Generate Binary Numbers from 1 to N: 3 Python and C++ Methods

Binary numbers from 1 to N appear in three different placement contexts: aptitude number-system MCQs, data-structures coding rounds that ask for the queue BFS approach, and general coding screens where bin(i)[2:] solves the problem in one line.

Each context expects a different answer. Queue BFS demonstrates breadth-first traversal on a string tree. The bit-shift loop shows bitwise reasoning. The format-string one-liner signals standard-library awareness. All three are worth understanding before placement season.

Why This Problem Appears in Placement Tests

Placement aptitude sections use binary conversion to test positional-notation fluency. Wikipedia’s article on the binary numeral system defines it as base-2 positional notation using only the digits 0 and 1. The decimal value of any binary string is the sum of each digit multiplied by its corresponding power of 2.

Coding rounds add a structural dimension. Interviewers in data-structures rounds ask for the queue BFS approach specifically because it requires knowing that a queue enables BFS. Asking to “print binary numbers” is a quieter way of testing whether a candidate understands breadth-first traversal.

The same discipline, generating output in a fixed sequential order from a methodical structure, recurs in related placement coding problems. The equilibrium index problem processes each array index in order against its prefix sum. Finding symmetric pairs in an array requires methodical pair enumeration with a hash map lookup at each step. The structural habits transfer across these problems.

Expected Output: Binary Numbers from 1 to 8

Before writing any code, verify the expected output from first principles. Each row below derives the binary string directly from powers of 2, independent of any algorithm:

n	Binary	Verification
1	1	1×1 = 1
2	10	1×2 + 0×1 = 2
3	11	1×2 + 1×1 = 3
4	100	1×4 + 0×2 + 0×1 = 4
5	101	1×4 + 0×2 + 1×1 = 5
6	110	1×4 + 1×2 + 0×1 = 6
7	111	1×4 + 1×2 + 1×1 = 7
8	1000	1×8 + 0×4 + 0×2 + 0×1 = 8

Every implementation below must produce ['1', '10', '11', '100', '101', '110', '111', '1000'] for n = 8. Test against this table before submitting in any round.

Method 1: Queue-Based BFS

The queue BFS method treats binary strings as nodes in a tree. The root is "1". Each node’s left child is formed by appending "0" to the current string; the right child is formed by appending "1". BFS visits nodes level by level. All k-bit binary strings, one level of the tree, are numerically smaller than all (k+1)-bit strings, so BFS automatically produces results in ascending numeric order without sorting.

Python implementation

from collections import deque

def generate_binary_bfs(n):
    result = []
    q = deque(["1"])
    while len(result) < n:
        front = q.popleft()
        result.append(front)
        q.append(front + "0")
        q.append(front + "1")
    return result

print(generate_binary_bfs(8))
# ['1', '10', '11', '100', '101', '110', '111', '1000']

Queue state trace for n = 8

Walking through the first four dequeue operations shows the tree structure clearly:

• Start: queue = ["1"]
• Dequeue "1", enqueue "10" and "11". Result so far: ["1"]
• Dequeue "10", enqueue "100" and "101". Result so far: ["1", "10"]
• Dequeue "11", enqueue "110" and "111". Result so far: ["1", "10", "11"]
• Dequeue "100", enqueue "1000" and "1001". Result so far: ["1", "10", "11", "100"]

The pattern continues until the result list reaches length n. The string "1000" enters the queue at step 4 and is dequeued at step 8, confirming the correct position for n = 8.

Method 2: Bit-Shift Iterative Loop

The bit-shift approach extracts bits from each integer one at a time. The expression x & 1 isolates the least significant bit (0 or 1), and x >>= 1 right-shifts x by one position, which is integer division by 2. Repeating these two operations collects bits from right to left, so the collected list must be reversed to produce the correct most-significant-bit-first string.

Python implementation

def generate_binary_bitshift(n):
    result = []
    for i in range(1, n + 1):
        bits = []
        x = i
        while x > 0:
            bits.append(x & 1)
            x >>= 1
        result.append(''.join(str(b) for b in reversed(bits)))
    return result

print(generate_binary_bitshift(8))
# ['1', '10', '11', '100', '101', '110', '111', '1000']

Manual verification for i = 6:

• x = 6: 6 & 1 = 0, shift to x = 3. Bits: [0]
• x = 3: 3 & 1 = 1, shift to x = 1. Bits: [0, 1]
• x = 1: 1 & 1 = 1, shift to x = 0. Bits: [0, 1, 1]
• Reversed: [1, 1, 0] gives string "110". Cross-check: 1×4 + 1×2 + 0×1 = 6. Correct.

C++ implementation

#include <iostream>
#include <string>
#include <vector>
using namespace std;

string toBinary(int n) {
    string result = "";
    while (n > 0) {
        result = char('0' + (n & 1)) + result;
        n >>= 1;
    }
    return result;
}

vector<string> generateBinary(int n) {
    vector<string> result;
    for (int i = 1; i <= n; i++) {
        result.push_back(toBinary(i));
    }
    return result;
}

int main() {
    auto v = generateBinary(8);
    for (const auto& s : v) cout << s << " ";
    // Output: 1 10 11 100 101 110 111 1000
    return 0;
}

The C++ toBinary function prepends each extracted bit to the front of the result string, so no separate reversal step is needed. Verify for n = 6: the loop extracts bit 0, then bit 1, then bit 1, prepending each, and the final string is "110".

Method 3: Built-in Format Conversion

Python’s standard library handles the conversion directly. The built-in bin() function returns a string prefixed with "0b", so slicing from index 2 gives the raw binary digits. The format(i, 'b') call produces identical output without the prefix and is equally readable.

def generate_binary_format(n):
    return [bin(i)[2:] for i in range(1, n + 1)]
    # Equivalent: [format(i, 'b') for i in range(1, n + 1)]

print(generate_binary_format(8))
# ['1', '10', '11', '100', '101', '110', '111', '1000']

This is the fastest implementation to write. Most competitive programming judges accept it, and interviewers who allow standard-library calls expect it. When demonstrating structural knowledge of BFS or bitwise logic matters more than solution brevity, reach for Method 1 or Method 2 instead.

Complexity Comparison

Method	Time	Space (result)	Best suited for
Queue BFS	O(N log N)	O(N log N)	Data-structures interview rounds
Bit-shift loop	O(N log N)	O(N log N)	No-built-ins constraint
Format string	O(N log N)	O(N log N)	Standard library allowed

All three share the same asymptotic complexity. Each number i from 1 to N has floor(log2(i)) + 1 bits. Summing those bit-lengths across all N numbers gives O(N log N) total. The queue BFS method holds up to O(N) strings in the queue at any one point; the total output storage is O(N log N) for all three.

The queue BFS approach is the expected answer in data-structures rounds because it shows the candidate sees a string-generation problem as a tree traversal. That same prefix-expansion logic, where each node generates its children by appending a symbol, appears in how language models generate token sequences one token at a time. If you want to see where placement-level DS theory connects to working model code, TinkerLLM covers that ground for ₹299 through exercises that tie tree and sequence concepts to actual model behaviour.