Most of the easy problems from tags like "constructive algorithms", "implementation", "brute force" etc. could be what you're looking for. Are you sure you checked this?

Please also include Interviewbit under interview questions / online sites to practice.

No, you store the results of first round in a Map, then for every result in second round, you search it in the map, O(n^2 * logn). Assuming, you didn't store the results in a map, but just in a array, to search for every a[p] ^ a[q] in second round, it would take O (n^3).

For B, The solution is simple. You anyhow have to check by XORring four elements at a time, i.e. a[i] ^ a[j] ^ a[p] ^ a[q]. But, instead of brute force'ing at one single time, you can divide the brute force into two. i.e. first compute a[i] ^ a[j], i.e. run brute force once in ( O(n^2) ) save the results, again brute force ( in O(n^2) again ) for a[p] ^ a[q] now either you can brute force again through previous results or just simply look for k ^ ( a[p] ^ a[q] ) in the previous results if you've stored them in a Map.

Why are people downvoting ? If you're here reading this post, then it is because you too were looking for an easy implementation of Suffix Tree, ryt? So, this is not fair.

Could you please also write a small function to search if a string exists as a substring in the original string ?

Only for Asia Pacific.