Given that the list is sorted, you can run two pointers through the list in O (n) time. Basically something like:
index1 = 0 index2 = 0 while index2 < size(array): if array[index2] - array[index1] == K: print both numbers and exit if array[index2] - array[index1] < K: index2++; else index1++;
In other words, if the difference between the numbers is too small, increase the larger number (make the difference larger), otherwise increase the smaller number (reduce the difference).
You can see this in action with the following Python program:
lst = [1,2,3,4,5,6,7,50,100,120,121,122,123,130,199,299,399] diff = 7 ix1 = 0 ix2 = 0 while ix2 < len (lst): print "Comparing [%d]=%d against [%d]=%d"%(ix1,lst[ix1],ix2,lst[ix2]) if lst[ix2] - lst[ix1] == diff: print lst[ix1], lst[ix2] break if lst[ix2] - lst[ix1] < diff: ix2 = ix2 + 1 else: ix1 = ix1 + 1
which outputs:
Comparing [0]=1 against [0]=1 Comparing [0]=1 against [1]=2 Comparing [0]=1 against [2]=3 Comparing [0]=1 against [3]=4 Comparing [0]=1 against [4]=5 Comparing [0]=1 against [5]=6 Comparing [0]=1 against [6]=7 Comparing [0]=1 against [7]=50 Comparing [1]=2 against [7]=50 Comparing [2]=3 against [7]=50 Comparing [3]=4 against [7]=50 Comparing [4]=5 against [7]=50 Comparing [5]=6 against [7]=50 Comparing [6]=7 against [7]=50 Comparing [7]=50 against [7]=50 Comparing [7]=50 against [8]=100 Comparing [8]=100 against [8]=100 Comparing [8]=100 against [9]=120 Comparing [9]=120 against [9]=120 Comparing [9]=120 against [10]=121 Comparing [9]=120 against [11]=122 Comparing [9]=120 against [12]=123 Comparing [9]=120 against [13]=130 Comparing [10]=121 against [13]=130 Comparing [11]=122 against [13]=130 Comparing [12]=123 against [13]=130 123 130