LeetCode #432 — HARD

All O`one Data Structure

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts.

Implement the AllOne class:

  • AllOne() Initializes the object of the data structure.
  • inc(String key) Increments the count of the string key by 1. If key does not exist in the data structure, insert it with count 1.
  • dec(String key) Decrements the count of the string key by 1. If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed that key exists in the data structure before the decrement.
  • getMaxKey() Returns one of the keys with the maximal count. If no element exists, return an empty string "".
  • getMinKey() Returns one of the keys with the minimum count. If no element exists, return an empty string "".

Note that each function must run in O(1) average time complexity.

Example 1:

Input
["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
[[], ["hello"], ["hello"], [], [], ["leet"], [], []]
Output
[null, null, null, "hello", "hello", null, "hello", "leet"]

Explanation
AllOne allOne = new AllOne();
allOne.inc("hello");
allOne.inc("hello");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "hello"
allOne.inc("leet");
allOne.getMaxKey(); // return "hello"
allOne.getMinKey(); // return "leet"

Constraints:

  • 1 <= key.length <= 10
  • key 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.
Patterns Used
🔄In-Place Reversal→🏗Design Patterns→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts. Implement the AllOne class: AllOne() Initializes the object of the data structure. inc(String key) Increments the count of the string key by 1. If key does not exist in the data structure, insert it with count 1. dec(String key) Decrements the count of the string key by 1. If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed that key exists in the data structure before the decrement. getMaxKey() Returns one of the keys with the maximal count. If no element exists, return an empty string "". getMinKey() Returns one of the keys with the minimum count. If no element exists, return an empty string "". Note that each function must run in O(1) average time complexity.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Linked List · Design

Example 1

["AllOne","inc","inc","getMaxKey","getMinKey","inc","getMaxKey","getMinKey"]
[[],["hello"],["hello"],[],[],["leet"],[],[]]
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #432: All O`one Data Structure
class AllOne {
    Node root = new Node();
    Map<String, Node> nodes = new HashMap<>();

    public AllOne() {
        root.next = root;
        root.prev = root;
    }

    public void inc(String key) {
        if (!nodes.containsKey(key)) {
            if (root.next == root || root.next.cnt > 1) {
                nodes.put(key, root.insert(new Node(key, 1)));
            } else {
                root.next.keys.add(key);
                nodes.put(key, root.next);
            }
        } else {
            Node curr = nodes.get(key);
            Node next = curr.next;
            if (next == root || next.cnt > curr.cnt + 1) {
                nodes.put(key, curr.insert(new Node(key, curr.cnt + 1)));
            } else {
                next.keys.add(key);
                nodes.put(key, next);
            }
            curr.keys.remove(key);
            if (curr.keys.isEmpty()) {
                curr.remove();
            }
        }
    }

    public void dec(String key) {
        Node curr = nodes.get(key);
        if (curr.cnt == 1) {
            nodes.remove(key);
        } else {
            Node prev = curr.prev;
            if (prev == root || prev.cnt < curr.cnt - 1) {
                nodes.put(key, prev.insert(new Node(key, curr.cnt - 1)));
            } else {
                prev.keys.add(key);
                nodes.put(key, prev);
            }
        }

        curr.keys.remove(key);
        if (curr.keys.isEmpty()) {
            curr.remove();
        }
    }

    public String getMaxKey() {
        return root.prev.keys.iterator().next();
    }

    public String getMinKey() {
        return root.next.keys.iterator().next();
    }
}

class Node {
    Node prev;
    Node next;
    int cnt;
    Set<String> keys = new HashSet<>();

    public Node() {
        this("", 0);
    }

    public Node(String key, int cnt) {
        this.cnt = cnt;
        keys.add(key);
    }

    public Node insert(Node node) {
        node.prev = this;
        node.next = this.next;
        node.prev.next = node;
        node.next.prev = node;
        return node;
    }

    public void remove() {
        this.prev.next = this.next;
        this.next.prev = this.prev;
    }
}

/**
 * Your AllOne object will be instantiated and called as such:
 * AllOne obj = new AllOne();
 * obj.inc(key);
 * obj.dec(key);
 * String param_3 = obj.getMaxKey();
 * String param_4 = obj.getMinKey();
 */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

COPY TO ARRAY
O(n) time
O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS
O(n) time
O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Related Problems
#146LRU Cachemedium→#355Design Twittermedium→#460LFU Cachehard→#705Design HashSeteasy→#706Design HashMapeasy→