##### Short Problem Definition:

Given N integers, count the number of pairs of integers whose difference is K.

##### Link

##### Complexity:

time complexity is `O(N\*log(N))`

space complexity is `O(N)`

##### Execution:

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.

##### Solution:

```
#!/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)
```