Valid Palindrome (LeetCode 125)
Call a phrase a palindrome when, after lowercasing every letter and stripping out anything that isn't a letter or digit, it reads identically forwards and backwards.
Given a string s, decide whether it forms such a palindrome, returning true or false.
Worked Examples
s = "A man, a plan, a canal: Panama"trues = "race a car"falses = " "true⚖️Formal Constraints & Bounds
1 <= s.length <= 2 * 105sconsists only of printable ASCII characters.
Core Insight
Skipping punctuation inward avoids allocating a cleaned string copy, preserving O(1) auxiliary space.
Real-World Scenario & Production Applications
DNA palindrome sequence recognition, search query normalization, and packet header verification.
Subproblems & Recurrence Decomposition3 Phases
Start pointers at both ends of the string: left = 0, right = len(s) - 1. They will converge toward the center, checking character equality along the way.
left, right = 0, len(s) - 1Step-by-Step Execution Trace Table
| Step | left char | right char | Action |
|---|---|---|---|
| 1 | 'A' (idx 0) | 'a' (idx 30) | Match ('a' == 'a') → left++, right-- |
| 2 | ' ' (skip) | 'm' | Skip non-alphanumeric |
| 3 | 'm' (idx 2) | 'm' (idx 28) | Match ('m' == 'm') → left++, right-- |
| 1 | left, right = 0, len(s) - 1 |
| 2 | while left < right: |
| 3 | while left < right and not s[left].isalnum(): |
| 4 | left += 1 |
| 5 | while left < right and not s[right].isalnum(): |
| 6 | right -= 1 |
| 7 | if s[left].lower() != s[right].lower(): |
| 8 | return False |
| 9 | left += 1 |
| 10 | right -= 1 |
| 11 | return True |
Conceptual Narrative
🧭 Conceptual Foundation & Pattern Intuition
A palindrome is symmetric: reading from the left forward yields the exact same sequence as reading from the right backward. The naive approach normalizes the entire string into a new array of alphanumeric characters and compares it with its reverse, requiring auxiliary heap memory. Two converging pointers achieve optimal space by filtering and comparing characters in-flight directly within the input string.
🏟️ The Analogy: Two Inspecting Guards at a Palace Gate
Imagine two palace guards walking inward from opposite ends of a long hallway lined with antique statues (letters) and construction debris (spaces and punctuation). The guards ignore the debris completely, kicking it aside. Whenever both guards stand before an actual statue, they radio each other. If the statues match, both guards take another step inward. If at any point the statues do not match, they immediately sound the alarm.
🔑 Why It Works
Notice the inner while left < right guard on each pointer increment. Without this guard, an input containing only punctuation (e.g. "...") would increment left past len(s), triggering an IndexError. The two pointers meet or cross in at most operations.
Strings with no alphanumeric characters must evaluate to
True: inputs like" "or",."have nothing to compare, soleft/rightcross without ever finding a mismatch — treat this as a valid palindrome, notFalse.Inner skipping loops must guard
left < right: the filter loops advancingleft/rightpast non-alphanumeric characters must checkleft < righton every step, or an all-punctuation string walks a pointer straight off the end of the string (IndexErrorin Python).
4-Phase Thought Process Model
Pattern Recognition Signals
The 10-second spot
Palindrome verification with punctuation/whitespace filtering and O(1) extra space constraint.
Formulating the Predicate & Invariants
Turning intuition into a boolean rule
Invariant: cleaned prefix s[0..left] is symmetric mirror of cleaned suffix s[right..n-1].
Silent Failure Traps & Edge Cases
Where confident candidates still lose points
Index out of bounds when skipping all non-alphanumerics: guard
while left < right and not s[left].isalnum():.
The 60-Second Interview Pitch
Say this out loud before you type a single line
"I initialize left=0 and right=len(s)-1. At each step, I advance left past non-alphanumerics and decrement right past non-alphanumerics. If s[left].lower() != s[right].lower(), return False. Runs in
O(N)time andO(1)space."
Complexity & Mathematical Proof
O(N)
Look at the execution steps: Left and right pointers converge inward, skipping non-alphanumeric characters and comparing lowercase values. At most N pointer movements.
O(1) Auxiliary
Look at memory allocations: Call stack uses O(1) Iterative. Pointers operate directly on input string in-place without copying filtered strings.
T(N) = N/2 comparisons = O(N)
Derivation Progression
while not isalnum(): L += 1
Both pointers skip punctuation in O(1) checks per character.
s[L].lower() == s[R].lower()
Compares corresponding symmetric characters.
≤ N total increments
Pointers cross in at most N steps.
Variable Definitions
Length of the string after alphanumeric filtering
Memory Architecture & Bounds
O(1) Iterative
O(1) Two index pointers
O(1) Boolean return
Boundary Best / Worst Cases
on early mismatch at boundaries.
.
on valid palindrome.
Pointer Invariant Transition Progression
Senior SWE Deconstruction & Hardware Caveats
Triggers: "reads same forward and backward", "converting all uppercase letters to lowercase and removing all non-alphanumeric characters". Bilateral inward convergence with filter skipping.
. Time budget single pass, auxiliary space without building an intermediate cleaned string.
Inner skipping loops must check boundary left < right to avoid indexing out of bounds.
Valid Palindrome (LeetCode 125)
Call a phrase a palindrome when, after lowercasing every letter and stripping out anything that isn't a letter or digit, it reads identically forwards and backwards.
Given a string s, decide whether it forms such a palindrome, returning true or false.
Worked Examples
s = "A man, a plan, a canal: Panama"trues = "race a car"falses = " "true⚖️Formal Constraints & Bounds
1 <= s.length <= 2 * 105sconsists only of printable ASCII characters.
Core Insight
Skipping punctuation inward avoids allocating a cleaned string copy, preserving O(1) auxiliary space.
Real-World Scenario & Production Applications
DNA palindrome sequence recognition, search query normalization, and packet header verification.
Subproblems & Recurrence Decomposition3 Phases
Start pointers at both ends of the string: left = 0, right = len(s) - 1. They will converge toward the center, checking character equality along the way.
left, right = 0, len(s) - 1Step-by-Step Execution Trace Table
| Step | left char | right char | Action |
|---|---|---|---|
| 1 | 'A' (idx 0) | 'a' (idx 30) | Match ('a' == 'a') → left++, right-- |
| 2 | ' ' (skip) | 'm' | Skip non-alphanumeric |
| 3 | 'm' (idx 2) | 'm' (idx 28) | Match ('m' == 'm') → left++, right-- |
| 1 | left, right = 0, len(s) - 1 |
| 2 | while left < right: |
| 3 | while left < right and not s[left].isalnum(): |
| 4 | left += 1 |
| 5 | while left < right and not s[right].isalnum(): |
| 6 | right -= 1 |
| 7 | if s[left].lower() != s[right].lower(): |
| 8 | return False |
| 9 | left += 1 |
| 10 | right -= 1 |
| 11 | return True |
Conceptual Narrative
🧭 Conceptual Foundation & Pattern Intuition
A palindrome is symmetric: reading from the left forward yields the exact same sequence as reading from the right backward. The naive approach normalizes the entire string into a new array of alphanumeric characters and compares it with its reverse, requiring auxiliary heap memory. Two converging pointers achieve optimal space by filtering and comparing characters in-flight directly within the input string.
🏟️ The Analogy: Two Inspecting Guards at a Palace Gate
Imagine two palace guards walking inward from opposite ends of a long hallway lined with antique statues (letters) and construction debris (spaces and punctuation). The guards ignore the debris completely, kicking it aside. Whenever both guards stand before an actual statue, they radio each other. If the statues match, both guards take another step inward. If at any point the statues do not match, they immediately sound the alarm.
🔑 Why It Works
Notice the inner while left < right guard on each pointer increment. Without this guard, an input containing only punctuation (e.g. "...") would increment left past len(s), triggering an IndexError. The two pointers meet or cross in at most operations.
Strings with no alphanumeric characters must evaluate to
True: inputs like" "or",."have nothing to compare, soleft/rightcross without ever finding a mismatch — treat this as a valid palindrome, notFalse.Inner skipping loops must guard
left < right: the filter loops advancingleft/rightpast non-alphanumeric characters must checkleft < righton every step, or an all-punctuation string walks a pointer straight off the end of the string (IndexErrorin Python).
4-Phase Thought Process Model
Pattern Recognition Signals
The 10-second spot
Palindrome verification with punctuation/whitespace filtering and O(1) extra space constraint.
Formulating the Predicate & Invariants
Turning intuition into a boolean rule
Invariant: cleaned prefix s[0..left] is symmetric mirror of cleaned suffix s[right..n-1].
Silent Failure Traps & Edge Cases
Where confident candidates still lose points
Index out of bounds when skipping all non-alphanumerics: guard
while left < right and not s[left].isalnum():.
The 60-Second Interview Pitch
Say this out loud before you type a single line
"I initialize left=0 and right=len(s)-1. At each step, I advance left past non-alphanumerics and decrement right past non-alphanumerics. If s[left].lower() != s[right].lower(), return False. Runs in
O(N)time andO(1)space."
Complexity & Mathematical Proof
O(N)
Look at the execution steps: Left and right pointers converge inward, skipping non-alphanumeric characters and comparing lowercase values. At most N pointer movements.
O(1) Auxiliary
Look at memory allocations: Call stack uses O(1) Iterative. Pointers operate directly on input string in-place without copying filtered strings.
T(N) = N/2 comparisons = O(N)
Derivation Progression
while not isalnum(): L += 1
Both pointers skip punctuation in O(1) checks per character.
s[L].lower() == s[R].lower()
Compares corresponding symmetric characters.
≤ N total increments
Pointers cross in at most N steps.
Variable Definitions
Length of the string after alphanumeric filtering
Memory Architecture & Bounds
O(1) Iterative
O(1) Two index pointers
O(1) Boolean return
Boundary Best / Worst Cases
on early mismatch at boundaries.
.
on valid palindrome.
Pointer Invariant Transition Progression
Senior SWE Deconstruction & Hardware Caveats
Triggers: "reads same forward and backward", "converting all uppercase letters to lowercase and removing all non-alphanumeric characters". Bilateral inward convergence with filter skipping.
. Time budget single pass, auxiliary space without building an intermediate cleaned string.
Inner skipping loops must check boundary left < right to avoid indexing out of bounds.