Regular Expression Matching
A hard Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
Given an input string and a pattern containing lowercase letters and wildcard characters '?' and '*', determine if the pattern matches the entire input string. '?' matches any single character, and '*' matches any sequence of characters including an empty sequence.
Example 1
- Input
- s = "aa", p = "a"
- Output
- false
- Why
- "a" cannot match the entire string "aa".
Example 2
- Input
- s = "aa", p = "*"
- Output
- true
- Why
- "*" matches any sequence including "aa".
Example 3
- Input
- s = "cb", p = "?a"
- Output
- false
- Why
- '?' matches 'c' but 'a' does not match 'b'.
Constraints
- 0 <= s.length, p.length <= 2000
- s contains only lowercase English letters
- p contains only lowercase English letters, '?' or '*'
How to think about it
Updated 2026-09-29The asterisk is greedy in possibility: it can match zero characters (inherit match status from pattern without asterisk) or absorb one or more characters (inherit match status from string prefix without the current character). An exact match or '?' steps both indices together.
Approaches, worst first
Recursive matching
time O(2^(m+n)) · space O(m + n)
Step through string and pattern. On '*', branch into matching 0 characters or consuming one character and keeping '*'. Repeated branching on multiple asterisks results in exponential time.
2D dynamic programming grid
time O(m * n) · space O(m * n)
dp[i][j] tracks whether s[0..i-1] matches p[0..j-1]. If p[j-1] is '*', dp[i][j] = dp[i][j-1] || dp[i-1][j]. If matching letter or '?', dp[i][j] = dp[i-1][j-1].
Greedy two pointers with asterisk backtrackWrite this one
time O(m * n) · space O(1)
Walk both strings. When encountering '*', record the star's index and current string index. On mismatch, backtrack to the last seen star, let it absorb one more character, and resume matching.
Where people lose marks · 3
- Failing to initialize leading asterisks in empty string matching. A pattern of '***' matches an empty string, so dp[0][j] must be true as long as p[j-1] is '*'.
- Strings can be empty (length 0). Handling 0-length inputs without 1-based indexing causes index underflow.
- Confusing wildcard '*' (which stands alone) with regex '.*' (which attaches to a preceding token). Here '*' matches any sequence on its own.
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.
More Dynamic Programming problems
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