LeetCode #1656 — EASY

Design an Ordered Stream

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

There is a stream of n (idKey, value) pairs arriving in an arbitrary order, where idKey is an integer between 1 and n and value is a string. No two pairs have the same id.

Design a stream that returns the values in increasing order of their IDs by returning a chunk (list) of values after each insertion. The concatenation of all the chunks should result in a list of the sorted values.

Implement the OrderedStream class:

OrderedStream(int n) Constructs the stream to take n values.
String[] insert(int idKey, String value) Inserts the pair (idKey, value) into the stream, then returns the largest possible chunk of currently inserted values that appear next in the order.

Example:

Input
["OrderedStream", "insert", "insert", "insert", "insert", "insert"]
[[5], [3, "ccccc"], [1, "aaaaa"], [2, "bbbbb"], [5, "eeeee"], [4, "ddddd"]]
Output
[null, [], ["aaaaa"], ["bbbbb", "ccccc"], [], ["ddddd", "eeeee"]]

Explanation
// Note that the values ordered by ID is ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"].
OrderedStream os = new OrderedStream(5);
os.insert(3, "ccccc"); // Inserts (3, "ccccc"), returns [].
os.insert(1, "aaaaa"); // Inserts (1, "aaaaa"), returns ["aaaaa"].
os.insert(2, "bbbbb"); // Inserts (2, "bbbbb"), returns ["bbbbb", "ccccc"].
os.insert(5, "eeeee"); // Inserts (5, "eeeee"), returns [].
os.insert(4, "ddddd"); // Inserts (4, "ddddd"), returns ["ddddd", "eeeee"].
// Concatentating all the chunks returned:
// [] + ["aaaaa"] + ["bbbbb", "ccccc"] + [] + ["ddddd", "eeeee"] = ["aaaaa", "bbbbb", "ccccc", "ddddd", "eeeee"]
// The resulting order is the same as the order above.

Constraints:

1 <= n <= 1000
1 <= id <= n
value.length == 5
value consists only of lowercase letters.
Each call to insert will have a unique id.
Exactly n calls will be made to insert.

Step 01

Brute Force Baseline

Problem summary: There is a stream of n (idKey, value) pairs arriving in an arbitrary order, where idKey is an integer between 1 and n and value is a string. No two pairs have the same id. Design a stream that returns the values in increasing order of their IDs by returning a chunk (list) of values after each insertion. The concatenation of all the chunks should result in a list of the sorted values. Implement the OrderedStream class: OrderedStream(int n) Constructs the stream to take n values. String[] insert(int idKey, String value) Inserts the pair (idKey, value) into the stream, then returns the largest possible chunk of currently inserted values that appear next in the order.

Baseline thinking

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

Pattern signal: Array · Hash Map · Design

Example 1

["OrderedStream","insert","insert","insert","insert","insert"]
[[5],[3,"ccccc"],[1,"aaaaa"],[2,"bbbbb"],[5,"eeeee"],[4,"ddddd"]]

Core Insight

What unlocks the optimal approach

Maintain the next id that should be outputted.
Maintain the ids that were inserted in the stream.
Per each insert, make a loop where you check if the id that has the turn has been inserted, and if so increment the id that has the turn and continue the loop, else break.

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 #1656: Design an Ordered Stream
class OrderedStream {
    private int ptr = 1;
    private String[] data;

    public OrderedStream(int n) {
        data = new String[n + 1];
    }

    public List<String> insert(int idKey, String value) {
        data[idKey] = value;
        List<String> ans = new ArrayList<>();
        while (ptr < data.length && data[ptr] != null) {
            ans.add(data[ptr++]);
        }
        return ans;
    }
}

/**
 * Your OrderedStream object will be instantiated and called as such:
 * OrderedStream obj = new OrderedStream(n);
 * List<String> param_1 = obj.insert(idKey,value);
 */

// Accepted solution for LeetCode #1656: Design an Ordered Stream
type OrderedStream struct {
	ptr  int
	data []string
}

func Constructor(n int) OrderedStream {
	return OrderedStream{
		ptr:  1,
		data: make([]string, n+1),
	}
}

func (this *OrderedStream) Insert(idKey int, value string) []string {
	this.data[idKey] = value
	var ans []string
	for this.ptr < len(this.data) && this.data[this.ptr] != "" {
		ans = append(ans, this.data[this.ptr])
		this.ptr++
	}
	return ans
}

/**
 * Your OrderedStream object will be instantiated and called as such:
 * obj := Constructor(n);
 * param_1 := obj.Insert(idKey,value);
 */

# Accepted solution for LeetCode #1656: Design an Ordered Stream
class OrderedStream:
    def __init__(self, n: int):
        self.ptr = 1
        self.data = [None] * (n + 1)

    def insert(self, idKey: int, value: str) -> List[str]:
        self.data[idKey] = value
        ans = []
        while self.ptr < len(self.data) and self.data[self.ptr]:
            ans.append(self.data[self.ptr])
            self.ptr += 1
        return ans


# Your OrderedStream object will be instantiated and called as such:
# obj = OrderedStream(n)
# param_1 = obj.insert(idKey,value)

// Accepted solution for LeetCode #1656: Design an Ordered Stream
struct OrderedStream {
    ptr: usize,
    data: Vec<Option<String>>,
}

impl OrderedStream {
    fn new(n: i32) -> Self {
        OrderedStream {
            ptr: 1,
            data: vec![None; (n + 1) as usize],
        }
    }

    fn insert(&mut self, id_key: i32, value: String) -> Vec<String> {
        self.data[id_key as usize] = Some(value);
        let mut ans = Vec::new();
        while self.ptr < self.data.len() && self.data[self.ptr].is_some() {
            ans.push(self.data[self.ptr].take().unwrap());
            self.ptr += 1;
        }
        ans
    }
}

// Accepted solution for LeetCode #1656: Design an Ordered Stream
class OrderedStream {
    private ptr: number;
    private data: string[];

    constructor(n: number) {
        this.ptr = 1;
        this.data = Array(n + 1);
    }

    insert(idKey: number, value: string): string[] {
        this.data[idKey] = value;
        const ans: string[] = [];
        while (this.data[this.ptr]) {
            ans.push(this.data[this.ptr++]);
        }
        return ans;
    }
}

/**
 * Your OrderedStream object will be instantiated and called as such:
 * var obj = new OrderedStream(n)
 * var param_1 = obj.insert(idKey,value)
 */

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(n)

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.