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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.

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Hi👋SpeedAlgo

Invariant-First Algorithmic Mastery

Arena
238Items
Theory Context•Backtracking & State Search
MediumLC 22

Generate Parentheses (LeetCode 22)

Target Frequency:MetaAmazonGoogleApple

Given n pairs of parentheses, produce every distinct combination of n '(' and n ')' characters that forms a well-formed string.

A string is well-formed when every '(' is eventually matched by a ')', and reading left to right, the running count of ')' never exceeds the running count of '('.

Worked Examples

Example 1
Input:n = 3
Output:["((()))","(()())","(())()","()(())","()()()"]
Explanation: These are all five balanced strings of three ( and three ).
Example 2
Input:n = 1
Output:["()"]
Explanation: With one pair, "()" is the only balanced string.

⚖️Formal Constraints & Bounds

  • 1 <= n <= 8

Deep-Dive & Conceptual Insights

Core Insight

Dual counters constrain the search space exclusively to valid paths, achieving optimal O(Catalan(N)) generation.

Real-World Scenario & Production Applications

Compiler parser syntax validation, mathematical expression token generation.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Open Parenthesis Branching

Valid to add an opening bracket '(' whenever the count of open brackets open_count < n.

Mathematical Recurrence / Code Invariant
if open_count < n:
    path.append("(")
    dfs(open_count + 1, close_count)
    path.pop()

Step-by-Step Execution Trace Table

Code / Blueprint
                                         "" (open=0, close=0, d=0)
                                            |
                                            | choose '('
                                            v
                                   "(" (open=1, close=0, d=1)
                                  /                          \
                     choose '('  /                            \  choose ')'
                                /                              \
                    "((" (open=2, close=0, d=2)            "()" (open=1, close=1, d=2)
                   /                    \                          |
       choose '(' /                      \ choose ')'              | choose '(' (close==open)
                 /                        \                        v
         "(((" (open=3, close=0, d=3)   "(()" (open=2, close=1, d=3) "()(" (open=2, close=1, d=3)
                 |                      /            \             /            \
      choose ')' |          choose '(' /   choose ')' \ choose '('/   choose ')' \
                 v                    /                \         /                \
        "((()" (d=4)          "(()(" (d=4)        "(())" (d=4)"()((" (d=4)        "()()" (d=4)
                 |                    |                |         |                |
      choose ')' |         choose ')' |     choose '(' |choose ')'|    choose '(' |
                 v                    v                v         v                v
       "((())" (d=5)         "(()()" (d=5)       "(())(" (d=5)"()(()" (d=5)       "()()(" (d=5)
                 |                    |                |         |                |
      choose ')' |         choose ')' |     choose ')' |choose ')'|    choose ')' |
                 v                    v                v         v                v
      "((()))" ✅            "(()())" ✅         "(())()" ✅  "()(())" ✅         "()()()" ✅
 
Step-by-Step Decision Trace (for n = 3)
  1. Step 1 (Root, d=0): path = "" (open=0, close=0). Rule: close < open is False (0 < 0), so must choose '(' first.
  2. Step 2 (Choose '(', d=1): path = "(" (open=1, close=0). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
  3. Step 3 (Branch Left: Choose '(', d=2): path = "((" (open=2, close=0). open < 3 is True →\rightarrow→ branch '(' first.
  4. Step 4 (Branch Left: Choose '(', d=3): path = "(((" (open=3, close=0). open == 3 quota reached →\rightarrow→ only legal choice is ')'.
  5. Step 5 (Choose ')', d=4): path = "((()" (open=3, close=1). close < open (1 < 3) →\rightarrow→ place ')'.
  6. Step 6 (Choose ')', d=5): path = "((())" (open=3, close=2). Place ')'.
  7. Step 7 (Choose ')', d=6): path = "((()))" (open=3, close=3). Base Case (len == 2*n = 6): Record "((()))" ✅ (#1).
  8. Step 8 (Backtrack / Pop ')', d=5): path = "((())" →\rightarrow→ ↩️ BACKTRACK / POP.
  9. Step 9 (Backtrack / Pop ')', d=4): path = "((()" →\rightarrow→ ↩️ BACKTRACK / POP.
  10. Step 10 (Backtrack / Pop ')', d=3): path = "(((" →\rightarrow→ ↩️ BACKTRACK / POP.
  11. Step 11 (Backtrack / Pop '(', d=2): path = "((" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
  12. Step 12 (Branch Right: Choose ')', d=3): path = "(()" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
  13. Step 13 (Branch Left: Choose '(', d=4): path = "(()(" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
  14. Step 14 (Choose ')', d=5): path = "(()()" (open=3, close=2). Place ')'.
  15. Step 15 (Choose ')', d=6): path = "(()())" (open=3, close=3). Base Case (len == 6): Record "(()())" ✅ (#2).
  16. Step 16 (Backtrack / Pop ')', d=5): path = "(()()" →\rightarrow→ ↩️ BACKTRACK / POP.
  17. Step 17 (Backtrack / Pop ')', d=4): path = "(()(" →\rightarrow→ ↩️ BACKTRACK / POP.
  18. Step 18 (Backtrack / Pop '(', d=3): path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
  19. Step 19 (Branch Right: Choose ')', d=4): path = "(())" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
  20. Step 20 (Choose '(', d=5): path = "(())(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
  21. Step 21 (Choose ')', d=6): path = "(())()" (open=3, close=3). Base Case (len == 6): Record "(())()" ✅ (#3).
  22. Step 22 (Backtrack / Pop ')', d=5): path = "(())(" →\rightarrow→ ↩️ BACKTRACK / POP.
  23. Step 23 (Backtrack / Pop '(', d=4): path = "(())" →\rightarrow→ ↩️ BACKTRACK / POP.
  24. Step 24 (Backtrack / Pop ')', d=3): path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP.
  25. Step 25 (Backtrack / Pop ')', d=2): path = "((" →\rightarrow→ ↩️ BACKTRACK / POP.
  26. Step 26 (Backtrack / Pop '(', d=1): path = "(" →\rightarrow→ ↩️ BACKTRACK / POP to explore main right branch: choose ')'.
  27. Step 27 (Branch Right: Choose ')', d=2): path = "()" (open=1, close=1). close == open →\rightarrow→ only legal choice is '('.
  28. Step 28 (Choose '(', d=3): path = "()(" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
  29. Step 29 (Branch Left: Choose '(', d=4): path = "()((" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
  30. Step 30 (Choose ')', d=5): path = "()(()" (open=3, close=2). Place ')'.
  31. Step 31 (Choose ')', d=6): path = "()(())" (open=3, close=3). Base Case (len == 6): Record "()(())" ✅ (#4).
  32. Step 32 (Backtrack / Pop ')', d=5): path = "()(()" →\rightarrow→ ↩️ BACKTRACK / POP.
  33. Step 33 (Backtrack / Pop ')', d=4): path = "()((" →\rightarrow→ ↩️ BACKTRACK / POP.
  34. Step 34 (Backtrack / Pop '(', d=3): path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
  35. Step 35 (Branch Right: Choose ')', d=4): path = "()()" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
  36. Step 36 (Choose '(', d=5): path = "()()(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
  37. Step 37 (Choose ')', d=6): path = "()()()" (open=3, close=3). Base Case (len == 6): Record "()()()" ✅ (#5).
  38. Step 38 (Backtrack / Pop ')', d=5): path = "()()(" →\rightarrow→ ↩️ BACKTRACK / POP.
  39. Step 39 (Backtrack / Pop '(', d=4): path = "()()" →\rightarrow→ ↩️ BACKTRACK / POP.
  40. Step 40 (Backtrack / Pop ')', d=3): path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP.
  41. Step 41 (Backtrack / Pop '(', d=2): path = "()" →\rightarrow→ ↩️ BACKTRACK / POP.
  42. Step 42 (Backtrack / Pop ')', d=1): path = "(" →\rightarrow→ ↩️ BACKTRACK / POP.
  43. Step 43 (Backtrack / Pop '(', d=0): path = "" →\rightarrow→ ↩️ BACKTRACK / POP to Root. All 5 Catalan combinations generated! Output: ["((()))", "(()())", "(())()", "()(())", "()()()"].
Full Walkthrough43 Steps
Input:n = 3Expected:["((()))","(()())","(())()","()(())","()()()"]
1🌱 ROOT(Root, d=0)
path = ""
Choices:( ✓ ) ✂️
path = "" (open=0, close=0). Rule: close < open is False (0 < 0), so must choose '(' first.
2🟢 CHOOSE(Choose '(', d=1)
path = "("
Choices:( ✓ )
path = "(" (open=1, close=0). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
3🟢 CHOOSE(Branch Left: Choose '(', d=2)
path = "(("
path = "((" (open=2, close=0). open < 3 is True →\rightarrow→ branch '(' first.
4🟢 CHOOSE(Branch Left: Choose '(', d=3)
path = "((("
path = "(((" (open=3, close=0). open == 3 quota reached →\rightarrow→ only legal choice is ')'.
5🟢 CHOOSE(Choose ')', d=4)
path = "((()"
Choices:( ) ✓
path = "((()" (open=3, close=1). close < open (1 < 3) →\rightarrow→ place ')'.
6🟢 CHOOSE(Choose ')', d=5)
path = "((())"
path = "((())" (open=3, close=2). Place ')'.
7✅ RECORD / GOAL(Choose ')', d=6)
path = "((()))"
path = "((()))" (open=3, close=3). Base Case (len == 2*n = 6): Record "((()))" ✅ (#1).
8↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "((())"
path = "((())" →\rightarrow→ ↩️ BACKTRACK / POP.
9↩️ BACKTRACK / POP(Backtrack / Pop ')', d=4)
path = "((()"
path = "((()" →\rightarrow→ ↩️ BACKTRACK / POP.
10↩️ BACKTRACK / POP(Backtrack / Pop ')', d=3)
path = "((("
path = "(((" →\rightarrow→ ↩️ BACKTRACK / POP.
11↩️ BACKTRACK / POP(Backtrack / Pop '(', d=2)
path = "(("
path = "((" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
12🟢 CHOOSE(Branch Right: Choose ')', d=3)
path = "(()"
Choices:( ) ✓
path = "(()" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
13🟢 CHOOSE(Branch Left: Choose '(', d=4)
path = "(()("
path = "(()(" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
14🟢 CHOOSE(Choose ')', d=5)
path = "(()()"
path = "(()()" (open=3, close=2). Place ')'.
15✅ RECORD / GOAL(Choose ')', d=6)
path = "(()())"
path = "(()())" (open=3, close=3). Base Case (len == 6): Record "(()())" ✅ (#2).
16↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "(()()"
path = "(()()" →\rightarrow→ ↩️ BACKTRACK / POP.
17↩️ BACKTRACK / POP(Backtrack / Pop ')', d=4)
path = "(()("
path = "(()(" →\rightarrow→ ↩️ BACKTRACK / POP.
18↩️ BACKTRACK / POP(Backtrack / Pop '(', d=3)
path = "(()"
path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
19🟢 CHOOSE(Branch Right: Choose ')', d=4)
path = "(())"
path = "(())" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
20🟢 CHOOSE(Choose '(', d=5)
path = "(())("
path = "(())(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
21✅ RECORD / GOAL(Choose ')', d=6)
path = "(())()"
path = "(())()" (open=3, close=3). Base Case (len == 6): Record "(())()" ✅ (#3).
22↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "(())("
path = "(())(" →\rightarrow→ ↩️ BACKTRACK / POP.
23↩️ BACKTRACK / POP(Backtrack / Pop '(', d=4)
path = "(())"
path = "(())" →\rightarrow→ ↩️ BACKTRACK / POP.
24↩️ BACKTRACK / POP(Backtrack / Pop ')', d=3)
path = "(()"
path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP.
25↩️ BACKTRACK / POP(Backtrack / Pop ')', d=2)
path = "(("
path = "((" →\rightarrow→ ↩️ BACKTRACK / POP.
26↩️ BACKTRACK / POP(Backtrack / Pop '(', d=1)
path = "("
path = "(" →\rightarrow→ ↩️ BACKTRACK / POP to explore main right branch: choose ')'.
27🟢 CHOOSE(Branch Right: Choose ')', d=2)
path = "()"
path = "()" (open=1, close=1). close == open →\rightarrow→ only legal choice is '('.
28🟢 CHOOSE(Choose '(', d=3)
path = "()("
Choices:( ✓ )
path = "()(" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
29🟢 CHOOSE(Branch Left: Choose '(', d=4)
path = "()(("
path = "()((" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
30🟢 CHOOSE(Choose ')', d=5)
path = "()(()"
path = "()(()" (open=3, close=2). Place ')'.
31✅ RECORD / GOAL(Choose ')', d=6)
path = "()(())"
path = "()(())" (open=3, close=3). Base Case (len == 6): Record "()(())" ✅ (#4).
32↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "()(()"
path = "()(()" →\rightarrow→ ↩️ BACKTRACK / POP.
33↩️ BACKTRACK / POP(Backtrack / Pop ')', d=4)
path = "()(("
path = "()((" →\rightarrow→ ↩️ BACKTRACK / POP.
34↩️ BACKTRACK / POP(Backtrack / Pop '(', d=3)
path = "()("
path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
35🟢 CHOOSE(Branch Right: Choose ')', d=4)
path = "()()"
path = "()()" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
36🟢 CHOOSE(Choose '(', d=5)
path = "()()("
path = "()()(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
37✅ RECORD / GOAL(Choose ')', d=6)
path = "()()()"
path = "()()()" (open=3, close=3). Base Case (len == 6): Record "()()()" ✅ (#5).
38↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "()()("
path = "()()(" →\rightarrow→ ↩️ BACKTRACK / POP.
39↩️ BACKTRACK / POP(Backtrack / Pop '(', d=4)
path = "()()"
path = "()()" →\rightarrow→ ↩️ BACKTRACK / POP.
40↩️ BACKTRACK / POP(Backtrack / Pop ')', d=3)
path = "()("
path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP.
41↩️ BACKTRACK / POP(Backtrack / Pop '(', d=2)
path = "()"
path = "()" →\rightarrow→ ↩️ BACKTRACK / POP.
42↩️ BACKTRACK / POP(Backtrack / Pop ')', d=1)
path = "("
path = "(" →\rightarrow→ ↩️ BACKTRACK / POP.
43🌱 ROOT(Backtrack / Pop '(', d=0)
path = ""
path = "" →\rightarrow→ ↩️ BACKTRACK / POP to Root. All 5 Catalan combinations generated! Output: ["((()))", "(()())", "(())()", "()(())", "()()()"].
Code Shape
Archetype Code Shape
Python
1results = []
2path = []
3def dfs(open_c, close_c):
4 if len(path) == 2 * n:
5 results.append(''.join(path))
6 return
7 if open_c < n:
8 path.append('(')
9 dfs(open_c + 1, close_c)
10 path.pop()
11 if close_c < open_c:
12 path.append(')')
13 dfs(open_c, close_c + 1)
14 path.pop()
15dfs(0, 0)
16return results

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

The naive approach — generate every possible string of ( and ), then filter for validity — wastes enormous effort building strings that were doomed from their very first character. This archetype never builds an invalid string in the first place.

⚖️ The Analogy: A Balance Meter That Refuses Illegal Moves

Imagine building a tower out of "open" and "close" blocks, watched by a balance meter that goes up by one for every open block and down by one for every close block. Two rules keep the meter legal at every single step, not just at the end: you can only add an open block if you haven't used up your total budget of n yet, and you can only add a close block if the meter is currently above zero — otherwise you'd be closing something that was never opened.

🔑 The Key Idea in the Code

Notice there's no separate "is this string valid?" check anywhere — validity is enforced move-by-move, through the two if guards, so every single path that reaches the base case is automatically well-formed.

  • Adding close parenthesis when close >= open: A closing parenthesis ')' is only valid if strictly fewer closing parentheses than opening parentheses have been placed (close < open). Violating this produces invalid expressions.

  • String immutability vs backtracking list: In Python, passing strings current + '(' creates new strings and requires no backtrack step. If using a mutable list, you must remember path.pop().

  • Base case check: The base case triggers when len(current) == 2 * n (both open == n and close == n).

  • Stagnant State in For-Loops. If you use a for option in options: loop for fixed-choice problems, do not use a generic dfs(start + 1) variable. Because the loop options don't dynamically shrink, your state variables (like open_count and close_count) must be passed into the recursive call, or you will trigger a Maximum Call Stack Exceeded error.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Generate all valid sequences subject to structural open/close pairing constraints. Notice that recursive state is NOT a single 'start' index, but dual tracking variables (open_count, close_count).

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

State Representation & Invariant: State is polymorphic — here, it is the multi-variable tuple (open_count, close_count). At every decision step, close_count <= open_count <= n guarantees partial Catalan validity.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

  • Assuming state is always a single 'start' index: In fixed-choice problems, state can be multiple tracking variables (open_count, close_count). Failing to pass these state counters into recursive calls leads to stagnant state.

  • Generating all 2^(2N) strings and filtering afterward: O(2^(2N)) vs optimal O(Catalan(N)).

  • Condition order: checking close < open (not close < n) ensures no unmatched closing brackets.

The 60-Second Interview Pitch

Say this out loud before you type a single line

"I maintain open_count and close_count as my polymorphic state. I add '(' when open_count < n, and add ')' when close_count < open_count. When len(path) == 2 * n, I record the valid combination. This explores ONLY valid Catalan paths in O(4^N / sqrt(N)) time and O(N) stack space."

Interview Tip: When under pressure, cognitive consistency is your best friend. Many engineers prefer to write out for option in ["(", ")"]: explicitly. Even though it is mathematically just two branches, mapping it perfectly 1:1 with the universal Backtracking template eliminates the mental friction of switching paradigms.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(4^N / sqrt(N))

Look at the execution steps: Number of valid parenthesizations of length 2N is the N-th Catalan number CNC_{N}CN​ = 1/(N+1) * binom(2N, N) ~ 4^N / (N^(3/2) * sqrt(pi)). Constructing each string of length 2N takes O(N).

SPACE COMPLEXITY

O(N) Call Stack

Recursion stack reaches depth 2N; path array stores 2N characters.

Formal Recurrence Relation

T(N) = CN · O(N) = O(4^N / √N)

Derivation Progression

Catalan Combinatorial Count

Catalan Count ~ 4^N / (N^1.5 · √π)

Exact number of well-formed parentheses combinations.

Pruning Conditions

open < N, close < open

Strictly prevents invalid prefix generation in O(1).

String Materialization

Total Time = O(4^N / √N)

Joins 2N characters per valid leaf.

Variable Definitions

NNN

Number of parentheses pairs (2N total characters in string)

CNC_NCN​

N-th Catalan number = (1 / (N+1)) · (2N choose N) ~ 4^N / (N^1.5 · √π)

Memory Architecture & Bounds

🟣 Call Stack

O(N) Call stack depth 2N

🔵 Auxiliary Heap

O(N) Mutable character list

🟢 Output Space

O(CNC_{N}CN​ · N) Result string list

Boundary Best / Worst Cases

Best Case

O(4N/√(N))O(4^N / √(N))O(4N/√(N))

Average Case

O(4N/√(N))O(4^N / √(N))O(4N/√(N))

Worst Case

O(4N/√(N))O(4^N / √(N))O(4N/√(N))

Decision & State-Space Search Tree

Decision & State-Space Search Tree
Synthesizing vector architecture diagram...
Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "generate all combinations of well-formed parentheses", "given n pairs". Balanced count tracking: an opening parenthesis can be placed if open<nopen < nopen<n; a closing parenthesis can be placed if close<openclose < openclose<open.

CONSTRAINTS & BOUNDS

1≤n≤81 \le n \le 81≤n≤8. Total valid combinations given by Catalan number C8=1430C_8 = 1430C8​=1430.

FAANG PRODUCTION TRAPS & EDGE CASES

Validating parentheses after building string (O(22N⋅N)O(2^{2N} \cdot N)O(22N⋅N)): pruning during construction reduces state space directly to Catalan numbers (O(Cn)O(C_n)O(Cn​)).

Theory Context•Backtracking & State Search
MediumLC 22

Generate Parentheses (LeetCode 22)

Target Frequency:MetaAmazonGoogleApple

Given n pairs of parentheses, produce every distinct combination of n '(' and n ')' characters that forms a well-formed string.

A string is well-formed when every '(' is eventually matched by a ')', and reading left to right, the running count of ')' never exceeds the running count of '('.

Worked Examples

Example 1
Input:n = 3
Output:["((()))","(()())","(())()","()(())","()()()"]
Explanation: These are all five balanced strings of three ( and three ).
Example 2
Input:n = 1
Output:["()"]
Explanation: With one pair, "()" is the only balanced string.

⚖️Formal Constraints & Bounds

  • 1 <= n <= 8

Deep-Dive & Conceptual Insights

Core Insight

Dual counters constrain the search space exclusively to valid paths, achieving optimal O(Catalan(N)) generation.

Real-World Scenario & Production Applications

Compiler parser syntax validation, mathematical expression token generation.

Subproblems & Recurrence Decomposition3 Phases

🧩Subproblem 1: Open Parenthesis Branching

Valid to add an opening bracket '(' whenever the count of open brackets open_count < n.

Mathematical Recurrence / Code Invariant
if open_count < n:
    path.append("(")
    dfs(open_count + 1, close_count)
    path.pop()

Step-by-Step Execution Trace Table

Code / Blueprint
                                         "" (open=0, close=0, d=0)
                                            |
                                            | choose '('
                                            v
                                   "(" (open=1, close=0, d=1)
                                  /                          \
                     choose '('  /                            \  choose ')'
                                /                              \
                    "((" (open=2, close=0, d=2)            "()" (open=1, close=1, d=2)
                   /                    \                          |
       choose '(' /                      \ choose ')'              | choose '(' (close==open)
                 /                        \                        v
         "(((" (open=3, close=0, d=3)   "(()" (open=2, close=1, d=3) "()(" (open=2, close=1, d=3)
                 |                      /            \             /            \
      choose ')' |          choose '(' /   choose ')' \ choose '('/   choose ')' \
                 v                    /                \         /                \
        "((()" (d=4)          "(()(" (d=4)        "(())" (d=4)"()((" (d=4)        "()()" (d=4)
                 |                    |                |         |                |
      choose ')' |         choose ')' |     choose '(' |choose ')'|    choose '(' |
                 v                    v                v         v                v
       "((())" (d=5)         "(()()" (d=5)       "(())(" (d=5)"()(()" (d=5)       "()()(" (d=5)
                 |                    |                |         |                |
      choose ')' |         choose ')' |     choose ')' |choose ')'|    choose ')' |
                 v                    v                v         v                v
      "((()))" ✅            "(()())" ✅         "(())()" ✅  "()(())" ✅         "()()()" ✅
 
Step-by-Step Decision Trace (for n = 3)
  1. Step 1 (Root, d=0): path = "" (open=0, close=0). Rule: close < open is False (0 < 0), so must choose '(' first.
  2. Step 2 (Choose '(', d=1): path = "(" (open=1, close=0). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
  3. Step 3 (Branch Left: Choose '(', d=2): path = "((" (open=2, close=0). open < 3 is True →\rightarrow→ branch '(' first.
  4. Step 4 (Branch Left: Choose '(', d=3): path = "(((" (open=3, close=0). open == 3 quota reached →\rightarrow→ only legal choice is ')'.
  5. Step 5 (Choose ')', d=4): path = "((()" (open=3, close=1). close < open (1 < 3) →\rightarrow→ place ')'.
  6. Step 6 (Choose ')', d=5): path = "((())" (open=3, close=2). Place ')'.
  7. Step 7 (Choose ')', d=6): path = "((()))" (open=3, close=3). Base Case (len == 2*n = 6): Record "((()))" ✅ (#1).
  8. Step 8 (Backtrack / Pop ')', d=5): path = "((())" →\rightarrow→ ↩️ BACKTRACK / POP.
  9. Step 9 (Backtrack / Pop ')', d=4): path = "((()" →\rightarrow→ ↩️ BACKTRACK / POP.
  10. Step 10 (Backtrack / Pop ')', d=3): path = "(((" →\rightarrow→ ↩️ BACKTRACK / POP.
  11. Step 11 (Backtrack / Pop '(', d=2): path = "((" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
  12. Step 12 (Branch Right: Choose ')', d=3): path = "(()" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
  13. Step 13 (Branch Left: Choose '(', d=4): path = "(()(" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
  14. Step 14 (Choose ')', d=5): path = "(()()" (open=3, close=2). Place ')'.
  15. Step 15 (Choose ')', d=6): path = "(()())" (open=3, close=3). Base Case (len == 6): Record "(()())" ✅ (#2).
  16. Step 16 (Backtrack / Pop ')', d=5): path = "(()()" →\rightarrow→ ↩️ BACKTRACK / POP.
  17. Step 17 (Backtrack / Pop ')', d=4): path = "(()(" →\rightarrow→ ↩️ BACKTRACK / POP.
  18. Step 18 (Backtrack / Pop '(', d=3): path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
  19. Step 19 (Branch Right: Choose ')', d=4): path = "(())" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
  20. Step 20 (Choose '(', d=5): path = "(())(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
  21. Step 21 (Choose ')', d=6): path = "(())()" (open=3, close=3). Base Case (len == 6): Record "(())()" ✅ (#3).
  22. Step 22 (Backtrack / Pop ')', d=5): path = "(())(" →\rightarrow→ ↩️ BACKTRACK / POP.
  23. Step 23 (Backtrack / Pop '(', d=4): path = "(())" →\rightarrow→ ↩️ BACKTRACK / POP.
  24. Step 24 (Backtrack / Pop ')', d=3): path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP.
  25. Step 25 (Backtrack / Pop ')', d=2): path = "((" →\rightarrow→ ↩️ BACKTRACK / POP.
  26. Step 26 (Backtrack / Pop '(', d=1): path = "(" →\rightarrow→ ↩️ BACKTRACK / POP to explore main right branch: choose ')'.
  27. Step 27 (Branch Right: Choose ')', d=2): path = "()" (open=1, close=1). close == open →\rightarrow→ only legal choice is '('.
  28. Step 28 (Choose '(', d=3): path = "()(" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
  29. Step 29 (Branch Left: Choose '(', d=4): path = "()((" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
  30. Step 30 (Choose ')', d=5): path = "()(()" (open=3, close=2). Place ')'.
  31. Step 31 (Choose ')', d=6): path = "()(())" (open=3, close=3). Base Case (len == 6): Record "()(())" ✅ (#4).
  32. Step 32 (Backtrack / Pop ')', d=5): path = "()(()" →\rightarrow→ ↩️ BACKTRACK / POP.
  33. Step 33 (Backtrack / Pop ')', d=4): path = "()((" →\rightarrow→ ↩️ BACKTRACK / POP.
  34. Step 34 (Backtrack / Pop '(', d=3): path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
  35. Step 35 (Branch Right: Choose ')', d=4): path = "()()" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
  36. Step 36 (Choose '(', d=5): path = "()()(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
  37. Step 37 (Choose ')', d=6): path = "()()()" (open=3, close=3). Base Case (len == 6): Record "()()()" ✅ (#5).
  38. Step 38 (Backtrack / Pop ')', d=5): path = "()()(" →\rightarrow→ ↩️ BACKTRACK / POP.
  39. Step 39 (Backtrack / Pop '(', d=4): path = "()()" →\rightarrow→ ↩️ BACKTRACK / POP.
  40. Step 40 (Backtrack / Pop ')', d=3): path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP.
  41. Step 41 (Backtrack / Pop '(', d=2): path = "()" →\rightarrow→ ↩️ BACKTRACK / POP.
  42. Step 42 (Backtrack / Pop ')', d=1): path = "(" →\rightarrow→ ↩️ BACKTRACK / POP.
  43. Step 43 (Backtrack / Pop '(', d=0): path = "" →\rightarrow→ ↩️ BACKTRACK / POP to Root. All 5 Catalan combinations generated! Output: ["((()))", "(()())", "(())()", "()(())", "()()()"].
Full Walkthrough43 Steps
Input:n = 3Expected:["((()))","(()())","(())()","()(())","()()()"]
1🌱 ROOT(Root, d=0)
path = ""
Choices:( ✓ ) ✂️
path = "" (open=0, close=0). Rule: close < open is False (0 < 0), so must choose '(' first.
2🟢 CHOOSE(Choose '(', d=1)
path = "("
Choices:( ✓ )
path = "(" (open=1, close=0). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
3🟢 CHOOSE(Branch Left: Choose '(', d=2)
path = "(("
path = "((" (open=2, close=0). open < 3 is True →\rightarrow→ branch '(' first.
4🟢 CHOOSE(Branch Left: Choose '(', d=3)
path = "((("
path = "(((" (open=3, close=0). open == 3 quota reached →\rightarrow→ only legal choice is ')'.
5🟢 CHOOSE(Choose ')', d=4)
path = "((()"
Choices:( ) ✓
path = "((()" (open=3, close=1). close < open (1 < 3) →\rightarrow→ place ')'.
6🟢 CHOOSE(Choose ')', d=5)
path = "((())"
path = "((())" (open=3, close=2). Place ')'.
7✅ RECORD / GOAL(Choose ')', d=6)
path = "((()))"
path = "((()))" (open=3, close=3). Base Case (len == 2*n = 6): Record "((()))" ✅ (#1).
8↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "((())"
path = "((())" →\rightarrow→ ↩️ BACKTRACK / POP.
9↩️ BACKTRACK / POP(Backtrack / Pop ')', d=4)
path = "((()"
path = "((()" →\rightarrow→ ↩️ BACKTRACK / POP.
10↩️ BACKTRACK / POP(Backtrack / Pop ')', d=3)
path = "((("
path = "(((" →\rightarrow→ ↩️ BACKTRACK / POP.
11↩️ BACKTRACK / POP(Backtrack / Pop '(', d=2)
path = "(("
path = "((" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
12🟢 CHOOSE(Branch Right: Choose ')', d=3)
path = "(()"
Choices:( ) ✓
path = "(()" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
13🟢 CHOOSE(Branch Left: Choose '(', d=4)
path = "(()("
path = "(()(" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
14🟢 CHOOSE(Choose ')', d=5)
path = "(()()"
path = "(()()" (open=3, close=2). Place ')'.
15✅ RECORD / GOAL(Choose ')', d=6)
path = "(()())"
path = "(()())" (open=3, close=3). Base Case (len == 6): Record "(()())" ✅ (#2).
16↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "(()()"
path = "(()()" →\rightarrow→ ↩️ BACKTRACK / POP.
17↩️ BACKTRACK / POP(Backtrack / Pop ')', d=4)
path = "(()("
path = "(()(" →\rightarrow→ ↩️ BACKTRACK / POP.
18↩️ BACKTRACK / POP(Backtrack / Pop '(', d=3)
path = "(()"
path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
19🟢 CHOOSE(Branch Right: Choose ')', d=4)
path = "(())"
path = "(())" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
20🟢 CHOOSE(Choose '(', d=5)
path = "(())("
path = "(())(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
21✅ RECORD / GOAL(Choose ')', d=6)
path = "(())()"
path = "(())()" (open=3, close=3). Base Case (len == 6): Record "(())()" ✅ (#3).
22↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "(())("
path = "(())(" →\rightarrow→ ↩️ BACKTRACK / POP.
23↩️ BACKTRACK / POP(Backtrack / Pop '(', d=4)
path = "(())"
path = "(())" →\rightarrow→ ↩️ BACKTRACK / POP.
24↩️ BACKTRACK / POP(Backtrack / Pop ')', d=3)
path = "(()"
path = "(()" →\rightarrow→ ↩️ BACKTRACK / POP.
25↩️ BACKTRACK / POP(Backtrack / Pop ')', d=2)
path = "(("
path = "((" →\rightarrow→ ↩️ BACKTRACK / POP.
26↩️ BACKTRACK / POP(Backtrack / Pop '(', d=1)
path = "("
path = "(" →\rightarrow→ ↩️ BACKTRACK / POP to explore main right branch: choose ')'.
27🟢 CHOOSE(Branch Right: Choose ')', d=2)
path = "()"
path = "()" (open=1, close=1). close == open →\rightarrow→ only legal choice is '('.
28🟢 CHOOSE(Choose '(', d=3)
path = "()("
Choices:( ✓ )
path = "()(" (open=2, close=1). Both open < 3 and close < open are True →\rightarrow→ branch '(' first.
29🟢 CHOOSE(Branch Left: Choose '(', d=4)
path = "()(("
path = "()((" (open=3, close=1). open == 3 quota reached →\rightarrow→ place ')'.
30🟢 CHOOSE(Choose ')', d=5)
path = "()(()"
path = "()(()" (open=3, close=2). Place ')'.
31✅ RECORD / GOAL(Choose ')', d=6)
path = "()(())"
path = "()(())" (open=3, close=3). Base Case (len == 6): Record "()(())" ✅ (#4).
32↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "()(()"
path = "()(()" →\rightarrow→ ↩️ BACKTRACK / POP.
33↩️ BACKTRACK / POP(Backtrack / Pop ')', d=4)
path = "()(("
path = "()((" →\rightarrow→ ↩️ BACKTRACK / POP.
34↩️ BACKTRACK / POP(Backtrack / Pop '(', d=3)
path = "()("
path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP to explore right branch: choose ')'.
35🟢 CHOOSE(Branch Right: Choose ')', d=4)
path = "()()"
path = "()()" (open=2, close=2). close == open →\rightarrow→ only legal choice is '('.
36🟢 CHOOSE(Choose '(', d=5)
path = "()()("
path = "()()(" (open=3, close=2). open == 3 quota reached →\rightarrow→ place ')'.
37✅ RECORD / GOAL(Choose ')', d=6)
path = "()()()"
path = "()()()" (open=3, close=3). Base Case (len == 6): Record "()()()" ✅ (#5).
38↩️ BACKTRACK / POP(Backtrack / Pop ')', d=5)
path = "()()("
path = "()()(" →\rightarrow→ ↩️ BACKTRACK / POP.
39↩️ BACKTRACK / POP(Backtrack / Pop '(', d=4)
path = "()()"
path = "()()" →\rightarrow→ ↩️ BACKTRACK / POP.
40↩️ BACKTRACK / POP(Backtrack / Pop ')', d=3)
path = "()("
path = "()(" →\rightarrow→ ↩️ BACKTRACK / POP.
41↩️ BACKTRACK / POP(Backtrack / Pop '(', d=2)
path = "()"
path = "()" →\rightarrow→ ↩️ BACKTRACK / POP.
42↩️ BACKTRACK / POP(Backtrack / Pop ')', d=1)
path = "("
path = "(" →\rightarrow→ ↩️ BACKTRACK / POP.
43🌱 ROOT(Backtrack / Pop '(', d=0)
path = ""
path = "" →\rightarrow→ ↩️ BACKTRACK / POP to Root. All 5 Catalan combinations generated! Output: ["((()))", "(()())", "(())()", "()(())", "()()()"].
Code Shape
Archetype Code Shape
Python
1results = []
2path = []
3def dfs(open_c, close_c):
4 if len(path) == 2 * n:
5 results.append(''.join(path))
6 return
7 if open_c < n:
8 path.append('(')
9 dfs(open_c + 1, close_c)
10 path.pop()
11 if close_c < open_c:
12 path.append(')')
13 dfs(open_c, close_c + 1)
14 path.pop()
15dfs(0, 0)
16return results

Conceptual Narrative

🧭 Conceptual Foundation & Pattern Intuition

The naive approach — generate every possible string of ( and ), then filter for validity — wastes enormous effort building strings that were doomed from their very first character. This archetype never builds an invalid string in the first place.

⚖️ The Analogy: A Balance Meter That Refuses Illegal Moves

Imagine building a tower out of "open" and "close" blocks, watched by a balance meter that goes up by one for every open block and down by one for every close block. Two rules keep the meter legal at every single step, not just at the end: you can only add an open block if you haven't used up your total budget of n yet, and you can only add a close block if the meter is currently above zero — otherwise you'd be closing something that was never opened.

🔑 The Key Idea in the Code

Notice there's no separate "is this string valid?" check anywhere — validity is enforced move-by-move, through the two if guards, so every single path that reaches the base case is automatically well-formed.

  • Adding close parenthesis when close >= open: A closing parenthesis ')' is only valid if strictly fewer closing parentheses than opening parentheses have been placed (close < open). Violating this produces invalid expressions.

  • String immutability vs backtracking list: In Python, passing strings current + '(' creates new strings and requires no backtrack step. If using a mutable list, you must remember path.pop().

  • Base case check: The base case triggers when len(current) == 2 * n (both open == n and close == n).

  • Stagnant State in For-Loops. If you use a for option in options: loop for fixed-choice problems, do not use a generic dfs(start + 1) variable. Because the loop options don't dynamically shrink, your state variables (like open_count and close_count) must be passed into the recursive call, or you will trigger a Maximum Call Stack Exceeded error.

Senior SWE Reasoning Architecture

4-Phase Thought Process Model

Pattern Recognition Signals

The 10-second spot

Generate all valid sequences subject to structural open/close pairing constraints. Notice that recursive state is NOT a single 'start' index, but dual tracking variables (open_count, close_count).

Formulating the Predicate & Invariants

Turning intuition into a boolean rule

State Representation & Invariant: State is polymorphic — here, it is the multi-variable tuple (open_count, close_count). At every decision step, close_count <= open_count <= n guarantees partial Catalan validity.

Silent Failure Traps & Edge Cases

Where confident candidates still lose points

  • Assuming state is always a single 'start' index: In fixed-choice problems, state can be multiple tracking variables (open_count, close_count). Failing to pass these state counters into recursive calls leads to stagnant state.

  • Generating all 2^(2N) strings and filtering afterward: O(2^(2N)) vs optimal O(Catalan(N)).

  • Condition order: checking close < open (not close < n) ensures no unmatched closing brackets.

The 60-Second Interview Pitch

Say this out loud before you type a single line

"I maintain open_count and close_count as my polymorphic state. I add '(' when open_count < n, and add ')' when close_count < open_count. When len(path) == 2 * n, I record the valid combination. This explores ONLY valid Catalan paths in O(4^N / sqrt(N)) time and O(N) stack space."

Interview Tip: When under pressure, cognitive consistency is your best friend. Many engineers prefer to write out for option in ["(", ")"]: explicitly. Even though it is mathematically just two branches, mapping it perfectly 1:1 with the universal Backtracking template eliminates the mental friction of switching paradigms.

Big-O Invariant Derivation

Complexity & Mathematical Proof

TIME COMPLEXITY

O(4^N / sqrt(N))

Look at the execution steps: Number of valid parenthesizations of length 2N is the N-th Catalan number CNC_{N}CN​ = 1/(N+1) * binom(2N, N) ~ 4^N / (N^(3/2) * sqrt(pi)). Constructing each string of length 2N takes O(N).

SPACE COMPLEXITY

O(N) Call Stack

Recursion stack reaches depth 2N; path array stores 2N characters.

Formal Recurrence Relation

T(N) = CN · O(N) = O(4^N / √N)

Derivation Progression

Catalan Combinatorial Count

Catalan Count ~ 4^N / (N^1.5 · √π)

Exact number of well-formed parentheses combinations.

Pruning Conditions

open < N, close < open

Strictly prevents invalid prefix generation in O(1).

String Materialization

Total Time = O(4^N / √N)

Joins 2N characters per valid leaf.

Variable Definitions

NNN

Number of parentheses pairs (2N total characters in string)

CNC_NCN​

N-th Catalan number = (1 / (N+1)) · (2N choose N) ~ 4^N / (N^1.5 · √π)

Memory Architecture & Bounds

🟣 Call Stack

O(N) Call stack depth 2N

🔵 Auxiliary Heap

O(N) Mutable character list

🟢 Output Space

O(CNC_{N}CN​ · N) Result string list

Boundary Best / Worst Cases

Best Case

O(4N/√(N))O(4^N / √(N))O(4N/√(N))

Average Case

O(4N/√(N))O(4^N / √(N))O(4N/√(N))

Worst Case

O(4N/√(N))O(4^N / √(N))O(4N/√(N))

Decision & State-Space Search Tree

Decision & State-Space Search Tree
Synthesizing vector architecture diagram...
Staff+ Engineering Perspective

Senior SWE Deconstruction & Hardware Caveats

ARCHITECTURAL SIGNALS & INTERVIEW TRIGGERS

Triggers: "generate all combinations of well-formed parentheses", "given n pairs". Balanced count tracking: an opening parenthesis can be placed if open<nopen < nopen<n; a closing parenthesis can be placed if close<openclose < openclose<open.

CONSTRAINTS & BOUNDS

1≤n≤81 \le n \le 81≤n≤8. Total valid combinations given by Catalan number C8=1430C_8 = 1430C8​=1430.

FAANG PRODUCTION TRAPS & EDGE CASES

Validating parentheses after building string (O(22N⋅N)O(2^{2N} \cdot N)O(22N⋅N)): pruning during construction reduces state space directly to Catalan numbers (O(Cn)O(C_n)O(Cn​)).

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