Consolidation - Python riddles

Solve the following using standard Python features and built-in functions.

If possible, work in the pair programming paradigm: work in pairs, with one person taking the role of the driver (writing the code) and one taking the role of the navigator (reading and understanding the documentation). Alternate the roles. Try to find solutions that are short (i.e. few line sof code) but easy to understand.

Palyndrome checker

• Riddle: A word is a palyndrome if it reads the same forwards and backwards. Write a function is_palindrome(s) that takes an object s, checks that it is a string and returns True if s is a palindrome and False otherwise.
• Test it with the following test cases: is_palindrome("racecar")-->True , is_palindrome("hello")-->False and is_palindrome(3)-->error

Hint: an object is a string if typ(s) returns str.

The Peak Finder

• Riddle: In a list of numbers, a “peak” is a number that is greater than both its neighbors. The first and last elements can only be peaks if they’re greater than their single neighbor. Write find_peaks(data) that returns a list of all peak values (not their positions, just the values).

  Example: In [1, 3, 2, 5, 4, 6, 1], the peaks are [3, 5, 6] because:

  • 3 > 1 and 3 > 2 ✓
  • 5 > 2 and 5 > 4 ✓
  • 6 > 4 and 6 > 1 ✓

• Test cases:

  • find_peaks([1, 3, 2, 5, 4, 6, 1]) → [3, 5, 6]
  • find_peaks([1, 2, 3, 4, 5]) → [5] (only the last element)
  • find_peaks([5, 4, 3, 2, 1]) → [5] (only the first element)
  • find_peaks([1, 1, 1]) → [] (no peaks - they must be strictly greater!)

Hint: Loop through indices 0 to len(data)-1 and check neighbors carefully at the boundaries.

The Frequency Detective

• Riddle: In a dataset, you want to find the value that appears most often (the “mode”). But here’s the twist: if there’s a tie, return the smallest value among the most frequent ones. Write find_mode(data) to solve this mystery.

  Example: In [1, 2, 2, 3, 3, 4], both 2 and 3 appear twice. Return 2 (smallest of the tied values).

• Test cases:

  • find_mode([1, 2, 2, 3, 3, 4]) → 2 (2 and 3 tied at 2 occurrences, but 2 is smaller)
  • find_mode([5, 5, 3, 3, 3, 1]) → 3 (appears 3 times, most frequent)
  • find_mode([7, 7, 7, 2, 2, 2]) → 2 (tied at 3 occurrences, 2 is smaller)
  • find_mode([4]) → 4 (single element)

Hint: Count occurrences with a dictionary, find the maximum count, then among all values with that count, return the minimum.