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I thought of one more solution , which is first to take all the elements and put them one by one into a hash table , and then run 2 loops , taking a[i] and a[j] and computing the value of sqrt(a[i]^2 + a[j]^2) and find the resulting elements in the hash table , if they are found they form the triplet , so it takes an overall time of
- banerjee.abhik.hcl January 25, 2011O(n) {to hash all the elements} + O(n^2) {to run the loop and check for the elements from the triplet} ...
Because I feel if u already do a square and then sort u might loose the triplets like 3,4,5 and 3,4,-5 ..
suggestions / modifications are welcome.