LeetCode #2349 — MEDIUM

Design a Number Container System

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode

The Problem

Problem Statement

Design a number container system that can do the following:

Insert or Replace a number at the given index in the system.
Return the smallest index for the given number in the system.

Implement the NumberContainers class:

NumberContainers() Initializes the number container system.
void change(int index, int number) Fills the container at index with the number. If there is already a number at that index, replace it.
int find(int number) Returns the smallest index for the given number, or -1 if there is no index that is filled by number in the system.

Example 1:

Input
["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]
[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]
Output
[null, -1, null, null, null, null, 1, null, 2]

Explanation
NumberContainers nc = new NumberContainers();
nc.find(10); // There is no index that is filled with number 10. Therefore, we return -1.
nc.change(2, 10); // Your container at index 2 will be filled with number 10.
nc.change(1, 10); // Your container at index 1 will be filled with number 10.
nc.change(3, 10); // Your container at index 3 will be filled with number 10.
nc.change(5, 10); // Your container at index 5 will be filled with number 10.
nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.
nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20. 
nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.

Constraints:

1 <= index, number <= 10⁹
At most 10⁵ calls will be made in total to change and find.

Step 01

Brute Force Baseline

Problem summary: Design a number container system that can do the following: Insert or Replace a number at the given index in the system. Return the smallest index for the given number in the system. Implement the NumberContainers class: NumberContainers() Initializes the number container system. void change(int index, int number) Fills the container at index with the number. If there is already a number at that index, replace it. int find(int number) Returns the smallest index for the given number, or -1 if there is no index that is filled by number in the system.

Baseline thinking

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

Pattern signal: Hash Map · Design · Segment Tree

Example 1

["NumberContainers","find","change","change","change","change","find","change","find"]
[[],[10],[2,10],[1,10],[3,10],[5,10],[10],[1,20],[10]]

Core Insight

What unlocks the optimal approach

Use a hash table to efficiently map each number to all of its indices in the container and to map each index to their current number.
In addition, you can use ordered set to store all of the indices for each number to solve the find method. Do not forget to update the ordered set according to the change method.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Upper-end input sizes

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 #2349: Design a Number Container System
class NumberContainers {
    private Map<Integer, Integer> d = new HashMap<>();
    private Map<Integer, TreeSet<Integer>> g = new HashMap<>();

    public NumberContainers() {
    }

    public void change(int index, int number) {
        if (d.containsKey(index)) {
            int oldNumber = d.get(index);
            g.get(oldNumber).remove(index);
        }
        d.put(index, number);
        g.computeIfAbsent(number, k -> new TreeSet<>()).add(index);
    }

    public int find(int number) {
        var ids = g.get(number);
        return ids == null || ids.isEmpty() ? -1 : ids.first();
    }
}

/**
 * Your NumberContainers object will be instantiated and called as such:
 * NumberContainers obj = new NumberContainers();
 * obj.change(index,number);
 * int param_2 = obj.find(number);
 */

// Accepted solution for LeetCode #2349: Design a Number Container System
type NumberContainers struct {
	d map[int]int
	g map[int]*redblacktree.Tree
}

func Constructor() NumberContainers {
	return NumberContainers{map[int]int{}, map[int]*redblacktree.Tree{}}
}

func (this *NumberContainers) Change(index int, number int) {
	if oldNumber, ok := this.d[index]; ok {
		this.g[oldNumber].Remove(index)
	}
	this.d[index] = number
	if _, ok := this.g[number]; !ok {
		this.g[number] = redblacktree.NewWithIntComparator()
	}
	this.g[number].Put(index, nil)
}

func (this *NumberContainers) Find(number int) int {
	if ids, ok := this.g[number]; ok && ids.Size() > 0 {
		return ids.Left().Key.(int)
	}
	return -1
}

/**
 * Your NumberContainers object will be instantiated and called as such:
 * obj := Constructor();
 * obj.Change(index,number);
 * param_2 := obj.Find(number);
 */

# Accepted solution for LeetCode #2349: Design a Number Container System
class NumberContainers:
    def __init__(self):
        self.d = {}
        self.g = defaultdict(SortedSet)

    def change(self, index: int, number: int) -> None:
        if index in self.d:
            old_number = self.d[index]
            self.g[old_number].remove(index)
        self.d[index] = number
        self.g[number].add(index)

    def find(self, number: int) -> int:
        ids = self.g[number]
        return ids[0] if ids else -1


# Your NumberContainers object will be instantiated and called as such:
# obj = NumberContainers()
# obj.change(index,number)
# param_2 = obj.find(number)

// Accepted solution for LeetCode #2349: Design a Number Container System
use std::collections::{BTreeSet, HashMap};

struct NumberContainers {
    d: HashMap<i32, i32>,
    g: HashMap<i32, BTreeSet<i32>>,
}

/**
 * `&self` means the method takes an immutable reference.
 * If you need a mutable reference, change it to `&mut self` instead.
 */
impl NumberContainers {
    fn new() -> Self {
        Self {
            d: HashMap::new(),
            g: HashMap::new(),
        }
    }

    fn change(&mut self, index: i32, number: i32) {
        if let Some(&old_number) = self.d.get(&index) {
            if let Some(set) = self.g.get_mut(&old_number) {
                set.remove(&index);
            }
        }
        self.d.insert(index, number);
        self.g
            .entry(number)
            .or_insert_with(BTreeSet::new)
            .insert(index);
    }

    fn find(&self, number: i32) -> i32 {
        match self.g.get(&number) {
            Some(set) if !set.is_empty() => *set.iter().next().unwrap(),
            _ => -1,
        }
    }
}

// Accepted solution for LeetCode #2349: Design a Number Container System
class NumberContainers {
    private d = new Map<number, number>();
    private g = new Map<number, TreeSet<number>>();
    constructor() {}

    change(index: number, number: number): void {
        if (this.d.has(index)) {
            const oldNumber = this.d.get(index)!;
            this.g.get(oldNumber)!.delete(index);
            if (!this.g.get(oldNumber)!.size()) {
                this.g.delete(oldNumber);
            }
        }
        this.d.set(index, number);
        if (!this.g.has(number)) {
            this.g.set(number, new TreeSet());
        }
        this.g.get(number)!.add(index);
    }

    find(number: number): number {
        return this.g.has(number) ? this.g.get(number)!.first()! : -1;
    }
}

type Compare<T> = (lhs: T, rhs: T) => number;

class RBTreeNode<T = number> {
    data: T;
    count: number;
    left: RBTreeNode<T> | null;
    right: RBTreeNode<T> | null;
    parent: RBTreeNode<T> | null;
    color: number;
    constructor(data: T) {
        this.data = data;
        this.left = this.right = this.parent = null;
        this.color = 0;
        this.count = 1;
    }

    sibling(): RBTreeNode<T> | null {
        if (!this.parent) return null; // sibling null if no parent
        return this.isOnLeft() ? this.parent.right : this.parent.left;
    }

    isOnLeft(): boolean {
        return this === this.parent!.left;
    }

    hasRedChild(): boolean {
        return (
            Boolean(this.left && this.left.color === 0) ||
            Boolean(this.right && this.right.color === 0)
        );
    }
}

class RBTree<T> {
    root: RBTreeNode<T> | null;
    lt: (l: T, r: T) => boolean;
    constructor(compare: Compare<T> = (l: T, r: T) => (l < r ? -1 : l > r ? 1 : 0)) {
        this.root = null;
        this.lt = (l: T, r: T) => compare(l, r) < 0;
    }

    rotateLeft(pt: RBTreeNode<T>): void {
        const right = pt.right!;
        pt.right = right.left;

        if (pt.right) pt.right.parent = pt;
        right.parent = pt.parent;

        if (!pt.parent) this.root = right;
        else if (pt === pt.parent.left) pt.parent.left = right;
        else pt.parent.right = right;

        right.left = pt;
        pt.parent = right;
    }

    rotateRight(pt: RBTreeNode<T>): void {
        const left = pt.left!;
        pt.left = left.right;

        if (pt.left) pt.left.parent = pt;
        left.parent = pt.parent;

        if (!pt.parent) this.root = left;
        else if (pt === pt.parent.left) pt.parent.left = left;
        else pt.parent.right = left;

        left.right = pt;
        pt.parent = left;
    }

    swapColor(p1: RBTreeNode<T>, p2: RBTreeNode<T>): void {
        const tmp = p1.color;
        p1.color = p2.color;
        p2.color = tmp;
    }

    swapData(p1: RBTreeNode<T>, p2: RBTreeNode<T>): void {
        const tmp = p1.data;
        p1.data = p2.data;
        p2.data = tmp;
    }

    fixAfterInsert(pt: RBTreeNode<T>): void {
        let parent = null;
        let grandParent = null;

        while (pt !== this.root && pt.color !== 1 && pt.parent?.color === 0) {
            parent = pt.parent;
            grandParent = pt.parent.parent;

            /*  Case : A
                Parent of pt is left child of Grand-parent of pt */
            if (parent === grandParent?.left) {
                const uncle = grandParent.right;

                /* Case : 1
                   The uncle of pt is also red
                   Only Recoloring required */
                if (uncle && uncle.color === 0) {
                    grandParent.color = 0;
                    parent.color = 1;
                    uncle.color = 1;
                    pt = grandParent;
                } else {
                    /* Case : 2
                       pt is right child of its parent
                       Left-rotation required */
                    if (pt === parent.right) {
                        this.rotateLeft(parent);
                        pt = parent;
                        parent = pt.parent;
                    }

                    /* Case : 3
                       pt is left child of its parent
                       Right-rotation required */
                    this.rotateRight(grandParent);
                    this.swapColor(parent!, grandParent);
                    pt = parent!;
                }
            } else {
                /* Case : B
               Parent of pt is right child of Grand-parent of pt */
                const uncle = grandParent!.left;

                /*  Case : 1
                    The uncle of pt is also red
                    Only Recoloring required */
                if (uncle != null && uncle.color === 0) {
                    grandParent!.color = 0;
                    parent.color = 1;
                    uncle.color = 1;
                    pt = grandParent!;
                } else {
                    /* Case : 2
                       pt is left child of its parent
                       Right-rotation required */
                    if (pt === parent.left) {
                        this.rotateRight(parent);
                        pt = parent;
                        parent = pt.parent;
                    }

                    /* Case : 3
                       pt is right child of its parent
                       Left-rotation required */
                    this.rotateLeft(grandParent!);
                    this.swapColor(parent!, grandParent!);
                    pt = parent!;
                }
            }
        }
        this.root!.color = 1;
    }

    delete(val: T): boolean {
        const node = this.find(val);
        if (!node) return false;
        node.count--;
        if (!node.count) this.deleteNode(node);
        return true;
    }

    deleteAll(val: T): boolean {
        const node = this.find(val);
        if (!node) return false;
        this.deleteNode(node);
        return true;
    }

    deleteNode(v: RBTreeNode<T>): void {
        const u = BSTreplace(v);

        // True when u and v are both black
        const uvBlack = (u === null || u.color === 1) && v.color === 1;
        const parent = v.parent!;

        if (!u) {
            // u is null therefore v is leaf
            if (v === this.root) this.root = null;
            // v is root, making root null
            else {
                if (uvBlack) {
                    // u and v both black
                    // v is leaf, fix double black at v
                    this.fixDoubleBlack(v);
                } else {
                    // u or v is red
                    if (v.sibling()) {
                        // sibling is not null, make it red"
                        v.sibling()!.color = 0;
                    }
                }
                // delete v from the tree
                if (v.isOnLeft()) parent.left = null;
                else parent.right = null;
            }
            return;
        }

        if (!v.left || !v.right) {
            // v has 1 child
            if (v === this.root) {
                // v is root, assign the value of u to v, and delete u
                v.data = u.data;
                v.left = v.right = null;
            } else {
                // Detach v from tree and move u up
                if (v.isOnLeft()) parent.left = u;
                else parent.right = u;
                u.parent = parent;
                if (uvBlack) this.fixDoubleBlack(u);
                // u and v both black, fix double black at u
                else u.color = 1; // u or v red, color u black
            }
            return;
        }

        // v has 2 children, swap data with successor and recurse
        this.swapData(u, v);
        this.deleteNode(u);

        // find node that replaces a deleted node in BST
        function BSTreplace(x: RBTreeNode<T>): RBTreeNode<T> | null {
            // when node have 2 children
            if (x.left && x.right) return successor(x.right);
            // when leaf
            if (!x.left && !x.right) return null;
            // when single child
            return x.left ?? x.right;
        }
        // find node that do not have a left child
        // in the subtree of the given node
        function successor(x: RBTreeNode<T>): RBTreeNode<T> {
            let temp = x;
            while (temp.left) temp = temp.left;
            return temp;
        }
    }

    fixDoubleBlack(x: RBTreeNode<T>): void {
        if (x === this.root) return; // Reached root

        const sibling = x.sibling();
        const parent = x.parent!;
        if (!sibling) {
            // No sibiling, double black pushed up
            this.fixDoubleBlack(parent);
        } else {
            if (sibling.color === 0) {
                // Sibling red
                parent.color = 0;
                sibling.color = 1;
                if (sibling.isOnLeft()) this.rotateRight(parent);
                // left case
                else this.rotateLeft(parent); // right case
                this.fixDoubleBlack(x);
            } else {
                // Sibling black
                if (sibling.hasRedChild()) {
                    // at least 1 red children
                    if (sibling.left && sibling.left.color === 0) {
                        if (sibling.isOnLeft()) {
                            // left left
                            sibling.left.color = sibling.color;
                            sibling.color = parent.color;
                            this.rotateRight(parent);
                        } else {
                            // right left
                            sibling.left.color = parent.color;
                            this.rotateRight(sibling);
                            this.rotateLeft(parent);
                        }
                    } else {
                        if (sibling.isOnLeft()) {
                            // left right
                            sibling.right!.color = parent.color;
                            this.rotateLeft(sibling);
                            this.rotateRight(parent);
                        } else {
                            // right right
                            sibling.right!.color = sibling.color;
                            sibling.color = parent.color;
                            this.rotateLeft(parent);
                        }
                    }
                    parent.color = 1;
                } else {
                    // 2 black children
                    sibling.color = 0;
                    if (parent.color === 1) this.fixDoubleBlack(parent);
                    else parent.color = 1;
                }
            }
        }
    }

    insert(data: T): boolean {
        // search for a position to insert
        let parent = this.root;
        while (parent) {
            if (this.lt(data, parent.data)) {
                if (!parent.left) break;
                else parent = parent.left;
            } else if (this.lt(parent.data, data)) {
                if (!parent.right) break;
                else parent = parent.right;
            } else break;
        }

        // insert node into parent
        const node = new RBTreeNode(data);
        if (!parent) this.root = node;
        else if (this.lt(node.data, parent.data)) parent.left = node;
        else if (this.lt(parent.data, node.data)) parent.right = node;
        else {
            parent.count++;
            return false;
        }
        node.parent = parent;
        this.fixAfterInsert(node);
        return true;
    }

    find(data: T): RBTreeNode<T> | null {
        let p = this.root;
        while (p) {
            if (this.lt(data, p.data)) {
                p = p.left;
            } else if (this.lt(p.data, data)) {
                p = p.right;
            } else break;
        }
        return p ?? null;
    }

    *inOrder(root: RBTreeNode<T> = this.root!): Generator<T, undefined, void> {
        if (!root) return;
        for (const v of this.inOrder(root.left!)) yield v;
        yield root.data;
        for (const v of this.inOrder(root.right!)) yield v;
    }

    *reverseInOrder(root: RBTreeNode<T> = this.root!): Generator<T, undefined, void> {
        if (!root) return;
        for (const v of this.reverseInOrder(root.right!)) yield v;
        yield root.data;
        for (const v of this.reverseInOrder(root.left!)) yield v;
    }
}

class TreeSet<T = number> {
    _size: number;
    tree: RBTree<T>;
    compare: Compare<T>;
    constructor(
        collection: T[] | Compare<T> = [],
        compare: Compare<T> = (l: T, r: T) => (l < r ? -1 : l > r ? 1 : 0),
    ) {
        if (typeof collection === 'function') {
            compare = collection;
            collection = [];
        }
        this._size = 0;
        this.compare = compare;
        this.tree = new RBTree(compare);
        for (const val of collection) this.add(val);
    }

    size(): number {
        return this._size;
    }

    has(val: T): boolean {
        return !!this.tree.find(val);
    }

    add(val: T): boolean {
        const successful = this.tree.insert(val);
        this._size += successful ? 1 : 0;
        return successful;
    }

    delete(val: T): boolean {
        const deleted = this.tree.deleteAll(val);
        this._size -= deleted ? 1 : 0;
        return deleted;
    }

    ceil(val: T): T | undefined {
        let p = this.tree.root;
        let higher = null;
        while (p) {
            if (this.compare(p.data, val) >= 0) {
                higher = p;
                p = p.left;
            } else {
                p = p.right;
            }
        }
        return higher?.data;
    }

    floor(val: T): T | undefined {
        let p = this.tree.root;
        let lower = null;
        while (p) {
            if (this.compare(val, p.data) >= 0) {
                lower = p;
                p = p.right;
            } else {
                p = p.left;
            }
        }
        return lower?.data;
    }

    higher(val: T): T | undefined {
        let p = this.tree.root;
        let higher = null;
        while (p) {
            if (this.compare(val, p.data) < 0) {
                higher = p;
                p = p.left;
            } else {
                p = p.right;
            }
        }
        return higher?.data;
    }

    lower(val: T): T | undefined {
        let p = this.tree.root;
        let lower = null;
        while (p) {
            if (this.compare(p.data, val) < 0) {
                lower = p;
                p = p.right;
            } else {
                p = p.left;
            }
        }
        return lower?.data;
    }

    first(): T | undefined {
        return this.tree.inOrder().next().value;
    }

    last(): T | undefined {
        return this.tree.reverseInOrder().next().value;
    }

    shift(): T | undefined {
        const first = this.first();
        if (first === undefined) return undefined;
        this.delete(first);
        return first;
    }

    pop(): T | undefined {
        const last = this.last();
        if (last === undefined) return undefined;
        this.delete(last);
        return last;
    }

    *[Symbol.iterator](): Generator<T, void, void> {
        for (const val of this.values()) yield val;
    }

    *keys(): Generator<T, void, void> {
        for (const val of this.values()) yield val;
    }

    *values(): Generator<T, undefined, void> {
        for (const val of this.tree.inOrder()) yield val;
        return undefined;
    }

    /**
     * Return a generator for reverse order traversing the set
     */
    *rvalues(): Generator<T, undefined, void> {
        for (const val of this.tree.reverseInOrder()) yield val;
        return undefined;
    }
}

/**
 * Your NumberContainers object will be instantiated and called as such:
 * var obj = new NumberContainers()
 * obj.change(index,number)
 * var param_2 = obj.find(number)
 */

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

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.