What have you tried? Pseudocode would be ...
for all keys in hash-1, where k-1 is current key
for all keys in hash-2, where k-2 is current key
put k-1 in store if k-1 = k-2
end-for
end-for
# "store" would have all the common keys in the end
Time passes. Oh yes, finding if any keys are present ...
common set to false
for all keys in hash-1, where k-1 is current key
for all keys in hash-2, where k-2 is current key
if k-1 = k-2,
common set to true
break out of both loops
end-for
end-for
# test common to check for commonality