Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/imperative-programming/1/b/ii/solution

There are candidate values in and exactly occurrences there. If , then
Thus the pigeonhole principle gives
This conclusion does not assert that every value on the lower side is present; duplicates may leave both sides deficient.

New to topics? Read the docs here!