Given two positive integers n
and x
, determine the number of ways n
can be expressed as the sum of the x
th power of unique positive integers. In other words, find the number of sets of unique integers [n1, n2, ..., nk]
such that n = n1x + n2x + ... + nkx
. Since the result can be very large, return it modulo 10^9 + 7
. Let's consider some examples:
n = 160
and x = 3
, one way to express n
is n = 2^3 + 3^3 + 5^3
.n = 10
and x = 2
, the output should be 1
because 10 = 3^2 + 1^2
is the only way to express 10 as the sum of the 2nd power of unique integers.n = 4
and x = 1
, the output should be 2
because n = 4^1
and n = 3^1 + 1^1
are the two possible ways.Write a function that takes integers n
and x
as input and returns the number of ways n
can be expressed as the sum of unique integers raised to the power of x
.
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force approach to this problem is like trying every possible combination of numbers raised to a power to see if they add up to the target number. We check each combination, no matter how inefficient, until we find all the valid solutions. It's like trying every possible lock combination until you find the right one.
Here's how the algorithm would work step-by-step:
def ways_to_express_integer_as_sum_of_powers(number, power):
count_of_ways = 0
def find_ways(remaining_number, current_number):
nonlocal count_of_ways
power_of_current_number = current_number ** power
# If the power is equal to remaining, we found a way
if power_of_current_number == remaining_number:
count_of_ways += 1
return
# Power greater than remaining, no more solutions from here
if power_of_current_number > remaining_number:
return
# Explore the possibility of excluding the current number.
find_ways(remaining_number, current_number + 1)
# Explore the possibility of including the current number.
find_ways(remaining_number - power_of_current_number, current_number + 1)
find_ways(number, 1)
return count_of_ways
The best way to solve this problem is to use a method called recursion combined with a clever trick to avoid doing the same calculations over and over. We explore possible solutions by making choices at each step, remembering the results of those choices so we don't have to recalculate them later.
Here's how the algorithm would work step-by-step:
def integer_sum_of_powers(number, power):
memo = {}
def count_ways(remaining_number, current_power):
if remaining_number == 0:
return 1
if remaining_number < 0 or current_power == 0:
return 0
key = (remaining_number, current_power)
if key in memo:
return memo[key]
# Calculate the largest power we can use
largest_power_value = int(remaining_number**(1/power))
# Consider using the current power
current_power_value = largest_power_value**power
# Explore ways using the current power.
use_power = count_ways(remaining_number - current_power_value, largest_power_value - 1)
# Explore ways without using the current power.
skip_power = count_ways(remaining_number, largest_power_value - 1)
total_ways = use_power + skip_power
memo[key] = total_ways
return total_ways
return count_ways(number, int(number**(1/power)))
Case | How to Handle |
---|---|
n is zero | If n is zero, the only way to express it is with no terms, so return 1 if x is 0 and 0 otherwise. |
x is zero | If x is zero, there's only one way if n is zero, and zero ways otherwise, return 1 if n is 0, 0 otherwise. |
x is one | If x is one, the number of ways is 1 if n is 1, otherwise it's 0. |
n is one | The only way to represent it is x^1, so return 1. |
x is negative | If x is negative, and n is even, result is same as positive x; if n is odd, it's not allowed as we deal with powers. |
Integer overflow when calculating powers | Use long data type or check for overflow before calculating the power to prevent incorrect results. |
x raised to n is greater than the target number | Prune the search space by stopping the recursion early, as any power higher than the target cannot be used. |
x and n are very large, leading to stack overflow with recursion | Convert the recursive solution to an iterative dynamic programming solution to avoid stack overflow errors. |