💻 Binary Number Generation with Queue

Explore binary generation using a queue — choose mode and watch the queue in action.

💻 Sneha's Binary Number Quest

🎯 The Challenge: (Problem Statement)

Sneha wants to generate the binary representations for the first $n$ positive integers ($1, 2, 3, \ldots, n$) using an efficient queue-based approach instead of standard arithmetic conversion.

📋 Specification:

  • Generate binary numbers using a **Queue (FIFO)** structure.
  • Mode A: generate binary representations for decimals $1$ up to $n$ (inclusive, stopping when decimal value $>$ $n$).
  • Mode B: generate the first $n$ binary strings ($1, 10, 11, 100, \ldots$) regardless of their decimal value.

Key Concept: Breadth-First Search (BFS) Tree Traversal

The sequence of binary numbers ($1, 10, 11, 100, 101, \ldots$) is simply a Breadth-First Search traversal of an implicit tree where each node $S$ has two children: $S + '0'$ and $S + '1'$. A queue is the perfect data structure for BFS.

Example: Decimal limit $n = 5$ (Mode A)

1
→
1
2
→
10
3
→
11
4
→
100
5
→
101

🔄 Queue-Based Strategy

Algorithm Steps

  1. **Initialize:** Start the queue with the first binary number, "1".
  2. **Loop:** While the stopping condition is not met (based on chosen mode):
  3. **Dequeue:** Remove the front string $S$ from the queue.
  4. **Output:** Record $S$ as the next result.
  5. **Enqueue Children:** Create two new strings: $S_0 = S + '0'$ and $S_1 = S + '1'$.
  6. **Filter (Mode A only):** If using **Decimal limit mode** ($1..n$), convert $S_0$ and $S_1$ to decimal. Only enqueue strings whose decimal value is $\le n$.
  7. **Enqueue (Mode B only):** If using **Count mode** (first $n$ binaries), always enqueue $S_0$ and $S_1$.

Stopping Rules

  • **Decimal limit (Mode A):** Stop when the queue is empty, or when the decimal value of the element being processed exceeds $n$. (The most robust stop is when $S_0$'s decimal value $> n$).
  • **First n binaries (Mode B):** Stop when the total number of recorded outputs reaches $n$.

Queue approach (BFS)

Time: $O(n \cdot L)$, where $L$ is the length of the binary string (approx. $\log_{2} n$). $O(n)$ additions/dequeues + string copy costs.

Naive Conversion (Iterative)

Convert each integer $i=1..n$ to binary: $O(\sum_{i=1}^{n} \log i) \approx O(n \log n)$ total time.

🔍 Step-by-Step Queue Demo

Ready. Use controls below to start demo.

Queue State (FIFO)

Generated Binary Numbers (Count: 0)

Click Start to run demo

Progress tracker

1. Initialize queue with "1"
2. Dequeue front element $S$
3. Check stopping condition/count
4. Print $S$ and increment count
5. Enqueue $S+'0'$ and $S+'1'$ (after filtering if in Decimal Mode)
6. Repeat until done

🎮 Interactive Practice

Enter n and press Generate

Examples to Consider

Minimal Case (n = 1)
Decimal Mode: "1" | Count Mode: "1"
n = 7
Decimal Mode: "1".."111" | Count Mode: first 7 binaries
n = 10
Decimal Mode: "1".."1010" | Count Mode: first 10 binaries

📊 Analysis & Optimization

Time Complexity

$O(n \cdot L)$

Where $L$ is the length of the $n$-th binary string ($L \approx \log_{2} n$). Dominated by string concatenations/copies.

Space Complexity

$O(n \cdot L)$

Queue stores approximately $O(n)$ binary strings, each of length $O(L)$.

Optimization Summary

The queue-based approach is often favored in interview settings due to its elegant use of **BFS** to naturally generate the numbers in increasing order.

  • **Trade-off:** If arithmetic conversion is available, $O(n \log n)$ total time might be faster in practice than $O(n \cdot L)$ due to overheads of string manipulation in the queue approach.
  • **Large $n$:** For very large $n$, if the problem required the sequence of numbers but not the full strings, we could store only integers in the queue and calculate the next two numbers ($2i$ and $2i+1$). However, this specific problem requires the *binary string* output.