Wildcard Matching
Time O(m * n) · Space O(1) · Official statement on LeetCode
Solutions
# Time: O(m + n) ~ O(m * n)
# Space: O(1)
# iterative solution with greedy
class Solution(object):
def isMatch(self, s, p):
"""
:type s: str
:type p: str
:rtype: bool
"""
count = 0 # used for complexity check
p_ptr, s_ptr, last_s_ptr, last_p_ptr = 0, 0, -1, -1
while s_ptr < len(s):
if p_ptr < len(p) and (s[s_ptr] == p[p_ptr] or p[p_ptr] == '?'):
s_ptr += 1
p_ptr += 1
elif p_ptr < len(p) and p[p_ptr] == '*':
p_ptr += 1
last_s_ptr = s_ptr
last_p_ptr = p_ptr
elif last_p_ptr != -1:
last_s_ptr += 1
s_ptr = last_s_ptr
p_ptr = last_p_ptr
else:
assert(count <= (len(p)+1) * (len(s)+1))
return False
count += 1 # used for complexity check
while p_ptr < len(p) and p[p_ptr] == '*':
p_ptr += 1
count += 1 # used for complexity check
assert(count <= (len(p)+1) * (len(s)+1))
return p_ptr == len(p)
# dp with rolling window
# Time: O(m * n)
# Space: O(n)
class Solution2(object):
# @return a boolean
def isMatch(self, s, p):
k = 2
result = [[False for j in xrange(len(p) + 1)] for i in xrange(k)]
result[0][0] = True
for i in xrange(1, len(p) + 1):
if p[i-1] == '*':
result[0][i] = result[0][i-1]
for i in xrange(1,len(s) + 1):
result[i % k][0] = False
for j in xrange(1, len(p) + 1):
if p[j-1] != '*':
result[i % k][j] = result[(i-1) % k][j-1] and (s[i-1] == p[j-1] or p[j-1] == '?')
else:
result[i % k][j] = result[i % k][j-1] or result[(i-1) % k][j]
return result[len(s) % k][len(p)]
# dp
# Time: O(m * n)
# Space: O(m * n)
class Solution3(object):
# @return a boolean
def isMatch(self, s, p):
result = [[False for j in xrange(len(p) + 1)] for i in xrange(len(s) + 1)]
result[0][0] = True
for i in xrange(1, len(p) + 1):
if p[i-1] == '*':
result[0][i] = result[0][i-1]
for i in xrange(1,len(s) + 1):
result[i][0] = False
for j in xrange(1, len(p) + 1):
if p[j-1] != '*':
result[i][j] = result[i-1][j-1] and (s[i-1] == p[j-1] or p[j-1] == '?')
else:
result[i][j] = result[i][j-1] or result[i-1][j]
return result[len(s)][len(p)]
# recursive, slowest, TLE
class Solution4(object):
# @return a boolean
def isMatch(self, s, p):
if not p or not s:
return not s and not p
if p[0] != '*':
if p[0] == s[0] or p[0] == '?':
return self.isMatch(s[1:], p[1:])
else:
return False
else:
while len(s) > 0:
if self.isMatch(s, p[1:]):
return True
s = s[1:]
return self.isMatch(s, p[1:])
Beginner Explanation
What is Wildcard Matching?
Wildcard Matching (LeetCode #44) is a Hard problem that primarily trains dynamic programming.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with greedy.
- Only then translate the idea into code.
Why this problem matters
Hard problems force you to combine patterns and prove complexity carefully — interview gold. Official solution notes mention: Greedy.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Wildcard Matching
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to greedy.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(m * n)) and space (O(1)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(m * n) time and O(1) space.
Pattern focus: greedy
Use the pattern as a checklist:
- greedy — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(m * n) |
| Space | O(1) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Wildcard Matching
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for greedy — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to greedy:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: dynamic programming.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Wildcard Matching in a second language (python).
- Drill 3–5 more problems tagged dynamic programming.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the greedy approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Wildcard Matching (#44) — Hard. Pattern: greedy. Complexity: O(m * n) time / O(1) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Wildcard Matching?+
The reference solutions aim for O(m * n) time and O(1) space. Always re-derive complexity from the code you write in the interview.
What pattern does Wildcard Matching use?+
It primarily maps to greedy, within the broader topic of dynamic programming.
Is Wildcard Matching good for interviews?+
Yes — as a Hard problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/wildcard-matching/