Solution Approach

The implementation follows a greedy traversal approach where we process each character position from left to right:

1. Convert string to list: First, we convert the input string s to a list cs = list(s) to make it mutable, since strings in Python are immutable and we need to modify characters.

2. Iterate through each position: We use enumerate(s) to get both the index i and character c1 at each position.

3. Try smaller characters: For each position, we iterate through all lowercase letters using ascii_lowercase (which gives 'a' to 'z'). We only consider characters c2 that are smaller than the current character c1:

   • The condition if c2 >= c1: break stops the iteration once we reach characters that are not smaller than c1

4. Calculate minimum distance: For each candidate character c2 that is smaller than c1, we calculate the minimum cyclic distance:

   d = min(ord(c1) - ord(c2), 26 - ord(c1) + ord(c2))

   • ord(c1) - ord(c2): Direct backward distance
   • 26 - ord(c1) + ord(c2): Forward distance with wrap-around

5. Make the change if affordable: If the distance d is within our budget (d <= k):

   • Update the character at position i: cs[i] = c2
   • Deduct the cost from our budget: k -= d
   • Break from the inner loop to move to the next position

6. Return the result: After processing all positions, we join the list back into a string using "".join(cs) and return it.

The algorithm ensures optimality because it greedily makes each position as small as possible from left to right, using the available budget efficiently. The time complexity is O(26n) where n is the length of the string, since for each character we potentially check up to 26 lowercase letters.

Example Walkthrough

Let's walk through the solution with s = "zbbz" and k = 3.

Initial state: cs = ['z', 'b', 'b', 'z'], k = 3

Position 0 (character 'z'):

• We try to change 'z' to smaller characters starting from 'a'
• For 'z' → 'a':
  • Direct backward distance: ord('z') - ord('a') = 25
  • Forward wrap-around distance: 26 - ord('z') + ord('a') = 26 - 25 + 0 = 1
  • Minimum distance: min(25, 1) = 1
  • Since 1 ≤ 3, we can make this change!
• Update: cs[0] = 'a', k = 3 - 1 = 2
• Result so far: ['a', 'b', 'b', 'z']

Position 1 (character 'b'):

• We try to change 'b' to 'a' (the only smaller character)
• For 'b' → 'a':
  • Direct backward distance: ord('b') - ord('a') = 1
  • Forward wrap-around distance: 26 - ord('b') + ord('a') = 26 - 1 + 0 = 25
  • Minimum distance: min(1, 25) = 1
  • Since 1 ≤ 2, we can make this change!
• Update: cs[1] = 'a', k = 2 - 1 = 1
• Result so far: ['a', 'a', 'b', 'z']

Position 2 (character 'b'):

• We try to change 'b' to 'a'
• For 'b' → 'a':
  • Minimum distance: 1 (as calculated above)
  • Since 1 ≤ 1, we can make this change!
• Update: cs[2] = 'a', k = 1 - 1 = 0
• Result so far: ['a', 'a', 'a', 'z']

Position 3 (character 'z'):

• We have k = 0, so we cannot make any more changes
• The character remains 'z'

Final result: "aaaz"

The algorithm successfully transformed "zbbz" into "aaaz" using exactly 3 units of distance, creating the lexicographically smallest possible string within the budget.

Time and Space Complexity

Time Complexity: O(n × |Σ|) where n is the length of the string s and |Σ| is the size of the character set (at most 26 for lowercase English letters).

The algorithm iterates through each character in the string s (outer loop runs n times). For each character position, it iterates through the lowercase alphabet checking for potential replacements (inner loop runs at most |Σ| = 26 times). The inner loop breaks as soon as it finds a valid replacement or when c2 >= c1, but in the worst case, it may check all 26 letters. All operations inside the nested loops (comparisons, arithmetic operations, and character replacement) take O(1) time.

Space Complexity: O(n) where n is the length of the string s.

The algorithm creates a list cs from the input string s, which requires O(n) space to store all characters. The variable ascii_lowercase is a constant string of 26 characters (O(1) space), and other variables like i, c1, c2, d, and k use O(1) space. The final string construction with "".join(cs) creates a new string of length n, but this is typically considered part of the output rather than auxiliary space.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Not Considering All Possible Replacements

The Issue: The current implementation only tries to replace characters with lexicographically smaller ones. However, this misses opportunities where we might need to replace a character with a lexicographically larger one at an earlier position to save budget for more impactful changes at later positions.

Example: Consider s = "zzzzzz" with k = 3. The optimal answer is "aaazzz" (changing first 3 'z's to 'a's, each costing 1 due to wrap-around). However, if we had a string like "lmnxyz" with limited k, we might want to keep early characters unchanged or even change them to slightly larger characters if it helps us make more significant improvements to later positions.

The Real Issue: Actually, the main pitfall is that the code stops looking for replacements once it finds a character >= the original. This means it won't consider changing a character to itself (distance 0) or to a larger character that might have a small cyclic distance.

Corrected Solution:

class Solution:  def getSmallestString(self, s: str, k: int) -> str:  from string import ascii_lowercase   char_list = list(s)   for i, original_char in enumerate(s):  best_char = original_char  best_cost = 0   # Try all possible characters, not just smaller ones  for replacement_char in ascii_lowercase:  # Calculate minimum cyclic distance  diff = abs(ord(original_char) - ord(replacement_char))  min_distance = min(diff, 26 - diff)   # If this replacement is affordable and better  if min_distance <= k:  # Choose the lexicographically smallest character within budget  if replacement_char < best_char:  best_char = replacement_char  best_cost = min_distance  # Break early if we found 'a' (can't do better)  if best_char == 'a':  break   # Apply the best replacement found  if best_char != original_char:  char_list[i] = best_char  k -= best_cost   return "".join(char_list)

Pitfall 2: Incorrect Distance Calculation for Wrap-Around

The Issue: When calculating wrap-around distance, the formula 26 - ord(c1) + ord(c2) assumes going from c1 forward to c2. This works when c2 < c1, but if we're considering all possible replacements, we need a more general formula.

Correct Approach: Use min(abs(ord(c1) - ord(c2)), 26 - abs(ord(c1) - ord(c2))) which correctly handles all cases regardless of which character is larger.