LeetCode #381 — HARD

Insert Delete GetRandom O(1) - Duplicates allowed

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

The Problem

Problem Statement

RandomizedCollection is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element.

Implement the RandomizedCollection class:

RandomizedCollection() Initializes the empty RandomizedCollection object.
bool insert(int val) Inserts an item val into the multiset, even if the item is already present. Returns true if the item is not present, false otherwise.
bool remove(int val) Removes an item val from the multiset if present. Returns true if the item is present, false otherwise. Note that if val has multiple occurrences in the multiset, we only remove one of them.
int getRandom() Returns a random element from the current multiset of elements. The probability of each element being returned is linearly related to the number of the same values the multiset contains.

You must implement the functions of the class such that each function works on average O(1) time complexity.

Note: The test cases are generated such that getRandom will only be called if there is at least one item in the RandomizedCollection.

Example 1:

Input
["RandomizedCollection", "insert", "insert", "insert", "getRandom", "remove", "getRandom"]
[[], [1], [1], [2], [], [1], []]
Output
[null, true, false, true, 2, true, 1]

Explanation
RandomizedCollection randomizedCollection = new RandomizedCollection();
randomizedCollection.insert(1);   // return true since the collection does not contain 1.
                                  // Inserts 1 into the collection.
randomizedCollection.insert(1);   // return false since the collection contains 1.
                                  // Inserts another 1 into the collection. Collection now contains [1,1].
randomizedCollection.insert(2);   // return true since the collection does not contain 2.
                                  // Inserts 2 into the collection. Collection now contains [1,1,2].
randomizedCollection.getRandom(); // getRandom should:
                                  // - return 1 with probability 2/3, or
                                  // - return 2 with probability 1/3.
randomizedCollection.remove(1);   // return true since the collection contains 1.
                                  // Removes 1 from the collection. Collection now contains [1,2].
randomizedCollection.getRandom(); // getRandom should return 1 or 2, both equally likely.

Constraints:

-2³¹ <= val <= 2³¹ - 1
At most 2 * 10⁵ calls in total will be made to insert, remove, and getRandom.
There will be at least one element in the data structure when getRandom is called.

Step 01

Brute Force Baseline

Problem summary: RandomizedCollection is a data structure that contains a collection of numbers, possibly duplicates (i.e., a multiset). It should support inserting and removing specific elements and also reporting a random element. Implement the RandomizedCollection class: RandomizedCollection() Initializes the empty RandomizedCollection object. bool insert(int val) Inserts an item val into the multiset, even if the item is already present. Returns true if the item is not present, false otherwise. bool remove(int val) Removes an item val from the multiset if present. Returns true if the item is present, false otherwise. Note that if val has multiple occurrences in the multiset, we only remove one of them. int getRandom() Returns a random element from the current multiset of elements. The probability of each element being returned is linearly related to the number of the same values the multiset

Baseline thinking

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

Pattern signal: Array · Hash Map · Math · Design

Example 1

["RandomizedCollection","insert","insert","insert","getRandom","remove","getRandom"]
[[],[1],[1],[2],[],[1],[]]

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

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

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 #381: Insert Delete GetRandom O(1) - Duplicates allowed
class RandomizedCollection {
    private Map<Integer, Set<Integer>> m;
    private List<Integer> l;
    private Random rnd;

    /** Initialize your data structure here. */
    public RandomizedCollection() {
        m = new HashMap<>();
        l = new ArrayList<>();
        rnd = new Random();
    }

    /**
     * Inserts a value to the collection. Returns true if the collection did not already contain
     * the specified element.
     */
    public boolean insert(int val) {
        m.computeIfAbsent(val, k -> new HashSet<>()).add(l.size());
        l.add(val);
        return m.get(val).size() == 1;
    }

    /**
     * Removes a value from the collection. Returns true if the collection contained the specified
     * element.
     */
    public boolean remove(int val) {
        if (!m.containsKey(val)) {
            return false;
        }
        Set<Integer> idxSet = m.get(val);
        int idx = idxSet.iterator().next();
        int lastIdx = l.size() - 1;
        l.set(idx, l.get(lastIdx));
        idxSet.remove(idx);

        Set<Integer> lastIdxSet = m.get(l.get(lastIdx));
        lastIdxSet.remove(lastIdx);
        if (idx < lastIdx) {
            lastIdxSet.add(idx);
        }
        if (idxSet.isEmpty()) {
            m.remove(val);
        }
        l.remove(lastIdx);
        return true;
    }

    /** Get a random element from the collection. */
    public int getRandom() {
        int size = l.size();
        return size == 0 ? -1 : l.get(rnd.nextInt(size));
    }
}

/**
 * Your RandomizedCollection object will be instantiated and called as such:
 * RandomizedCollection obj = new RandomizedCollection();
 * boolean param_1 = obj.insert(val);
 * boolean param_2 = obj.remove(val);
 * int param_3 = obj.getRandom();
 */

// Accepted solution for LeetCode #381: Insert Delete GetRandom O(1) - Duplicates allowed
// Auto-generated Go example from java.
func exampleSolution() {
}
// Reference (java):
// // Accepted solution for LeetCode #381: Insert Delete GetRandom O(1) - Duplicates allowed
// class RandomizedCollection {
//     private Map<Integer, Set<Integer>> m;
//     private List<Integer> l;
//     private Random rnd;
// 
//     /** Initialize your data structure here. */
//     public RandomizedCollection() {
//         m = new HashMap<>();
//         l = new ArrayList<>();
//         rnd = new Random();
//     }
// 
//     /**
//      * Inserts a value to the collection. Returns true if the collection did not already contain
//      * the specified element.
//      */
//     public boolean insert(int val) {
//         m.computeIfAbsent(val, k -> new HashSet<>()).add(l.size());
//         l.add(val);
//         return m.get(val).size() == 1;
//     }
// 
//     /**
//      * Removes a value from the collection. Returns true if the collection contained the specified
//      * element.
//      */
//     public boolean remove(int val) {
//         if (!m.containsKey(val)) {
//             return false;
//         }
//         Set<Integer> idxSet = m.get(val);
//         int idx = idxSet.iterator().next();
//         int lastIdx = l.size() - 1;
//         l.set(idx, l.get(lastIdx));
//         idxSet.remove(idx);
// 
//         Set<Integer> lastIdxSet = m.get(l.get(lastIdx));
//         lastIdxSet.remove(lastIdx);
//         if (idx < lastIdx) {
//             lastIdxSet.add(idx);
//         }
//         if (idxSet.isEmpty()) {
//             m.remove(val);
//         }
//         l.remove(lastIdx);
//         return true;
//     }
// 
//     /** Get a random element from the collection. */
//     public int getRandom() {
//         int size = l.size();
//         return size == 0 ? -1 : l.get(rnd.nextInt(size));
//     }
// }
// 
// /**
//  * Your RandomizedCollection object will be instantiated and called as such:
//  * RandomizedCollection obj = new RandomizedCollection();
//  * boolean param_1 = obj.insert(val);
//  * boolean param_2 = obj.remove(val);
//  * int param_3 = obj.getRandom();
//  */

# Accepted solution for LeetCode #381: Insert Delete GetRandom O(1) - Duplicates allowed
class RandomizedCollection:
    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.m = {}
        self.l = []

    def insert(self, val: int) -> bool:
        """
        Inserts a value to the collection. Returns true if the collection did not already contain the specified element.
        """
        idx_set = self.m.get(val, set())
        idx_set.add(len(self.l))
        self.m[val] = idx_set
        self.l.append(val)
        return len(idx_set) == 1

    def remove(self, val: int) -> bool:
        """
        Removes a value from the collection. Returns true if the collection contained the specified element.
        """
        if val not in self.m:
            return False
        idx_set = self.m[val]
        idx = list(idx_set)[0]
        last_idx = len(self.l) - 1
        self.l[idx] = self.l[last_idx]
        idx_set.remove(idx)

        last_idx_set = self.m[self.l[last_idx]]
        if last_idx in last_idx_set:
            last_idx_set.remove(last_idx)
        if idx < last_idx:
            last_idx_set.add(idx)
        if not idx_set:
            self.m.pop(val)
        self.l.pop()
        return True

    def getRandom(self) -> int:
        """
        Get a random element from the collection.
        """
        return -1 if len(self.l) == 0 else random.choice(self.l)


# Your RandomizedCollection object will be instantiated and called as such:
# obj = RandomizedCollection()
# param_1 = obj.insert(val)
# param_2 = obj.remove(val)
# param_3 = obj.getRandom()

// Accepted solution for LeetCode #381: Insert Delete GetRandom O(1) - Duplicates allowed
use rand::prelude::*;
use std::collections::BinaryHeap;
use std::collections::HashMap;

#[derive(Debug)]
struct RandomizedCollection {
    rng: ThreadRng,
    indexes: HashMap<i32, BinaryHeap<usize>>,
    choices: Vec<i32>,
}

impl RandomizedCollection {
    fn new() -> Self {
        let rng = thread_rng();
        let indexes = HashMap::new();
        let choices = vec![];
        RandomizedCollection {
            rng,
            indexes,
            choices,
        }
    }

    fn insert(&mut self, val: i32) -> bool {
        let ids = self.indexes.entry(val).or_default();
        ids.push(self.choices.len());
        self.choices.push(val);
        ids.len() == 1
    }

    fn remove(&mut self, val: i32) -> bool {
        if let Some(ids) = self.indexes.get_mut(&val) {
            if let Some(index) = ids.pop() {
                let last_index = self.choices.len() - 1;
                let last_value = self.choices[last_index];
                if last_value != val {
                    let other_ids = self.indexes.get_mut(&last_value).unwrap();
                    other_ids.pop();
                    other_ids.push(index);
                    self.choices.swap(index, last_index);
                }
                self.choices.pop();
                true
            } else {
                false
            }
        } else {
            false
        }
    }

    fn get_random(&mut self) -> i32 {
        self.choices[self.rng.gen_range(0, self.choices.len())]
    }
}

#[test]
fn test() {
    let mut obj = RandomizedCollection::new();
    assert_eq!(obj.remove(0), false);
    // assert_eq!(obj.insert(1), false);
    // assert_eq!(obj.insert(2), true);
    // assert_eq!(obj.insert(2), false);
    // assert_eq!(obj.insert(2), false);
    // assert_eq!(obj.remove(1), true);
    // assert_eq!(obj.remove(1), true);
    // assert_eq!(obj.remove(2), true);
    // assert_eq!(obj.insert(1), true);
    // assert_eq!(obj.remove(2), true);
}

// Accepted solution for LeetCode #381: Insert Delete GetRandom O(1) - Duplicates allowed
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #381: Insert Delete GetRandom O(1) - Duplicates allowed
// class RandomizedCollection {
//     private Map<Integer, Set<Integer>> m;
//     private List<Integer> l;
//     private Random rnd;
// 
//     /** Initialize your data structure here. */
//     public RandomizedCollection() {
//         m = new HashMap<>();
//         l = new ArrayList<>();
//         rnd = new Random();
//     }
// 
//     /**
//      * Inserts a value to the collection. Returns true if the collection did not already contain
//      * the specified element.
//      */
//     public boolean insert(int val) {
//         m.computeIfAbsent(val, k -> new HashSet<>()).add(l.size());
//         l.add(val);
//         return m.get(val).size() == 1;
//     }
// 
//     /**
//      * Removes a value from the collection. Returns true if the collection contained the specified
//      * element.
//      */
//     public boolean remove(int val) {
//         if (!m.containsKey(val)) {
//             return false;
//         }
//         Set<Integer> idxSet = m.get(val);
//         int idx = idxSet.iterator().next();
//         int lastIdx = l.size() - 1;
//         l.set(idx, l.get(lastIdx));
//         idxSet.remove(idx);
// 
//         Set<Integer> lastIdxSet = m.get(l.get(lastIdx));
//         lastIdxSet.remove(lastIdx);
//         if (idx < lastIdx) {
//             lastIdxSet.add(idx);
//         }
//         if (idxSet.isEmpty()) {
//             m.remove(val);
//         }
//         l.remove(lastIdx);
//         return true;
//     }
// 
//     /** Get a random element from the collection. */
//     public int getRandom() {
//         int size = l.size();
//         return size == 0 ? -1 : l.get(rnd.nextInt(size));
//     }
// }
// 
// /**
//  * Your RandomizedCollection object will be instantiated and called as such:
//  * RandomizedCollection obj = new RandomizedCollection();
//  * boolean param_1 = obj.insert(val);
//  * boolean param_2 = obj.remove(val);
//  * int param_3 = obj.getRandom();
//  */

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.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.