Minimum Insertions/Deletions to Convert String
A medium Dynamic Programming problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Dynamic Programming
- Sheets
- 1
- Core for
- 9 roles
- Platform
- LeetCode
The problem
Given two strings, find the minimum number of operations (insertions and deletions) needed to make the two strings equal.
Example 1
- Input
- word1 = "sea", word2 = "eat"
- Output
- 2
- Why
- Delete 's' from "sea" and delete 't' from "eat" to get "ea" for both.
Example 2
- Input
- word1 = "leetcode", word2 = "etco"
- Output
- 4
- Why
- Delete 4 characters total to make the strings equal.
Example 3
- Input
- word1 = "abc", word2 = "abc"
- Output
- 0
- Why
- Strings are already equal, no operations needed.
Constraints
- 1 <= word1.length, word2.length <= 500
- word1 and word2 consist of lowercase English letters
How to think about it
Updated 2026-09-09The largest common part that requires zero operations to retain is their Longest Common Subsequence. Any character in word1 not in the LCS must be deleted, and any character in word2 not in the LCS must be inserted. The answer is (len1 - lcs) + (len2 - lcs).
Approaches, worst first
Direct edit distance modification
time O(m * n) · space O(m * n)
Run classic LeetCode edit distance DP but disable substitutions by giving substitution an infinite cost, leaving only deletion and insertion at cost 1 each.
LCS reductionWrite this one
time O(m * n) · space O(min(m, n))
Compute the length of the Longest Common Subsequence between word1 and word2 using a 2D table. Return word1.length + word2.length - 2 * lcs_length.
Where people lose marks · 3
- Confusing this problem with Edit Distance (which permits character replacements at cost 1). Here replacement requires 1 deletion plus 1 insertion, costing 2.
- Failing to handle identical strings where answer is 0.
- Allocating full m x n matrix when memory is limited; a 1D rolling array handles LCS reduction readily.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
What each operation costs
| Operation | Time |
|---|---|
| fill dynamic programming table of n states | O(n) |
| solve two-dimensional grid of m by n states | O(m * n) |
| space-optimized state transition keeping one row | O(n) |
What usually goes wrong with Dynamic Programming
- Filling a bottom-up table in an order where the current cell needs values that have not been computed yet, reading uninitialized zeros.
- Failing to initialize base cases properly, such as filling a minimization table with zeros instead of infinity, which traps the answer at zero.
- Overwriting values in a 1D space-optimized knapsack array by scanning in the wrong direction, allowing the same item to be chosen multiple times.
Which roles need this problem
Dynamic Programming is a core topic for these 9 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Game Developer, Cryptography Engineer, Performance Engineer.
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