Definition
n = k³
Includes positive, negative, and zero values.
A perfect cube is an integer n = k3 for some integer k. This tutorial covers float estimate + verify, pure integer search, binary search, a live preview, worked Python examples, edge cases, and complexity.
n = k³
Includes positive, negative, and zero values.
round(³√)
Round the cube root, then check k * k * k == n.
No float
Grow k until k³ ≥ |n|, then test equality.
O(log |n|)
Faster integer route for large magnitudes.
Try any n
Classify positive, negative, and zero instantly.
Always verify
Never trust a floating estimate alone.
Perfect cubes (cube numbers) are integers of the form n = k3. Classic examples: 0, 1, 8, 27, 64 and negatives like -8, -27.
A practical check estimates an integer cube root, then verifies with exact integer multiplication. You can also avoid floating point entirely with a linear or binary search on k.
It drills integer roots, sign handling, floating-point caution, and search-based number checks.
n equals some integer cubed.
Final check: k * k * k == n.
Odd powers keep negatives as cubes.
Both are perfect cubes.
In short: find a candidate integer root k, then accept n only if k³ == n exactly.
Given an integer n, decide whether n = k3 for some integer k.
# 27 = 3**3 → yes
# 28 sits between 3**3 and 4**3 → no
# -8 = (-2)**3 → yes | Item | Type | Description |
|---|---|---|
n | int | Any integer (positive, negative, or zero). |
| Return / print | bool / text | True if n is a perfect cube. |
function is_perfect_cube(n):
x = abs(n)
k = 0
while k * k * k < x:
k = k + 1
return k * k * k == x | Method | Idea | Notes |
|---|---|---|
| Root + verify | round(|n|^(1/3)), then k³ == n | Fast; watch float precision |
| Linear scan | Increment k until k³ ≥ |n| | No float; O(|n|^(1/3)) |
| Binary search | Search k on [0, |n|] | O(log |n|) integer route |
| Goal | Pattern |
|---|---|
| Estimate root | k = round(abs(n) ** (1 / 3)) |
| Apply sign | if n < 0: k = -k |
| Exact verify | k * k * k == n |
| Integer scan | while k * k * k < x: k += 1 |
| Classic yes | 27 = 3³, -8 = (-2)³ |
| Classic no | 28 (between 27 and 64) |
Same yes/no answer — different precision and speed trade-offs.
round + verifyShort; must verify in integer arithmetic
grow kNo float; simple for interviews
O(log |n|)Best pure-integer route for large n
verify alwaysSay why floating estimate alone is unsafe
Reach for perfect-cube checks when integer roots and precision matter.
Tests roots, signs, and float-vs-integer reasoning.
Same pattern as perfect-square problems, with odd powers.
List cubes in 1…N for small classroom ranges.
Great prompt for “why verify after float math?”
Perfect-cube interviews usually mean integer n and integer k.
Key benefit: one short boolean check that covers roots, signs, float caution, and search complexity.
JavaScript safe integers. Uses integer scan on absolute value, then applies sign logic.
Three complete Python programs — float estimate + verify, integer range scan, and binary search. Click View Output to reveal sample console results.
Estimate an integer root, then verify exactly.
Fast for one value. Round the floating cube root, then confirm with integer multiplication.
def is_cube(number: int) -> bool:
k = round(abs(number) ** (1 / 3))
if number < 0:
k = -k
return k * k * k == number
input_number = 27
if is_cube(input_number):
print(f"{input_number} is a cube number.")
else:
print(f"{input_number} is not a cube number.") The rounded root is only a candidate. The final check k * k * k == number is what guarantees correctness, including for negatives.
No floating point — scan and filter a small interval.
Grow k until k³ ≥ num; equality means a perfect cube.
def is_cube_number(num: int) -> bool:
k = 0
while k * k * k < num:
k += 1
return k * k * k == num
print("Cube numbers in the range 1 to 50:")
for i in range(1, 51):
if is_cube_number(i):
print(i, end=" ") The loop finds the smallest k with k³ ≥ num. From 1 to 50 the cubes are exactly 1, 8, and 27.
Binary search on k for large magnitudes.
Search k on [0, |n|] and verify equality; apply sign for negatives.
def is_cube_binary(n: int) -> bool:
x = abs(n)
lo, hi = 0, x
while lo <= hi:
mid = (lo + hi) // 2
cube = mid * mid * mid
if cube == x:
return True
if cube < x:
lo = mid + 1
else:
hi = mid - 1
return False
for value in (27, 28, -8, 0, 64):
print(f"{value}: {'cube' if is_cube_binary(value) else 'not a cube'}") Because x ≥ 0, searching nonnegative k is enough; odd powers mean negatives are cubes whenever |n| is. Each step halves the search range — O(log |n|) comparisons.
Work with x = |n| (0 is immediately a cube).
Estimate with float, scan linearly, or binary-search.
Accept only if k * k * k equals |n| (then restore sign conceptually).
Equality → perfect cube; otherwise not.
n = 28Trace the integer scan. Cubes nearby: 27 = 3³, 64 = 4³.
| k | k³ | Compare to 28 | Action |
|---|---|---|---|
0 | 0 | < 28 | Increment |
1 | 1 | < 28 | Increment |
2 | 8 | < 28 | Increment |
3 | 27 | < 28 | Increment |
4 | 64 | > 28 | Stop; 64 ≠ 28 |
Final: 28 is not a perfect cube.
Where perfect-cube checks show up beyond the interview prompt.
Roots, signs, and verify-after-estimate in one problem.
Example: write is_cube(n).
Makes 1, 8, 27, 64… memorable with counterexamples.
Example: chalkboard 27 vs 28.
Shows why estimates need an exact check.
Example: “is round enough?”
Print cubes in a classroom interval.
Example: 1 to 50 → 1 8 27.
Monotone k³ is a clean search predicate.
Example: find integer cube root.
Exponents divisible by 3 ↔ perfect cube.
Example: prime-factor argument.
Pro Tip: always say “estimate, then verify with integer cube” — that sentence scores well in interviews.
Why this pattern works well in interviews and classwork.
n = k³ is easy to state and test.
Float estimate, linear scan, or binary search all work.
A few integers suffice — O(1) extra space.
Negatives, zero, and float precision give structured follow-ups.
Pro Tip: prefer k * k * k over k ** 3 in interviews when discussing overflow in fixed-width languages.
Small habits that keep cube checks interview-ready.
Never accept a floating estimate without k³ == n.
Negate k when n is negative for the estimate method.
0 = 0³ — include it in tests.
Assert 27 and -8 are cubes; 28 is not.
Upgrade from linear scan when |n| can be huge.
Pro Tip: for the scan method, searching |n| is enough — odd powers make negatives automatic.
Mistakes that commonly break perfect-cube solutions.
Near-boundary roots can round wrong for large n.
→ Always verify with integer cubing.
-8 and -27 are perfect cubes.
→ Preserve sign or search on absolute value.
0 = 0³ is a valid cube.
→ Include n = 0 in your tests.
Checking root ** 3 == n in float can fail.
→ Cubing must happen with integers.
O(|n|^(1/3)) can be slow for large magnitudes.
→ Offer binary search as an upgrade.
Check these inputs before calling the solution done.
Negative perfect cubes exist (-8, -27). Preserve sign correctly.
n = 0Always a cube because 0 = 0³.
For very large magnitudes, floating-point root may be slightly off.
Neighbors of 27 are not cubes — good counterexamples.
n = 11 = 1³ — smallest positive cube.
Grows with |n|^(1/3). Use binary search if needed.
Handy follow-ups interviewers sometimes ask.
Try these variations to lock in the pattern.
Quick Takeaway: find a candidate integer k, then accept n only if k * k * k == n.
| Method | Time | Extra space |
|---|---|---|
| Root estimate + verify | O(1) style | O(1) |
| Linear integer scan | O(|n|^(1/3)) | O(1) |
| Binary search on k | O(log |n|) | O(1) |
For scan-based methods, n denotes input magnitude and root growth is about |n|^(1/3).
A perfect cube is an integer n = k³. Estimate a root if you like, but always verify with exact integer cubing — or skip floats with a scan / binary search.
Practice the three examples above, then continue to decimal-to-binary for another classic conversion warm-up.
Handle negatives and zero, never trust float alone, and mention binary search for large |n|.
Check perfect cubes the interview-friendly way.
n = k³
Definitionk*k*k == n
GuardNegatives OK
Math0 is a cube
EdgeBinary O(log)
AnalysisA nonzero integer n is a perfect cube iff in its prime factorization every exponent is a multiple of three. Also, 0 and 1 are perfect cubes.
Learn how to convert decimal integers to binary with loops and built-ins.
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