-a[j] = a[i]+i
4. use map to store number of indeces having value x = a[i]+i before j and calculate the answer.