Short Problem Definition:
Given N integers, count the number of pairs of integers whose difference is K.
time complexity is
space complexity is
The solution is pretty straight-forward, just read the code :). The runtime complexity is calculated with log(N) access times for tree-based sets (not the case in Python).
Note: The problem specification has a contraint on K: 0<K<10^9. If K was allowed to be 0, this challenge would become slightly more difficult.
#!/usr/bin/py def pairs(a,k): answer = 0 s = set(a) for v in s: if v+k in s: answer += 1 return answer if __name__ == '__main__': n, k = map(int, raw_input().split()) b = map(int, raw_input().split()) print pairs(b, k)