This problem requires calculating the minimum edits needed to convert word1 into word2 using insertions, deletions, and replacements. A state transition dynamic programming approach efficiently captures overlapping subproblems by storing intermediate results. Tracking these transitions ensures correct computation even for strings up to 500 characters, avoiding redundant recursion and making the solution practical for interview scenarios.
Problem Statement
Given two strings, word1 and word2, determine the minimum number of operations required to convert word1 into word2. Allowed operations include inserting a character, deleting a character, or replacing a character.
Return the minimum total operations needed. For example, converting 'horse' to 'ros' involves replacing 'h' with 'r', then deleting 'r' and 'e', yielding 3 operations. Input strings have lengths up to 500 and contain only lowercase English letters.
Examples
Example 1
Input: word1 = "horse", word2 = "ros"
Output: 3
horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')
Example 2
Input: word1 = "intention", word2 = "execution"
Output: 5
intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')
Constraints
- 0 <= word1.length, word2.length <= 500
- word1 and word2 consist of lowercase English letters.
Solution Approach
Define the DP State
Let dp[i][j] represent the minimum operations to convert the first i characters of word1 to the first j characters of word2. This state captures all previous edits and sets up the framework for state transitions.
State Transition Formula
For each dp[i][j], if word1[i-1] equals word2[j-1], then dp[i][j] = dp[i-1][j-1]; otherwise, dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]), corresponding to deletion, insertion, or replacement.
Implementation and Space Optimization
Iteratively fill the dp table from base cases: dp[0][j] = j and dp[i][0] = i. For large strings, reduce space from O(m*n) to O(min(m,n)) by storing only the previous row, maintaining correctness while saving memory.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(mn) because each subproblem dp[i][j] is computed once, iterating through all i x j combinations. Space complexity is O(mn) for the full DP table, or O(min(m,n)) if optimized to store only one previous row, which is critical for long strings.
What Interviewers Usually Probe
- Asks how the DP table represents overlapping subproblems and why it's necessary.
- Checks if you can optimize space for large input strings up to 500 characters.
- Inquires about handling equal characters and how it affects the state transition.
Common Pitfalls or Variants
Common pitfalls
- Failing to initialize base cases correctly, leading to off-by-one errors in dp[0][j] or dp[i][0].
- Confusing insertion and deletion operations, which can produce incorrect minimal edits.
- Attempting recursive solutions without memoization, causing exponential time complexity and timeouts.
Follow-up variants
- Compute edit distance with only insertions and deletions allowed, ignoring replacements.
- Find all sequences of operations that achieve the minimum edit distance.
- Apply edit distance to detect approximate string matching in a text or DNA sequence.
How GhostInterview Helps
- GhostInterview provides step-by-step DP table construction to track state transitions efficiently.
- It highlights the minimum operation choice at each step, avoiding common missteps in insertion, deletion, and replacement handling.
- The solver visualizes space optimization options and shows their impact on large inputs, ensuring practical interview solutions.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main pattern behind the Edit Distance problem?
The main pattern is state transition dynamic programming, where each subproblem depends on previous states representing partial string conversions.
Can the space complexity be reduced for Edit Distance?
Yes, by storing only the previous DP row, space can be reduced from O(m*n) to O(min(m,n)) without affecting correctness.
How do insertions, deletions, and replacements relate in the DP formula?
They correspond to dp[i-1][j] for deletion, dp[i][j-1] for insertion, and dp[i-1][j-1] for replacement, ensuring each choice is counted correctly.
What are common mistakes in implementing Edit Distance?
Typical mistakes include off-by-one errors, confusing insertion with deletion, and using recursive approaches without memoization.
Is Edit Distance suitable for strings of length 500?
Yes, the DP solution handles up to 500 characters efficiently, especially when optimized for space using only one previous row.
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