0937MediumMathBinary Search

Whole Numbers Whose Power Lands in Range

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Trains the technique from

LeetCode 3932Count K-th Roots in a Range

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Given l, r and k, count the whole numbers x at or above zero for which x raised to the power k is at least l and at most r.

Examples

Example 1

Input: l = 17, r = 4913, k = 3
Output: 15

The cubes landing between 17 and 4913 run from 3 cubed, which is 27, up to 17 cubed, which is 4913, so fifteen bases qualify.

Example 2

Input: l = 100, r = 121, k = 2
Output: 2

The squares in range are 100 and 121, from the bases 10 and 11.

Example 3

Input: l = 101, r = 120, k = 2
Output: 0

No square lands between 101 and 120, since the nearest are 100 just below and 121 just above.

Constraints

0 <= l <= r <= 10^9
1 <= k <= 30

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_kth_roots(l: int, r: int, k: int) -> int:

Java

public int countKthRoots(int l, int r, int k)

The intended approach

Time O(k log r): a halving search over the bases with the power worked out each step.Space O(1).

Raising to a fixed power never decreases as the base grows, so the numbers whose power lands in the range form one unbroken run. Its ends are the largest `x` with `x` to the `k` at most `r`, and the smallest `x` with `x` to the `k` at least `l`, and the count is the difference plus one, or nothing when the run turns out empty. Both ends come from halving the range of candidate bases, comparing `x` to the `k` against the bound. Working the power out in whole numbers and stopping early once it passes the bound keeps the arithmetic small however large `k` is, since a base of two already passes a billion by the thirtieth power.

Pythonentry point: count_kth_roots

def count_kth_roots(l, r, k):
    def power_at_most(x, cap):
        # is x to the k at most cap, worked out without overflowing past the cap
        total = 1
        for _ in range(k):
            total *= x
            if total > cap:
                return False
        return True

    # largest base whose power stays at or below r
    low = 0
    high = r + 1
    while low < high:
        mid = (low + high + 1) // 2
        if power_at_most(mid, r):
            low = mid
        else:
            high = mid - 1
    top = low
    # smallest base whose power reaches l
    low = 0
    high = r + 1
    while low < high:
        mid = (low + high) // 2
        if power_at_most(mid, l - 1):
            low = mid + 1
        else:
            high = mid
    bottom = low
    if bottom > top:
        return 0
    return top - bottom + 1

The obvious version

Walk the bases upward from zero, work out each one's power, and count those landing in the range, stopping once the power passes the top of it. Costs a step per base rather than a halving search.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_kth_roots(l, r, k):
    total = 0
    x = 0
    while True:
        value = 1
        for _ in range(k):
            value *= x
            if value > r:
                break
        if value > r:
            break
        if value >= l:
            total += 1
        x += 1
    return total

How this problem was checked

Reference solution executed against 13 test cases — 13 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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