Wildcard Matching
A hard Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 1
- 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. This variant uses '?' for any single character and '*' for any sequence including empty.
Example 1
- Input
- s = "aa", p = "*"
- Output
- true
- Why
- "*" matches any sequence including "aa".
Example 2
- Input
- s = "cb", p = "?a"
- Output
- false
- Why
- '?' matches 'c' but 'a' does not equal 'b'.
Example 3
- Input
- s = "adceb", p = "*a*b"
- Output
- true
- Why
- The first '*' matches empty, then 'a' matches 'a', second '*' matches 'dce', 'b' matches '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 pattern advances synchronously with the string on literal matches and '?', but '*' acts as a flexible bridge. If a mismatch occurs later in the string, backtrack to the most recent '*' and let it absorb one more character, restarting matching from that point.
Approaches, worst first
Recursive search
time O(2^(m+n)) · space O(m + n)
Recurse over indices (i, j). When pattern character is '*', branch into matching zero characters or one character from the string. Suffers exponential blowup on multiple asterisks.
2D table
time O(m * n) · space O(m * n)
Matrix dp[i][j] where dp[0][0] = true. If p[j-1] == '*', dp[i][j] = dp[i-1][j] || dp[i][j-1]. If characters match or p[j-1] == '?', dp[i][j] = dp[i-1][j-1].
Two pointers with star backtrackWrite this one
time O(m * n) · space O(1)
Track sIdx, pIdx, lastStarIdx, and sHistory. On '*', record lastStarIdx and sHistory = sIdx. On mismatch with an active star, advance sHistory, reset sIdx = sHistory and pIdx = lastStarIdx + 1.
Where people lose marks · 3
- Unmatched trailing pattern characters: once the string is exhausted, remaining pattern characters must all be '*' for the match to succeed.
- Empty string matching: an empty string matches pattern only if every character in pattern is '*'.
- Overlooking s length 0 or p length 0 edge cases during pointer initialization.
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.
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