Booby Trap

U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Q
  • What is the desired outcome?
    • To determine if it's possible to remove exactly one letter from the code so that the frequency of all remaining letters is equal.
  • What input is provided?
    • A string code consisting of only lowercase English letters.

P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Calculate the frequency of each letter in the string and then analyze whether removing one letter can result in equal frequencies for all remaining letters.

1) Calculate the frequency of each character in the `code`.
2) Calculate the frequency of these frequencies.
3) If there are more than two different frequencies, it's impossible to balance the code by removing one letter.
4) If there is exactly one frequency, the code is already balanced.
5) If there are exactly two different frequencies, check the following conditions:
   - One frequency should be exactly 1 more than the other, and it should appear only once.
   - Or one frequency should be 1 and it should appear only once.
6) Return `True` if any of the conditions hold; otherwise, return `False`.

⚠️ Common Mistakes

• Not considering edge cases where the string is already balanced.
• Incorrectly handling cases with more than two different frequencies.

I-mplement

def is_balanced(code):
    # Step 1: Calculate frequencies of each character
    freq = {}
    for char in code:
        if char in freq:
            freq[char] += 1
        else:
            freq[char] = 1
    
    # Step 2: Calculate frequencies of these frequencies
    freq_of_freq = {}
    for count in freq.values():
        if count in freq_of_freq:
            freq_of_freq[count] += 1
        else:
            freq_of_freq[count] = 1
    
    # There can be at most two different frequencies to consider valid transformation
    if len(freq_of_freq) > 2:
        return False
    
    # Analyze the two cases where we might be able to equalize the frequencies
    if len(freq_of_freq) == 1:
        # All characters already have the same frequency
        return True

    # If there are exactly two different frequencies, we need to check further conditions
    f1, f2 = list(freq_of_freq.keys())
    if (freq_of_freq[f1] == 1 and (f1 - 1 == f2 or f1 == 1)) or (freq_of_freq[f2] == 1 and (f2 - 1 == f1 or f2 == 1)):
        return True
    
    return False

Example Usage: code1 = "abcc" code2 = "aazz" print(is_balanced(code1)) # Output: True print(is_balanced(code2)) # Output: False