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easyBasicsSometimes asked

Average of an Array

Basics · 6 min

Given a non-empty integer array nums, return the average (mean) of its elements as a floating-point number.

In plain English

Sum the elements, then divide by the count. Do the division in floating point to keep the fractional part.

Examples

Example 1

Input: nums = [1, 2, 3, 4]
Output: 2.5
Note: 10 / 4.

Example 2

Input: nums = [5, 5, 5]
Output: 5.0
Note: Mean is 5.

Example 3

Input: nums = [2, 3]
Output: 2.5
Note: 5 / 2.

Constraints

1 ≤ nums.length ≤ 10^5
-10^9 ≤ nums[i] ≤ 10^9

Clarifying questions2First question: Should the result be a float or rounded?Explore questions

Q1.Should the result be a float or rounded?
I'll return a double mean; rounding is a separate, easy step if needed.
Q2.Can the array be empty?
If so I need an agreed return for division by zero.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Sum with a `long`, then divide by `nums.length` as a `double` so you don't lose the fraction.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you recognise integer-division truncation and cast at the right moment.

Where candidates slip

Computing `total / n` in ints and losing the fraction.

How to stand out

Sum in long, cast to double before dividing, and mention the empty-array guard.

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

Sum then divide

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Mean is sum over count — what type must the division happen in?

2Hint 2Explore hint

Integer division truncates: 10/4 becomes 2, not 2.5.

3Hint 3Explore hint

Cast one operand to double BEFORE dividing; sum in long to avoid overflow.

A · Core idea

The mean is sum / count. The whole subtlety is floating-point division: you must divide in a real-number type or the fraction is lost.

Why cast before, not after. In Java, total / nums.length with two integers does integer division first — 10 / 4 = 2 — and only then would a later cast widen 2 to 2.0. You must cast an operand before the division: (double) total / nums.length makes the whole division floating-point, yielding 2.5.

Why sum in `long`. The numerator can overflow int just like array-sum, so accumulate in long, then cast to double for the divide.

Empty input. Dividing by a zero length is undefined; guard it (return 0 or throw) if the contract allows empty arrays. Here it's guaranteed non-empty.

B · Visual dry run

Interactive dry run

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

nums = [1, 2, 3, 4]

sum

1+2+3+4

Statetotal = 10

Final result2.5

C · Code with teaching comments

solution.javaJava

Sum then divide — optimal

1 teaching note

1
public double average(int[] nums) {
2
    long total = 0;
3
    for (int x : nums) total += x;
4
    // Cast to double BEFORE dividing so the fraction survives.
5
    return (double) total / nums.length;
6
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.6 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

(double) total / n is the fix — cast before the division, not after.

ComplexityO(n) time, O(1) space — one pass to sum plus a single division — linear time, constant space.

E · Edge cases and pitfalls5nums = [5, 5, 5] — Returns 5.0; whole-number mean still works.Explore edge cases

Edge cases handled

nums = [5, 5, 5]Returns 5.0; whole-number mean still works.
nums = [2, 3]5 / 2 = 2.5 — the case that exposes integer division.
nums = []Division by zero — guard if empties are allowed.

Pitfalls

Integer division truncating the fractional part.
Dividing by zero on an empty array.

Practice

Try the solution yourself

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

Editable code previewpublic double average(int[] nums) { long total = 0; for (int x : nums) total += x; return (double) total / nums.length; }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.

Integer division

`total / nums.length` with two ints truncates before any cast; 10/4 yields 2, not 2.5.

Explore the code comparison

Don't write this

double avg = total / nums.length;        // truncates to 2

Write this instead

double avg = (double) total / nums.length;

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Mean = sum / count

Code pattern to memorize

return (double) sum / n;

Rules to remember

1Sum in long.
2Cast to double before dividing.
3Guard against empty input.

Use this pattern when

Averages/ratios from integer counts
Anywhere integer division would truncate

Interview takeaway

Cast before dividing, not after — order matters.

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: Mean = sum / count. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Sum of Array Elements
The numerator; same overflow care.

easyBasicsSometimes asked

Average of an Array

Basics · 6 min

Given a non-empty integer array nums, return the average (mean) of its elements as a floating-point number.

In plain English

Sum the elements, then divide by the count. Do the division in floating point to keep the fractional part.

Examples

Example 1

Input: nums = [1, 2, 3, 4]
Output: 2.5
Note: 10 / 4.

Example 2

Input: nums = [5, 5, 5]
Output: 5.0
Note: Mean is 5.

Example 3

Input: nums = [2, 3]
Output: 2.5
Note: 5 / 2.

Constraints

1 ≤ nums.length ≤ 10^5
-10^9 ≤ nums[i] ≤ 10^9

Clarifying questions2First question: Should the result be a float or rounded?Explore questions

Q1.Should the result be a float or rounded?
I'll return a double mean; rounding is a separate, easy step if needed.
Q2.Can the array be empty?
If so I need an agreed return for division by zero.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Sum with a `long`, then divide by `nums.length` as a `double` so you don't lose the fraction.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you recognise integer-division truncation and cast at the right moment.

Where candidates slip

Computing `total / n` in ints and losing the fraction.

How to stand out

Sum in long, cast to double before dividing, and mention the empty-array guard.

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

Sum then divide

TimeO(n)SpaceO(1)

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

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Mean is sum over count — what type must the division happen in?

2Hint 2Explore hint

Integer division truncates: 10/4 becomes 2, not 2.5.

3Hint 3Explore hint

Cast one operand to double BEFORE dividing; sum in long to avoid overflow.

A · Core idea

The mean is sum / count. The whole subtlety is floating-point division: you must divide in a real-number type or the fraction is lost.

Why sum in `long`. The numerator can overflow int just like array-sum, so accumulate in long, then cast to double for the divide.

Empty input. Dividing by a zero length is undefined; guard it (return 0 or throw) if the contract allows empty arrays. Here it's guaranteed non-empty.

B · Visual dry run

Interactive dry run

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

nums = [1, 2, 3, 4]

sum

1+2+3+4

Statetotal = 10

Final result2.5

C · Code with teaching comments

solution.javaJava

Sum then divide — optimal

1 teaching note

1
public double average(int[] nums) {
2
    long total = 0;
3
    for (int x : nums) total += x;
4
    // Cast to double BEFORE dividing so the fraction survives.
5
    return (double) total / nums.length;
6
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.6 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

(double) total / n is the fix — cast before the division, not after.

ComplexityO(n) time, O(1) space — one pass to sum plus a single division — linear time, constant space.

E · Edge cases and pitfalls5nums = [5, 5, 5] — Returns 5.0; whole-number mean still works.Explore edge cases

Edge cases handled

nums = [5, 5, 5]Returns 5.0; whole-number mean still works.
nums = [2, 3]5 / 2 = 2.5 — the case that exposes integer division.
nums = []Division by zero — guard if empties are allowed.

Pitfalls

Integer division truncating the fractional part.
Dividing by zero on an empty array.

Practice

Try the solution yourself

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

Editable code previewpublic double average(int[] nums) { long total = 0; for (int x : nums) total += x; return (double) total / nums.length; }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.

Integer division

`total / nums.length` with two ints truncates before any cast; 10/4 yields 2, not 2.5.

Explore the code comparison

Don't write this

double avg = total / nums.length;        // truncates to 2

Write this instead

double avg = (double) total / nums.length;

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Mean = sum / count

Code pattern to memorize

return (double) sum / n;

Rules to remember

1Sum in long.
2Cast to double before dividing.
3Guard against empty input.

Use this pattern when

Averages/ratios from integer counts
Anywhere integer division would truncate

Interview takeaway

Cast before dividing, not after — order matters.

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: Mean = sum / count. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Sum of Array Elements
The numerator; same overflow care.