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Challenge 205: Perfect Cube Count

curriculum/challenges/english/blocks/daily-coding-challenges-python/698a1a73ade5ac0e19180fa8.md

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--description--

Given two integers, determine how many perfect cubes exist in the range between and including the two numbers.

  • The lower number is not guaranteed to be the first argument.
  • A number is a perfect cube if there exists an integer (n) where n * n * n = number. For example, 27 is a perfect cube because 3 * 3 * 3 = 27.

--hints--

count_perfect_cubes(3, 30) should return 2.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_perfect_cubes(3, 30), 2)`)
}})

count_perfect_cubes(1, 30) should return 3.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_perfect_cubes(1, 30), 3)`)
}})

count_perfect_cubes(30, 0) should return 4.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_perfect_cubes(30, 0), 4)`)
}})

count_perfect_cubes(-64, 64) should return 9.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_perfect_cubes(-64, 64), 9)`)
}})

count_perfect_cubes(9214, -8127) should return 41.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_perfect_cubes(9214, -8127), 41)`)
}})

--seed--

--seed-contents--

py
def count_perfect_cubes(a, b):

    return a

--solutions--

py
def count_perfect_cubes(a, b):
    start, end = min(a, b), max(a, b)
    count = 0

    n = 0
    while n ** 3 <= end:
        if n ** 3 >= start:
            count += 1
        n += 1

    n = -1
    while n ** 3 >= start:
        if n ** 3 <= end:
            count += 1
        n -= 1

    return count