All O`one Data Structure (LeetCode 432)
Track a running count per string, and be able to report a string currently at the maximum count and one at the minimum count at any moment.
Build an AllOne class:
AllOne()starts with an empty structure.inc(String key)bumpskey's count by1, inserting it at count1if it wasn't tracked yet.dec(String key)lowerskey's count by1, removingkeyentirely once its count hits0.keyis always guaranteed to be present before adeccall.getMaxKey()returns any one key currently at the maximum count, or""if the structure is empty.getMinKey()returns any one key currently at the minimum count, or""if the structure is empty.
Every one of these four operations must run in O(1) average time.
Worked Examples
["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
[[], ["hello"], ["hello"], [], [], ["leet"], [], []][null, null, null, "hello", "hello", null, "hello", "leet"]⚖️Formal Constraints & Bounds
1 <= key.length <= 10key consists of lowercase English letters.It is guaranteed that for each call to dec, key is existing in the data structure.At most 5 * 104 calls will be made to inc, dec, getMaxKey, and getMinKey.
Core Insight
A count only ever moves by exactly +-1 per operation, so a doubly-linked chain of buckets sorted by count (each bucket a set of same-count keys), with sentinels whose immediate neighbors are always the true min/max, gives O(1) inc, dec, getMaxKey, and getMinKey without ever scanning or resorting.
Real-World Scenario & Production Applications
Real-time leaderboard "who's currently most/least active" queries, autocomplete frequency ranking, rate-limiter hot-key/cold-key tracking.
Subproblems & Recurrence Decomposition3 Phases
A hash map key_bucket resolves any key directly to the bucket node it currently occupies in O(1), so inc/dec never need to search for a key's current count.
self.key_bucket = {} # key -> the KeySet bucket node it sits inStep-by-Step Execution Trace Table
| Step | Operation | Key Counts Map | Doubly-Linked Bucket List (Head <-> ... <-> Tail) | getMaxKey() | getMinKey() |
|---|---|---|---|---|---|
| 1 | inc("hello") | {"hello": 1} | [Head] <-> [cnt=1: {"hello"}] <-> [Tail] | "hello" | "hello" |
| 2 | inc("hello") | {"hello": 2} | [Head] <-> [cnt=2: {"hello"}] <-> [Tail] (bucket 1 deleted) | "hello" | "hello" |
| 3 | inc("world") | {"hello": 2, "world": 1} | [Head] <-> [cnt=1: {"world"}] <-> [cnt=2: {"hello"}] <-> [Tail] | "hello" | "world" |
| 4 | inc("leet") | {"hello": 2, "world": 1, "leet": 1} | [Head] <-> [cnt=1: {"world", "leet"}] <-> [cnt=2: {"hello"}] <-> [Tail] | "hello" | "world" or "leet" |
| 5 | dec("hello") | {"hello": 1, "world": 1, "leet": 1} | [Head] <-> [cnt=1: {"world", "leet", "hello"}] <-> [Tail] (bucket 2 deleted) | any of 3 | any of 3 |
| 1 | self.key_bucket = {} # key -> bucket node it currently sits in |
| 2 | head <-> [count=1 bucket] <-> ... <-> [count=max bucket] <-> tail |
| 3 | inc(key) -> move key to bucket at count+1 (create if needed) |
| 4 | dec(key) -> move key to bucket at count-1 (remove key entirely if count was 1) |
| 5 | getMaxKey() -> any key in tail.prev's set |
| 6 | getMinKey() -> any key in head.next's set |
Conceptual Narrative
LFU Cache needed frequency buckets to evict the minimum. All Oone must report BOTH the minimum and maximum keys by count in O(1)time, while supportingO(1)` increments and decrements.
🧭 Conceptual Foundation & Pattern Intuition
All Oone Data Structure (LeetCode 432) demands four methods running in strict $O(1)$ time: inc(key), dec(key), getMaxKey(), and getMinKey()`.
Because keys must be queryable by their extremes, and counts only change by per operation, we maintain a Doubly-Linked List of Count Buckets:
- Each Bucket node in the DLL holds an integer
countand a hash setkeyscontaining all strings that currently have that exact count. - The DLL is strictly sorted by
count:head <-> Bucket(1) <-> Bucket(3) <-> Bucket(7) <-> tail. key_to_bucketmaps each string to its current Bucket node in the DLL.- When
inc(key)is called, the key moves from bucket to bucket . If an adjacent bucket with count already exists in the DLL, the key is added to it; otherwise, a new bucket node is spliced in immediately after bucket . - When a key departs a bucket and that bucket's
keysset becomes empty, that bucket is unlinked from the DLL immediately. getMaxKey()reads any key fromtail.prev.keys;getMinKey()reads any key fromhead.next.keys. Both run in strict time!
🪜 The Analogy: The Staircase of Contestant Leaderboard Podiums
Imagine a staircase where step holds a box containing all game contestants who currently have exactly points. When a contestant scores a point, they step out of box and step into box . If box doesn't exist on the staircase yet, a carpenter instantly slides a new wooden step into the staircase right above step . If box becomes completely empty, the carpenter removes that step so no gap remains. When the judges ask "Who is in first place?", they look at the very top step of the staircase. When they ask "Who is in last place?", they look at the very bottom step.
🔑 Why It Works
Because the list only ever contains non-empty buckets, head.next is guaranteed to be the global minimum and tail.prev is guaranteed to be the global maximum at all times.
Empty bucket cleanup omission: When the last key is removed from a count bucket, that bucket node must be unlinked from the doubly-linked list of buckets. Leaving empty buckets causes
getMaxKeyorgetMinKeyto return keys from empty buckets.Inserting new key at front without count 1 bucket: When inserting a brand-new key with count 1, if the bucket adjacent to
headdoes not have count 1, a new bucket node must be created and linked immediately afterhead.Returning empty string when data structure contains zero keys: Both
getMaxKey()andgetMinKey()must return""when the data structure contains no keys.
4-Phase Thought Process Model
Pattern Recognition Signals
The 10-second spot
Need O(1) increment/decrement of a per-key counter AND O(1) retrieval of any key currently at the max count AND any key at the min count.
Formulating the Predicate & Invariants
Turning intuition into a boolean rule
Invariant: a key's count changes by exactly +-1 per call, so it always moves to the immediately adjacent bucket; sentinels guarantee head.next is always the true min-count occupied bucket and tail.prev the true max-count occupied bucket.
Silent Failure Traps & Edge Cases
Where confident candidates still lose points
Failing to unlink a bucket the instant its key set becomes empty, which would let a stale empty bucket masquerade as the true min or max.
Reusing an existing adjacent bucket instead of blindly creating a new one when inc/dec lands on a count that already has an occupied bucket -- otherwise two buckets end up representing the same count.
Forgetting dec() must delete the key entirely (not move it to a count-0 bucket) when its count was exactly 1.
The 60-Second Interview Pitch
Say this out loud before you type a single line
"I keep a doubly-linked list of buckets strictly ordered by count, with head and tail sentinels, plus a hash map from key to the bucket node it currently occupies. inc(key) moves the key to the bucket at count+1 (creating it via
O(1)pointer splice if it doesn't already exist as the immediate neighbor), and dec(key) moves it to count-1 or deletes the key outright if its count was 1. Any bucket that loses its last key is unlinked immediately. Because the chain is always contiguous and sorted, getMaxKey reads any key from tail.prev's set and getMinKey reads any key from head.next's set, bothO(1), with all mutations alsoO(1)."
Complexity & Mathematical Proof
O(1) Average
Look at inc(key), dec(key), getMaxKey(), getMinKey(): key_bucket gives O(1) average lookup of the bucket a key currently sits in. _insert_after either reuses the immediately adjacent bucket (O(1) check) or splices a brand-new bucket node between two existing ones (O(1) pointer mutation). Removing a key from its old bucket's set and adding to the new bucket's set are O(1) average hash-set operations. getMaxKey/getMinKey simply peek at tail.prev/head.next's key set, O(1). Every operation is strictly O(1) average time.
O(K) Auxiliary
Look at memory allocation: K = number of distinct keys ever inserted and currently present. key_bucket holds O(K) entries. The bucket chain holds at most O(K) bucket nodes (bounded by the number of distinct counts, which cannot exceed K), and the union of all buckets' key sets is exactly K keys. Total auxiliary space is O(K).
Derivation Progression
key_bucket locates the exact bucket node a key currently occupies in O(1) average time.
_insert_after either reuses bucket.next (already at the target count) or wires in a brand-new bucket node between two existing chain links.
head.next and tail.prev are always genuine non-empty extremes because emptied buckets are unlinked immediately, so no scanning is ever needed.
Variable Definitions
Number of distinct keys currently tracked by the structure
Total number of inc/dec/getMaxKey/getMinKey calls performed
Memory Architecture & Bounds
O(1) Non-recursive method calls
O(K) key_bucket map + O(K) bucket nodes across the chain, key sets summing to K entries
O(1) Scalar return values
Boundary Best / Worst Cases
for every operation.
standard hash table distribution.
amortized average ( in rare catastrophic hash collision degeneration).
Pointer Invariant Transition Progression
Senior SWE Deconstruction & Hardware Caveats
Triggers: "inc(key), dec(key), getMaxKey(), getMinKey() in O(1) time". Maintaining sorted frequencies with max and min lookups without sorting. A Senior SWE uses a Doubly Linked List of Frequency Buckets, where each bucket contains a set of keys sharing that frequency.
Up to operations. Strictly runtime per method call.
Pruning empty buckets: when the last key leaves a bucket, remove that bucket from the DLL in . When incrementing or decrementing, check if an adjacent bucket with target count already exists; if not, insert a new bucket in .
All O`one Data Structure (LeetCode 432)
Track a running count per string, and be able to report a string currently at the maximum count and one at the minimum count at any moment.
Build an AllOne class:
AllOne()starts with an empty structure.inc(String key)bumpskey's count by1, inserting it at count1if it wasn't tracked yet.dec(String key)lowerskey's count by1, removingkeyentirely once its count hits0.keyis always guaranteed to be present before adeccall.getMaxKey()returns any one key currently at the maximum count, or""if the structure is empty.getMinKey()returns any one key currently at the minimum count, or""if the structure is empty.
Every one of these four operations must run in O(1) average time.
Worked Examples
["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
[[], ["hello"], ["hello"], [], [], ["leet"], [], []][null, null, null, "hello", "hello", null, "hello", "leet"]⚖️Formal Constraints & Bounds
1 <= key.length <= 10key consists of lowercase English letters.It is guaranteed that for each call to dec, key is existing in the data structure.At most 5 * 104 calls will be made to inc, dec, getMaxKey, and getMinKey.
Core Insight
A count only ever moves by exactly +-1 per operation, so a doubly-linked chain of buckets sorted by count (each bucket a set of same-count keys), with sentinels whose immediate neighbors are always the true min/max, gives O(1) inc, dec, getMaxKey, and getMinKey without ever scanning or resorting.
Real-World Scenario & Production Applications
Real-time leaderboard "who's currently most/least active" queries, autocomplete frequency ranking, rate-limiter hot-key/cold-key tracking.
Subproblems & Recurrence Decomposition3 Phases
A hash map key_bucket resolves any key directly to the bucket node it currently occupies in O(1), so inc/dec never need to search for a key's current count.
self.key_bucket = {} # key -> the KeySet bucket node it sits inStep-by-Step Execution Trace Table
| Step | Operation | Key Counts Map | Doubly-Linked Bucket List (Head <-> ... <-> Tail) | getMaxKey() | getMinKey() |
|---|---|---|---|---|---|
| 1 | inc("hello") | {"hello": 1} | [Head] <-> [cnt=1: {"hello"}] <-> [Tail] | "hello" | "hello" |
| 2 | inc("hello") | {"hello": 2} | [Head] <-> [cnt=2: {"hello"}] <-> [Tail] (bucket 1 deleted) | "hello" | "hello" |
| 3 | inc("world") | {"hello": 2, "world": 1} | [Head] <-> [cnt=1: {"world"}] <-> [cnt=2: {"hello"}] <-> [Tail] | "hello" | "world" |
| 4 | inc("leet") | {"hello": 2, "world": 1, "leet": 1} | [Head] <-> [cnt=1: {"world", "leet"}] <-> [cnt=2: {"hello"}] <-> [Tail] | "hello" | "world" or "leet" |
| 5 | dec("hello") | {"hello": 1, "world": 1, "leet": 1} | [Head] <-> [cnt=1: {"world", "leet", "hello"}] <-> [Tail] (bucket 2 deleted) | any of 3 | any of 3 |
| 1 | self.key_bucket = {} # key -> bucket node it currently sits in |
| 2 | head <-> [count=1 bucket] <-> ... <-> [count=max bucket] <-> tail |
| 3 | inc(key) -> move key to bucket at count+1 (create if needed) |
| 4 | dec(key) -> move key to bucket at count-1 (remove key entirely if count was 1) |
| 5 | getMaxKey() -> any key in tail.prev's set |
| 6 | getMinKey() -> any key in head.next's set |
Conceptual Narrative
LFU Cache needed frequency buckets to evict the minimum. All Oone must report BOTH the minimum and maximum keys by count in O(1)time, while supportingO(1)` increments and decrements.
🧭 Conceptual Foundation & Pattern Intuition
All Oone Data Structure (LeetCode 432) demands four methods running in strict $O(1)$ time: inc(key), dec(key), getMaxKey(), and getMinKey()`.
Because keys must be queryable by their extremes, and counts only change by per operation, we maintain a Doubly-Linked List of Count Buckets:
- Each Bucket node in the DLL holds an integer
countand a hash setkeyscontaining all strings that currently have that exact count. - The DLL is strictly sorted by
count:head <-> Bucket(1) <-> Bucket(3) <-> Bucket(7) <-> tail. key_to_bucketmaps each string to its current Bucket node in the DLL.- When
inc(key)is called, the key moves from bucket to bucket . If an adjacent bucket with count already exists in the DLL, the key is added to it; otherwise, a new bucket node is spliced in immediately after bucket . - When a key departs a bucket and that bucket's
keysset becomes empty, that bucket is unlinked from the DLL immediately. getMaxKey()reads any key fromtail.prev.keys;getMinKey()reads any key fromhead.next.keys. Both run in strict time!
🪜 The Analogy: The Staircase of Contestant Leaderboard Podiums
Imagine a staircase where step holds a box containing all game contestants who currently have exactly points. When a contestant scores a point, they step out of box and step into box . If box doesn't exist on the staircase yet, a carpenter instantly slides a new wooden step into the staircase right above step . If box becomes completely empty, the carpenter removes that step so no gap remains. When the judges ask "Who is in first place?", they look at the very top step of the staircase. When they ask "Who is in last place?", they look at the very bottom step.
🔑 Why It Works
Because the list only ever contains non-empty buckets, head.next is guaranteed to be the global minimum and tail.prev is guaranteed to be the global maximum at all times.
Empty bucket cleanup omission: When the last key is removed from a count bucket, that bucket node must be unlinked from the doubly-linked list of buckets. Leaving empty buckets causes
getMaxKeyorgetMinKeyto return keys from empty buckets.Inserting new key at front without count 1 bucket: When inserting a brand-new key with count 1, if the bucket adjacent to
headdoes not have count 1, a new bucket node must be created and linked immediately afterhead.Returning empty string when data structure contains zero keys: Both
getMaxKey()andgetMinKey()must return""when the data structure contains no keys.
4-Phase Thought Process Model
Pattern Recognition Signals
The 10-second spot
Need O(1) increment/decrement of a per-key counter AND O(1) retrieval of any key currently at the max count AND any key at the min count.
Formulating the Predicate & Invariants
Turning intuition into a boolean rule
Invariant: a key's count changes by exactly +-1 per call, so it always moves to the immediately adjacent bucket; sentinels guarantee head.next is always the true min-count occupied bucket and tail.prev the true max-count occupied bucket.
Silent Failure Traps & Edge Cases
Where confident candidates still lose points
Failing to unlink a bucket the instant its key set becomes empty, which would let a stale empty bucket masquerade as the true min or max.
Reusing an existing adjacent bucket instead of blindly creating a new one when inc/dec lands on a count that already has an occupied bucket -- otherwise two buckets end up representing the same count.
Forgetting dec() must delete the key entirely (not move it to a count-0 bucket) when its count was exactly 1.
The 60-Second Interview Pitch
Say this out loud before you type a single line
"I keep a doubly-linked list of buckets strictly ordered by count, with head and tail sentinels, plus a hash map from key to the bucket node it currently occupies. inc(key) moves the key to the bucket at count+1 (creating it via
O(1)pointer splice if it doesn't already exist as the immediate neighbor), and dec(key) moves it to count-1 or deletes the key outright if its count was 1. Any bucket that loses its last key is unlinked immediately. Because the chain is always contiguous and sorted, getMaxKey reads any key from tail.prev's set and getMinKey reads any key from head.next's set, bothO(1), with all mutations alsoO(1)."
Complexity & Mathematical Proof
O(1) Average
Look at inc(key), dec(key), getMaxKey(), getMinKey(): key_bucket gives O(1) average lookup of the bucket a key currently sits in. _insert_after either reuses the immediately adjacent bucket (O(1) check) or splices a brand-new bucket node between two existing ones (O(1) pointer mutation). Removing a key from its old bucket's set and adding to the new bucket's set are O(1) average hash-set operations. getMaxKey/getMinKey simply peek at tail.prev/head.next's key set, O(1). Every operation is strictly O(1) average time.
O(K) Auxiliary
Look at memory allocation: K = number of distinct keys ever inserted and currently present. key_bucket holds O(K) entries. The bucket chain holds at most O(K) bucket nodes (bounded by the number of distinct counts, which cannot exceed K), and the union of all buckets' key sets is exactly K keys. Total auxiliary space is O(K).
Derivation Progression
key_bucket locates the exact bucket node a key currently occupies in O(1) average time.
_insert_after either reuses bucket.next (already at the target count) or wires in a brand-new bucket node between two existing chain links.
head.next and tail.prev are always genuine non-empty extremes because emptied buckets are unlinked immediately, so no scanning is ever needed.
Variable Definitions
Number of distinct keys currently tracked by the structure
Total number of inc/dec/getMaxKey/getMinKey calls performed
Memory Architecture & Bounds
O(1) Non-recursive method calls
O(K) key_bucket map + O(K) bucket nodes across the chain, key sets summing to K entries
O(1) Scalar return values
Boundary Best / Worst Cases
for every operation.
standard hash table distribution.
amortized average ( in rare catastrophic hash collision degeneration).
Pointer Invariant Transition Progression
Senior SWE Deconstruction & Hardware Caveats
Triggers: "inc(key), dec(key), getMaxKey(), getMinKey() in O(1) time". Maintaining sorted frequencies with max and min lookups without sorting. A Senior SWE uses a Doubly Linked List of Frequency Buckets, where each bucket contains a set of keys sharing that frequency.
Up to operations. Strictly runtime per method call.
Pruning empty buckets: when the last key leaves a bucket, remove that bucket from the DLL in . When incrementing or decrementing, check if an adjacent bucket with target count already exists; if not, insert a new bucket in .