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easyBasicsSometimes asked

Decimal to Binary

Basics · 6 min

Given a non-negative integer n, return its binary representation as a string (no leading zeros, "0" for zero).

In plain English

Repeatedly take n%2 for the next bit and n/2 to shrink, building the string from least to most significant.

Examples

Example 1

Input: n = 5
Output: "101"
Note: 4+1.

Example 2

Input: n = 0
Output: "0"
Note: Special case.

Example 3

Input: n = 8
Output: "1000"
Note: 2³.

Constraints

0 ≤ n ≤ 2^31 - 1

Clarifying questions1First question: Fixed width or minimal?Explore questions

Q1.Fixed width or minimal?
Minimal with no leading zeros; padding is a small change.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

If n == 0 return "0". Else, while n > 0, prepend `n % 2` and set `n /= 2`. (Built-ins: Integer.toBinaryString / bin(n).)

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you know the repeated-division base-conversion and handle 0/ordering.

Where candidates slip

Empty string for 0, or not reversing.

How to stand out

Generalise the scheme to any base.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Repeated division by 2

TimeO(log n)SpaceO(log n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What does n % 2 give you, and in which order?

2Hint 2Explore hint

Collect remainders, then reverse (or prepend).

3Hint 3Explore hint

What must you return for n = 0?

A · Core idea

Convert a non-negative integer to its binary string by repeated division by 2: each n % 2 is the next bit (least significant first) and n / 2 shrinks the number. Collect the bits and reverse.

Why bits come out LSB-first. The first remainder is the units (2⁰) bit, the next is the 2¹ bit, etc. — so you generate the binary digits in reverse order. Reverse the collected bits (or prepend each) to get the conventional MSB-first string.

Why handle 0 explicitly. For n = 0 the while (n > 0) loop never runs and would return an empty string; the convention is "0", so special-case it.

General base conversion. The same mod-then-divide scheme converts to any base (use base b instead of 2). Built-ins exist (Integer.toBinaryString, bin(n)) but the algorithm is the point.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

n = 5

5 % 2

1, n=2

Statebits: 1

Final result"101"

C · Code with teaching comments

solution.javaJava

Repeated division by 2 — optimal

2 teaching notes

1
public String toBinary(int n) {
2
    if (n == 0) return "0";
3
    StringBuilder sb = new StringBuilder();
4
    while (n > 0) {
5
        sb.append(n % 2);
6
        n /= 2;
7
    }
8
    // Bits were collected LSB-first, so reverse.
9
    return sb.reverse().toString();
10
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.10 lines

D · Why it works and complexityO(log n) timeO(log n) spaceOpen the invariant and proofExplore the proof

ComplexityO(log n) time, O(log n) space — one division per bit (≈ log₂ n), producing a string of that length.

E · Edge cases and pitfalls5n = 0 — Special-cased → "0".Explore edge cases

Edge cases handled

n = 0Special-cased → "0".
n = 8→ "1000".
n = 1→ "1".

Pitfalls

Returning empty string for n == 0.
Forgetting to reverse the collected bits.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic String toBinary(int n) { if (n == 0) return "0"; StringBuilder sb = new StringBuilder(); while (n > 0) { sb.append(n % 2);Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Returning empty string for n == 0.
Forgetting to reverse the collected bits.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Repeated mod/div base conversion

Code pattern to memorize

while n>0: bits.append(n%2); n//=2; reverse  (handle 0)

Rules to remember

1bit = n%2.
2n /= 2.
3Reverse at the end; handle 0.

Use this pattern when

Base conversion to base b

Interview takeaway

%2 and /2 convert decimal to binary, LSB first.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Repeated mod/div base conversion. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Binary to Decimal
The inverse via Horner's method.

Same pattern

easyBasicsSometimes asked

Decimal to Binary

Basics · 6 min

Given a non-negative integer n, return its binary representation as a string (no leading zeros, "0" for zero).

In plain English

Repeatedly take n%2 for the next bit and n/2 to shrink, building the string from least to most significant.

Examples

Example 1

Input: n = 5
Output: "101"
Note: 4+1.

Example 2

Input: n = 0
Output: "0"
Note: Special case.

Example 3

Input: n = 8
Output: "1000"
Note: 2³.

Constraints

0 ≤ n ≤ 2^31 - 1

Clarifying questions1First question: Fixed width or minimal?Explore questions

Q1.Fixed width or minimal?
Minimal with no leading zeros; padding is a small change.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

If n == 0 return "0". Else, while n > 0, prepend `n % 2` and set `n /= 2`. (Built-ins: Integer.toBinaryString / bin(n).)

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you know the repeated-division base-conversion and handle 0/ordering.

Where candidates slip

Empty string for 0, or not reversing.

How to stand out

Generalise the scheme to any base.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Repeated division by 2

TimeO(log n)SpaceO(log n)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

What does n % 2 give you, and in which order?

2Hint 2Explore hint

Collect remainders, then reverse (or prepend).

3Hint 3Explore hint

What must you return for n = 0?

A · Core idea

Convert a non-negative integer to its binary string by repeated division by 2: each n % 2 is the next bit (least significant first) and n / 2 shrinks the number. Collect the bits and reverse.

Why handle 0 explicitly. For n = 0 the while (n > 0) loop never runs and would return an empty string; the convention is "0", so special-case it.

General base conversion. The same mod-then-divide scheme converts to any base (use base b instead of 2). Built-ins exist (Integer.toBinaryString, bin(n)) but the algorithm is the point.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

n = 5

5 % 2

1, n=2

Statebits: 1

Final result"101"

C · Code with teaching comments

solution.javaJava

Repeated division by 2 — optimal

2 teaching notes

1
public String toBinary(int n) {
2
    if (n == 0) return "0";
3
    StringBuilder sb = new StringBuilder();
4
    while (n > 0) {
5
        sb.append(n % 2);
6
        n /= 2;
7
    }
8
    // Bits were collected LSB-first, so reverse.
9
    return sb.reverse().toString();
10
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.10 lines

D · Why it works and complexityO(log n) timeO(log n) spaceOpen the invariant and proofExplore the proof

ComplexityO(log n) time, O(log n) space — one division per bit (≈ log₂ n), producing a string of that length.

E · Edge cases and pitfalls5n = 0 — Special-cased → "0".Explore edge cases

Edge cases handled

n = 0Special-cased → "0".
n = 8→ "1000".
n = 1→ "1".

Pitfalls

Returning empty string for n == 0.
Forgetting to reverse the collected bits.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic String toBinary(int n) { if (n == 0) return "0"; StringBuilder sb = new StringBuilder(); while (n > 0) { sb.append(n % 2);Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Returning empty string for n == 0.
Forgetting to reverse the collected bits.

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Repeated mod/div base conversion

Code pattern to memorize

while n>0: bits.append(n%2); n//=2; reverse  (handle 0)

Rules to remember

1bit = n%2.
2n /= 2.
3Reverse at the end; handle 0.

Use this pattern when

Base conversion to base b

Interview takeaway

%2 and /2 convert decimal to binary, LSB first.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Repeated mod/div base conversion. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Binary to Decimal
The inverse via Horner's method.

Same pattern