Google Wordle Color String

Learning Outcome

Turn the Google Wordle Color String interview variant into a clear brute-force baseline, optimized pattern, and implementation plan.

Original Interview Statement

Given secret and guess of equal length, return b/y/g feedback with one secret character used at most once.

Examples

Brute Force Approach

For every guess character, search secret for an unused same character. This is O(n^2) and easy to get wrong with duplicates.

Optimized Approach

First mark greens, then count remaining secret letters. Use those counts to assign yellows exactly once.

Exact Pseudocode

mark all greens
count non-green secret chars
for each non-green guess char:
  if count exists: mark yellow and decrement
  else mark black
  

Reference Code

from collections import Counter

def color(secret, guess):
    n = len(secret)
    ans = ['b'] * n
    left = Counter()
    for i in range(n):
        if guess[i] == secret[i]:
            ans[i] = 'g'
        else:
            left[secret[i]] += 1
    for i in range(n):
        if ans[i] == 'g':
            continue
        if left[guess[i]] > 0:
            ans[i] = 'y'
            left[guess[i]] -= 1
    return ''.join(ans)
  

Complexity

Edge Cases

• Duplicate letters
• All greens
• No matches
• Guess char appears as green elsewhere

Follow-ups

• Support Unicode letters
• Return counts of each color

Nearest Practice References

• Wordle
• Bulls and Cows

Common Mistakes

• Copying the nearest LeetCode solution without checking the changed rule.
• Skipping duplicate or boundary cases.
• Not stating the brute force before the optimized approach.