Hi👋SpeedAlgo • Deliberate Practice & Cognitive Ergonomics for Software Engineers

An interactive algorithm mastery and technical interview preparation platform published by Hi👋WebEnterprise. Built for senior and staff software engineers preparing for rigorous coding screens at top tech companies (FAANG/MAMAA).

13 Core Algorithmic Patterns & 225 Practice Problems

  • 1. Two Pointers (16 Paradigms, 43 Problems): Opposite ends, sorted pair sums, palindrome verification, trapping rain water, 3Sum, container with most water — includes the Sliding Window and Fast & Slow Pointers paradigms (contiguous subarray invariants, longest substrings, Floyd's cycle detection, linked list middle).
  • 2. Binary Search (9 Paradigms, 14 Problems): Monotonic predicate partitioning (feasible(x)), boundary search, rotated sorted arrays, square roots, minimum ship capacity, median of two sorted arrays, matrix median on value range.
  • 3. Bit Manipulation (5 Paradigms, 11 Problems): Counting set bits, XOR cancellation, bit reversal, bitmask pairing, maximum XOR with a bitwise trie.
  • 4. Math & Geometry (8 Paradigms, 14 Problems): Sieve of Eratosthenes, integer/roman conversions, modular arithmetic, geometric simulation.
  • 5. Tree/Graph Depth-First Search (15 Paradigms, 24 Problems): Path sums, lowest common ancestor, tree diameter, validating BSTs, tree DP (House Robber III, Binary Tree Cameras).
  • 6. Tree/Graph Breadth-First Search (5 Paradigms, 12 Problems): Level-order traversals, shortest path in unweighted graphs, zig-zag traversals, rotting oranges, word ladders.
  • 7. Graphs (12 Paradigms, 22 Problems): Topological sort, cycle detection, Dijkstra's algorithm, bipartite graph validation, network delay.
  • 8. Backtracking (6 Paradigms, 14 Problems): Subsets, permutations, combinations, constraint satisfaction, pruning, N-Queens, Sudoku solver.
  • 9. Dynamic Programming (13 Paradigms, 28 Problems): Memoization vs tabulation, 0/1 knapsack, unbounded knapsack, coin change, edit distance, longest common subsequence.
  • 10. Heap / Priority Queue (9 Paradigms, 11 Problems): Running medians, top-k elements, interval scheduling, IPO, k-way merges.
  • 11. Advanced Data Structures (7 Paradigms, 17 Problems): Trie (prefix tree), Union-Find (Disjoint Set Union), LRU Cache, LFU Cache, range-sum queries with updates.
  • 12. Stack (5 Paradigms, 9 Problems): Next greater element, largest rectangle in histogram, remove k digits, decode string, min stack.
  • 13. Intervals & Greedy (2 Paradigms, 6 Problems): Merge intervals, interval scheduling, meeting rooms by sweep line, jump game.

4-Stage Deliberate Practice Framework

  1. Stage 1 (Compare & Learn): Read the pattern's mental model, loop invariant and Big-O proof, compare the pattern's template with the problem's solution side by side, and use the step-by-step debugger. The solution can also be shown in C#, Java, TypeScript, C++, Go or Rust for almost every problem.
  2. Stage 2 (Active Recall): Rebuild the solution from memory in a code editor, with an invariant checklist and optional progressive hints, and check it against a judged test suite.
  3. Stage 3 (AI Interview Coach, coming soon): A mock-interview stage still in development; for now, students go straight from Stage 2 to Stage 4.
  4. Stage 4 (Solve on Your Own): Solve a new problem on your own against a 15-minute timer; your code is graded against a test suite. Python and TypeScript run in your browser and C# runs on our server (design problems grade in Python and TypeScript).

Equipped with a spaced-repetition review hub on a fixed 7-step schedule (1 day up to 90 days), a two-pane Studio workspace, and study notes.

Pricing, Access & Commercial Terms

  • Core Curriculum: 100% Free. No credit card required.
  • Coins: New accounts get 40 free coins once, the daily claim adds +20 coins once a day, and inviting a friend earns you +25 coins when they complete their first practice step. These are the only ways to earn coins.
  • AI Coach: Each AI coach question costs 1-5 coins, priced by the server according to its size.
  • BYOK (Bring Your Own Key): AI coaching is free with your own Google Gemini, OpenAI or OpenRouter API key (BYOK).
  • Extras: Coins also unlock extras for 24 hours: a Recall hint level (1 coin), a Solve solution reveal (2), a LeetCode problem card's solutions (1) and a pattern's extra canonical problems (1).
  • Payments & Refunds: Nothing on the site is sold for money: coins can only be earned, never bought. There are no subscriptions or charges, so there is nothing to refund.
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Hi👋SpeedAlgo

Invariant-First Algorithmic Mastery

Arena
238Items
Theory Context•Two Pointers & Sliding Window
EasyLC 26

Remove Duplicates from Sorted Array (LeetCode 26)

Target Frequency:MetaAmazonMicrosoft

You're given an already-sorted (smallest to largest) integer array nums. Rewrite it in-place so every value appears just once, keeping the surviving values in their original relative order.

Say k is the count of distinct values in nums. Once the duplicates are gone, hand back that count k, and make sure the first k slots of nums hold those distinct values in sorted order — whatever sits past index k - 1 is never inspected.

Custom Judge:

We compare the k you return with the length of the expected trimmed array, then check that each of the first k slots of nums holds the expected value for that position. Anything after slot k - 1 is ignored.

Worked Examples

Example 1
Input:nums = [1,1,2]
Output:2, nums = [1,2,_]
Explanation: k comes back as 2, with slots 0 and 1 of nums holding 1 and 2 in order. Whatever is left past index k-1 (shown as an underscore) is never checked.
Example 2
Input:nums = [0,0,1,1,1,2,2,3,3,4]
Output:5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: k comes back as 5, with the first five slots of nums holding 0, 1, 2, 3, and 4 in order. Whatever is left past index k-1 (shown as underscores) is never checked.

⚖️Formal Constraints & Bounds

  • 1 <= nums.length <= 3 * 104

  • -100 <= nums[i] <= 100

  • nums is sorted in non-decreasing order.

Deep-Dive & Conceptual Insights

Core Insight

Sortedness ensures duplicate elements are strictly adjacent, enabling single-pass linear filtering in O(1) space.

Real-World Scenario & Production Applications

Memory compaction algorithms, columnar database run-length deduplication, and zero-allocation stream filters.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Slow/Fast Write-Read Pointer Setup

slow tracks the next write position for a unique value; fast scans ahead reading every element. Since the array is sorted, duplicates are always adjacent.

Mathematical Recurrence / Code Invariant
slow = 1  # first element is trivially unique

Step-by-Step Execution Trace Table

Stepreadernums[reader]nums[writer-1]ActionMutated Array Prefixwriter
1100Duplicate (0 == 0) → skip[0, ...]1
2210Unique (1 != 0) → nums[1]=1, writer++[0, 1, ...]2
3311Duplicate (1 == 1) → skip[0, 1, ...]2
4411Duplicate (1 == 1) → skip[0, 1, ...]2
5521Unique (2 != 1) → nums[2]=2, writer++[0, 1, 2, ...]3
6622Duplicate (2 == 2) → skip[0, 1, 2, ...]3
7732Unique (3 != 2) → nums[3]=3, writer++[0, 1, 2, 3, ...]4
8833Duplicate (3 == 3) → skip[0, 1, 2, 3, ...]4
9943Unique (4 != 3) → nums[4]=4, writer++[0, 1, 2, 3, 4]5
Code Shape
Archetype Code Shape
Python
1if not nums:
2 return 0
3writer = 1
4for reader in range(1, len(nums)):
5 if nums[reader] != nums[writer - 1]:
6 nums[writer] = nums[reader]
7 writer += 1
8return writer

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

Two pointers moving through the same array, one always at or behind the other, is the foundational shape of this entire pattern — worth slowing down on before it becomes muscle memory.

🧹 The Analogy: A Librarian Re-Shelving Duplicates

Picture a librarian walking down a single shelf of books already sorted alphabetically, holding a "next open slot" marker (writer) in one hand. They scan forward with their eyes (reader), one book at a time. Whenever they spot a book different from the last one they kept, they physically slide it back into the open slot and advance the marker. Duplicate books get scanned past and simply never get a slot — they're silently overwritten by whatever comes after them.

🔑 The Key Idea in the Code

Because the shelf is already sorted, every duplicate of a book sits immediately next to it — so "is this different from the last book I kept?" is a complete duplicate check, with no need to remember anything beyond the single most recent kept value.

🤔 Common Point of Confusion: Why writer - 1 Instead of reader - 1?

Comprehending the comparison on the first iteration can feel unnatural: we are comparing index 1 with index 0, and writer - 1 forces you to juggle two moving boundaries at once.

Since the array is sorted, duplicates are always adjacent, so you could also write:

Code / Blueprint
# The "Compare with Previous" Approach:
if nums[reader] != nums[reader - 1]:
nums[writer] = nums[reader]
writer += 1
 

Why this feels intuitive:

  • Separation of Concerns: The reader is solely responsible for discovering new unique elements by checking the card it just held (reader - 1). The writer strictly acts as the "next open slot" marker.
  • Mental Model: Like flipping through a deck of sorted cards, you just ask: "Is this card different from the card in my hand one second ago?"
🏛️ Architectural Insight: Which Version Is Canonically "Correct" and Why?

While nums[reader] != nums[reader - 1] is convenient for LC 26, nums[reader] != nums[writer - 1] is the canonically correct, architecturally superior version for the general pattern:

  1. Trust the "Kept" Data, Not the "Raw" Data: The core invariant of compaction is that the prefix nums[0..writer-1] represents your finalized, verified, processed data (the "Kept-Prefix Boundary"). The right side (nums[reader..]) is untrusted, raw input. When deciding whether to keep an element, you should compare against the last element you actually decided to keep (writer - 1), not just whatever you looked at a moment ago.

  2. Generalizability to Harder Problems (The Main Advantage): Comparing against writer allows you to solve advanced variants with almost zero code changes. Consider LC 80 (Remove Duplicates II), where you can keep at most 2 duplicates (or generally at most kkk duplicates).

  • With reader - 1, you need messy state flags and counters to track consecutive duplicate counts.
  • With the writer pattern, the code remains identical—you simply look kkk slots back in your trusted data:
Code / Blueprint
def removeDuplicatesTwice(nums: list[int]) -> int:
if len(nums) <= 2:
return len(nums)
writer = 2
for reader in range(2, len(nums)):
# Is this number different from what I kept TWO slots ago?
if nums[reader] != nums[writer - 2]:
nums[writer] = nums[reader]
writer += 1
return writer
 
  • Prematurely advancing writer before writing: The writer index marks the boundary where the next unique element belongs. Overwriting nums[writer] = nums[reader] before incrementing writer += 1 is required, or else you leave gaps.

  • Comparing reader against reader - 1 instead of trusted writer - 1: In standard deduplication, comparing with reader - 1 works for single duplicates, but fails to generalize to kkk-duplicates. The canonical boundary-of-truth compares with writer - 1.

  • Returning len(nums) instead of writer: The array is compacted in-place to index writer - 1; the total count of unique preserved elements is writer, not the original array length.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Sorted array + in-place modification requirement + O(1) extra memory constraint.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Invariant: nums[0..writer-1] contains the sorted unique elements. If nums[reader] != nums[writer-1], nums[writer] = nums[reader] and writer increments.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

  • Comparing against nums[writer] (the next free slot) instead of nums[writer - 1] (the last kept value): on [0,0,1,1,1,2,2,3,3,4] it keeps 7 values instead of 5.

The 60-Second Interview Pitch

Say this out loud before you type a single line

"I'd use Read-Write Pointers. The array is sorted, so equal values sit next to each other, and a value is new exactly when it differs from the last value I kept. writer marks the next free slot and starts at 1, because nums[0] is always kept; reader walks from index 1 to the end. When nums[reader] differs from nums[writer - 1], I copy it to nums[writer] and move writer forward. writer never passes reader, so I only ever overwrite a slot I have already read, and nums[0] to nums[writer - 1] always holds the distinct values in order. The trap is comparing with nums[writer] instead of nums[writer - 1]: that slot is the next free one and still holds an old value, so on [1, 2, 2] the second 2 is lost and I return 1 instead of 2. It's O(N) time and O(1) extra space, and I return writer."

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

Look at the loop: for reader in range(1, len(nums)) visits each element once and does one comparison with nums[writer - 1]; writer only moves when a new value is copied, so each slot is written at most once. N - 1 iterations of O(1) work: O(N).

SPACE COMPLEXITY

O(1) Auxiliary

Look at memory allocations: Call stack uses O(1) Iterative. In-place array manipulation overwriting indices in nums using two integer indices.

Formal Recurrence Relation

T(N) = ∑_{j=1}^N O(1) = O(N)

Derivation Progression

Reader Scan

for reader ∈ [1, N-1]

reader visits every element after the first exactly once.

Conditional Write

if nums[reader] != nums[writer - 1]: nums[writer] = nums[reader]; writer += 1

writer advances only when a new value is copied to the next free slot.

Linear Convergence

O(N) total steps

Array is partitioned in-place in exactly N iterations.

Variable Definitions

NNN

Total number of elements in the sorted array

Memory Architecture & Bounds

🟣 Call Stack

O(1) Iterative

🔵 Auxiliary Heap

O(1) Zero heap memory

🟢 Output Space

O(1) Returns integer count

Boundary Best / Worst Cases

Best Case

O(N)O(N)O(N)

Average Case

O(N)O(N)O(N)

Worst Case

O(N)O(N)O(N)

Pointer Invariant Transition Progression

Pointer Invariant Transition Progression
Synthesizing vector architecture diagram...
Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "remove duplicates in-place from sorted array", "return number of unique elements k", "do not allocate extra space". The sorted property guarantees duplicates are strictly adjacent. Reader-writer pointer compaction pattern partitions array into [0..write-1] (unique elements) and [write..N-1] (garbage/unprocessed).

CONSTRAINTS & BOUNDS

1≤N≤3×1041 \le N \le 3 \times 10^41≤N≤3×104. Requires strictly O(N)O(N)O(N) time and O(1)O(1)O(1) auxiliary space. Boundary choice: write pointer initializes to index 1 because the first element is unconditionally unique.

FAANG PRODUCTION TRAPS & EDGE CASES

N=1N=1N=1 base case (loop does not execute, returns 1 correctly). Overwriting candidate elements before inspection (avoided by write <= read). Comparing candidate with nums[read-1] instead of nums[write-1] (both work when sorted, but write-1 generalizes to k-duplicate compaction).

Theory Context•Two Pointers & Sliding Window
EasyLC 26

Remove Duplicates from Sorted Array (LeetCode 26)

Target Frequency:MetaAmazonMicrosoft

You're given an already-sorted (smallest to largest) integer array nums. Rewrite it in-place so every value appears just once, keeping the surviving values in their original relative order.

Say k is the count of distinct values in nums. Once the duplicates are gone, hand back that count k, and make sure the first k slots of nums hold those distinct values in sorted order — whatever sits past index k - 1 is never inspected.

Custom Judge:

We compare the k you return with the length of the expected trimmed array, then check that each of the first k slots of nums holds the expected value for that position. Anything after slot k - 1 is ignored.

Worked Examples

Example 1
Input:nums = [1,1,2]
Output:2, nums = [1,2,_]
Explanation: k comes back as 2, with slots 0 and 1 of nums holding 1 and 2 in order. Whatever is left past index k-1 (shown as an underscore) is never checked.
Example 2
Input:nums = [0,0,1,1,1,2,2,3,3,4]
Output:5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: k comes back as 5, with the first five slots of nums holding 0, 1, 2, 3, and 4 in order. Whatever is left past index k-1 (shown as underscores) is never checked.

⚖️Formal Constraints & Bounds

  • 1 <= nums.length <= 3 * 104

  • -100 <= nums[i] <= 100

  • nums is sorted in non-decreasing order.

Deep-Dive & Conceptual Insights

Core Insight

Sortedness ensures duplicate elements are strictly adjacent, enabling single-pass linear filtering in O(1) space.

Real-World Scenario & Production Applications

Memory compaction algorithms, columnar database run-length deduplication, and zero-allocation stream filters.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Slow/Fast Write-Read Pointer Setup

slow tracks the next write position for a unique value; fast scans ahead reading every element. Since the array is sorted, duplicates are always adjacent.

Mathematical Recurrence / Code Invariant
slow = 1  # first element is trivially unique

Step-by-Step Execution Trace Table

Stepreadernums[reader]nums[writer-1]ActionMutated Array Prefixwriter
1100Duplicate (0 == 0) → skip[0, ...]1
2210Unique (1 != 0) → nums[1]=1, writer++[0, 1, ...]2
3311Duplicate (1 == 1) → skip[0, 1, ...]2
4411Duplicate (1 == 1) → skip[0, 1, ...]2
5521Unique (2 != 1) → nums[2]=2, writer++[0, 1, 2, ...]3
6622Duplicate (2 == 2) → skip[0, 1, 2, ...]3
7732Unique (3 != 2) → nums[3]=3, writer++[0, 1, 2, 3, ...]4
8833Duplicate (3 == 3) → skip[0, 1, 2, 3, ...]4
9943Unique (4 != 3) → nums[4]=4, writer++[0, 1, 2, 3, 4]5
Code Shape
Archetype Code Shape
Python
1if not nums:
2 return 0
3writer = 1
4for reader in range(1, len(nums)):
5 if nums[reader] != nums[writer - 1]:
6 nums[writer] = nums[reader]
7 writer += 1
8return writer

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

Two pointers moving through the same array, one always at or behind the other, is the foundational shape of this entire pattern — worth slowing down on before it becomes muscle memory.

🧹 The Analogy: A Librarian Re-Shelving Duplicates

Picture a librarian walking down a single shelf of books already sorted alphabetically, holding a "next open slot" marker (writer) in one hand. They scan forward with their eyes (reader), one book at a time. Whenever they spot a book different from the last one they kept, they physically slide it back into the open slot and advance the marker. Duplicate books get scanned past and simply never get a slot — they're silently overwritten by whatever comes after them.

🔑 The Key Idea in the Code

Because the shelf is already sorted, every duplicate of a book sits immediately next to it — so "is this different from the last book I kept?" is a complete duplicate check, with no need to remember anything beyond the single most recent kept value.

🤔 Common Point of Confusion: Why writer - 1 Instead of reader - 1?

Comprehending the comparison on the first iteration can feel unnatural: we are comparing index 1 with index 0, and writer - 1 forces you to juggle two moving boundaries at once.

Since the array is sorted, duplicates are always adjacent, so you could also write:

Code / Blueprint
# The "Compare with Previous" Approach:
if nums[reader] != nums[reader - 1]:
nums[writer] = nums[reader]
writer += 1
 

Why this feels intuitive:

  • Separation of Concerns: The reader is solely responsible for discovering new unique elements by checking the card it just held (reader - 1). The writer strictly acts as the "next open slot" marker.
  • Mental Model: Like flipping through a deck of sorted cards, you just ask: "Is this card different from the card in my hand one second ago?"
🏛️ Architectural Insight: Which Version Is Canonically "Correct" and Why?

While nums[reader] != nums[reader - 1] is convenient for LC 26, nums[reader] != nums[writer - 1] is the canonically correct, architecturally superior version for the general pattern:

  1. Trust the "Kept" Data, Not the "Raw" Data: The core invariant of compaction is that the prefix nums[0..writer-1] represents your finalized, verified, processed data (the "Kept-Prefix Boundary"). The right side (nums[reader..]) is untrusted, raw input. When deciding whether to keep an element, you should compare against the last element you actually decided to keep (writer - 1), not just whatever you looked at a moment ago.

  2. Generalizability to Harder Problems (The Main Advantage): Comparing against writer allows you to solve advanced variants with almost zero code changes. Consider LC 80 (Remove Duplicates II), where you can keep at most 2 duplicates (or generally at most kkk duplicates).

  • With reader - 1, you need messy state flags and counters to track consecutive duplicate counts.
  • With the writer pattern, the code remains identical—you simply look kkk slots back in your trusted data:
Code / Blueprint
def removeDuplicatesTwice(nums: list[int]) -> int:
if len(nums) <= 2:
return len(nums)
writer = 2
for reader in range(2, len(nums)):
# Is this number different from what I kept TWO slots ago?
if nums[reader] != nums[writer - 2]:
nums[writer] = nums[reader]
writer += 1
return writer
 
  • Prematurely advancing writer before writing: The writer index marks the boundary where the next unique element belongs. Overwriting nums[writer] = nums[reader] before incrementing writer += 1 is required, or else you leave gaps.

  • Comparing reader against reader - 1 instead of trusted writer - 1: In standard deduplication, comparing with reader - 1 works for single duplicates, but fails to generalize to kkk-duplicates. The canonical boundary-of-truth compares with writer - 1.

  • Returning len(nums) instead of writer: The array is compacted in-place to index writer - 1; the total count of unique preserved elements is writer, not the original array length.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Sorted array + in-place modification requirement + O(1) extra memory constraint.

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

Invariant: nums[0..writer-1] contains the sorted unique elements. If nums[reader] != nums[writer-1], nums[writer] = nums[reader] and writer increments.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

  • Comparing against nums[writer] (the next free slot) instead of nums[writer - 1] (the last kept value): on [0,0,1,1,1,2,2,3,3,4] it keeps 7 values instead of 5.

The 60-Second Interview Pitch

Say this out loud before you type a single line

"I'd use Read-Write Pointers. The array is sorted, so equal values sit next to each other, and a value is new exactly when it differs from the last value I kept. writer marks the next free slot and starts at 1, because nums[0] is always kept; reader walks from index 1 to the end. When nums[reader] differs from nums[writer - 1], I copy it to nums[writer] and move writer forward. writer never passes reader, so I only ever overwrite a slot I have already read, and nums[0] to nums[writer - 1] always holds the distinct values in order. The trap is comparing with nums[writer] instead of nums[writer - 1]: that slot is the next free one and still holds an old value, so on [1, 2, 2] the second 2 is lost and I return 1 instead of 2. It's O(N) time and O(1) extra space, and I return writer."

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(N)

Look at the loop: for reader in range(1, len(nums)) visits each element once and does one comparison with nums[writer - 1]; writer only moves when a new value is copied, so each slot is written at most once. N - 1 iterations of O(1) work: O(N).

SPACE COMPLEXITY

O(1) Auxiliary

Look at memory allocations: Call stack uses O(1) Iterative. In-place array manipulation overwriting indices in nums using two integer indices.

Formal Recurrence Relation

T(N) = ∑_{j=1}^N O(1) = O(N)

Derivation Progression

Reader Scan

for reader ∈ [1, N-1]

reader visits every element after the first exactly once.

Conditional Write

if nums[reader] != nums[writer - 1]: nums[writer] = nums[reader]; writer += 1

writer advances only when a new value is copied to the next free slot.

Linear Convergence

O(N) total steps

Array is partitioned in-place in exactly N iterations.

Variable Definitions

NNN

Total number of elements in the sorted array

Memory Architecture & Bounds

🟣 Call Stack

O(1) Iterative

🔵 Auxiliary Heap

O(1) Zero heap memory

🟢 Output Space

O(1) Returns integer count

Boundary Best / Worst Cases

Best Case

O(N)O(N)O(N)

Average Case

O(N)O(N)O(N)

Worst Case

O(N)O(N)O(N)

Pointer Invariant Transition Progression

Pointer Invariant Transition Progression
Synthesizing vector architecture diagram...
Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "remove duplicates in-place from sorted array", "return number of unique elements k", "do not allocate extra space". The sorted property guarantees duplicates are strictly adjacent. Reader-writer pointer compaction pattern partitions array into [0..write-1] (unique elements) and [write..N-1] (garbage/unprocessed).

CONSTRAINTS & BOUNDS

1≤N≤3×1041 \le N \le 3 \times 10^41≤N≤3×104. Requires strictly O(N)O(N)O(N) time and O(1)O(1)O(1) auxiliary space. Boundary choice: write pointer initializes to index 1 because the first element is unconditionally unique.

FAANG PRODUCTION TRAPS & EDGE CASES

N=1N=1N=1 base case (loop does not execute, returns 1 correctly). Overwriting candidate elements before inspection (avoided by write <= read). Comparing candidate with nums[read-1] instead of nums[write-1] (both work when sorted, but write-1 generalizes to k-duplicate compaction).

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