The 6-Step Method — A Worked Example

The idea, in one line

Don't rush to code. Understand the problem, try a slow-but-simple idea, make it faster, then write it — in that order.

1 · Restate

Say it back in your own words: given text, return the first character (reading left to right) whose count in the whole string is exactly one. In: a string. Out: one character, or a signal if there is none.

2 · Examples

Write a few by hand, including a tricky one:

• "swiss" → "w" (s repeats; w is the first that doesn't)
• "aabb" → none (nothing is unique) ← the edge case worth writing down
• "z" → "z"

3 · Brute force

Reach for the obvious slow idea first: for each character, scan the whole string and count how many times it appears; return the first one whose count is 1. It works — even though it re-scans a lot.

4 · Optimize

All that re-scanning is wasted effort. Instead, count every character once into a dict, then make a second left-to-right pass and return the first character whose stored count is 1. Two passes instead of many.

5 · Code

def first_unique(text):
    # Step 4: count each character once
    counts = {}
    for ch in text:
        counts[ch] = counts.get(ch, 0) + 1

    # Second pass, left to right: first with count 1 wins
    for ch in text:
        if counts[ch] == 1:
            return ch
    return None            # edge case: nothing was unique

print(first_unique("swiss"))   # w
print(first_unique("aabb"))    # None
print(first_unique("z"))       # z

6 · Trace

Check it by hand. For "swiss" the counts are s→3, w→1, i→1. Second pass: s (3) skip, w (1) → return "w". It matches the example you wrote in step 2 — that match is your proof.

The house method — restate, examples, brute force, optimize, code, trace