L3 · FAANGDynamic programming~22 min · 6 tests

Edit Distance (Levenshtein)

Compute the minimum insertions, deletions and substitutions to turn one word into another with a 2-D DP in O(m·n), then cut space to one row. FAANG hard.

The problem

Return the minimum number of single-character insertions, deletions or substitutions needed to turn a into b.

Examples

Example 1
Input
```
edit_distance('horse', 'ros')
```
Expected output
```
3
```

Example 2

Input

edit_distance('intention', 'execution')

Expected output

+ 4 hidden tests on Submit.

Edge cases to ask about

Empty strings
Identical strings

Hints

0/3

How an interviewer scores this

0/9

1. Understand the problemRestate it, ask about edge cases and input size.
2. Write a brute-force solutionCorrect first, clever later.
3. Get the correct outputautoAll visible examples pass.
4. Analyse time complexityautoCommit to a Big-O in the Complexity Lab.
5. Analyse space complexityautoExtra memory, not the input.
6. Identify the bottleneckautoMeasure your code — where does it bend?
7. OptimiseUsually: a set or dict instead of a scan.
8. Explain why it worksSay it out loud to Solvi.
9. Test the edge casesautoSubmit passes every hidden test.

Python 3.13 · edit_distance

def edit_distance(a, b):
    """Minimum insert/delete/replace operations to turn a into b."""
    # Write your solution here
    pass

⌘/Ctrl + Enter runs the examples

Your code runs in real CPython inside your browser — nothing is sent anywhere. The first run downloads the interpreter (about 6 MB, once). Your code is saved on this device as you type.

Complexity Lab

What does this cost as n grows?

Interviewers score the analysis as much as the code. Commit to an answer first — then check it, and measure your code against the optimal one at growing input sizes.

Time complexity of the optimal solution

Space complexity (extra memory)

Pick both to reveal the answer.

Measure it

Runs the function on inputs of size 250 up to 16,000 and records the time and peak memory. Slow solutions stop early — a short curve is itself the answer.

From brute force to optimal

The progression an interviewer wants to hear, one step at a time.

Approach	Time	Space	Idea
Plain recursion	O(3^(m+n))	O(m + n)
2-D DP table	O(m · n)	O(m · n)	dp[i][j] = distance between prefixes a[:i] and b[:j].
bestRolling one row	O(m · n)	O(n)	Each row only needs the previous one.

Walkthrough of the optimal approach (try it yourself first)

dp[i][j] = distance between the first i characters of a and the first j of b. Base cases: turning a prefix into the empty string takes i deletions (dp[i][0] = i, dp[0][j] = j). If the last characters match, carry dp[i−1][j−1]; otherwise take 1 + the best of delete, insert and replace.

Each row depends only on the row above, so two rows (O(n) space) are enough.

Complexity: O(m · n) time, O(n) space. Every (i, j) prefix pair is computed once; only the previous row is kept.

Reveal the reference solution

def edit_distance(a, b):
    prev = list(range(len(b) + 1))
    for i in range(1, len(a) + 1):
        cur = [i] + [0] * len(b)
        for j in range(1, len(b) + 1):
            if a[i - 1] == b[j - 1]:
                cur[j] = prev[j - 1]
            else:
                cur[j] = 1 + min(prev[j],        # delete
                                 cur[j - 1],     # insert
                                 prev[j - 1])    # replace
        prev = cur
    return prev[-1]

Follow-ups interviewers ask

Reconstruct the actual edit operations.
Only insertions and deletions allowed.

Frequently asked interview questions

Core interview concepts, complexities, and follow-ups scored by hiring teams.

What is the time complexity of Edit Distance (Levenshtein) in Python?

The optimal solution runs in O(m · n) time and O(n) auxiliary space. Every (i, j) prefix pair is computed once; only the previous row is kept.

What is the brute-force approach, and how do you optimise it?

Plain recursion: O(3^(m+n)) time, O(m + n) space. 2-D DP table: O(m · n) time, O(m · n) space. dp[i][j] = distance between prefixes a[:i] and b[:j]. Rolling one row: O(m · n) time, O(n) space. Each row only needs the previous one.

What follow-up questions do interviewers ask about Edit Distance (Levenshtein)?

Reconstruct the actual edit operations. Only insertions and deletions allowed.