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Ways to Express an Integer as Sum of Powers

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Dynamic ProgrammingRecursion

Given two positive integers n and x, determine the number of ways n can be expressed as the sum of the xth 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:

  1. If n = 160 and x = 3, one way to express n is n = 2^3 + 3^3 + 5^3.
  2. If 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.
  3. If 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.

Solution


Clarifying Questions

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:

  1. What is the range of values for `n` and `power`? Are they guaranteed to be non-negative integers?
  2. If `n` cannot be expressed as a sum of `power`th powers, what should the function return?
  3. Are we looking for distinct combinations of powers, or can the same base number be raised to the power multiple times in a single sum?
  4. Can `n` be zero? If so, and `power` is also non-zero, what is the expected result?
  5. Could you provide a few examples to illustrate the expected output for different combinations of `n` and `power`?

Brute Force Solution

Approach

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:

  1. Start with the smallest possible number raised to the given power.
  2. Check if this number is equal to the target number. If it is, you've found one way to express the integer.
  3. If the number is less than the target number, subtract it from the target number and remember that you used this number.
  4. Repeat this process of finding the smallest possible number raised to the power and subtracting it from the remaining value, but now you also have the choice of NOT including that number in the sum.
  5. Keep track of all the numbers you've considered using and how much remains of the target number.
  6. If at any point the remaining value becomes zero, that means you have found a way to express the original integer as the sum of powers.
  7. If you ever reach a point where the smallest power is greater than the remaining value, then you can't use that power or any powers bigger than that. In that case, you go back to the previous decision to see if you can still use that power or skip it.
  8. Continue trying different combinations until you've exhausted all possibilities, and then count the number of valid combinations you found.

Code Implementation

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

Big(O) Analysis

Time Complexity
O(target^(1/power) * 2^(target^(1/power)))The described brute force approach explores all possible combinations of numbers raised to the given power that could sum up to the target. In the worst-case scenario, we consider all numbers from 1 up to target^(1/power) (since any number larger than this, when raised to the power, will exceed the target). For each number, we have two choices: either include it in the sum or exclude it. This creates a binary decision tree. The height of this tree is proportional to target^(1/power), and the number of leaves (representing all possible combinations) is 2 raised to the power of target^(1/power). Therefore, the time complexity is proportional to target^(1/power) * 2^(target^(1/power)).
Space Complexity
O(N)The described brute force approach utilizes recursion to explore all possible combinations. The depth of the recursion depends on the target number N, as each recursive call considers whether to include a power of a number. In the worst-case, the recursion depth could reach N (or some function of N like the Nth root if optimized). Therefore, the maximum space used by the call stack can be proportional to N, resulting in O(N) auxiliary space complexity.

Optimal Solution

Approach

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:

  1. Start by figuring out the largest power that is less than or equal to the number you're trying to express as a sum.
  2. Consider using that power in the sum. Then, find out how many ways you can express the remaining part of the number using powers less than or equal to the power you just used.
  3. Also, consider *not* using that power in the sum. Then, find out how many ways you can express the original number using powers strictly less than the power you skipped.
  4. Add the number of ways you found in the previous two steps. This is the total number of ways to express the number as a sum of powers.
  5. Here's the trick to make it fast: whenever you calculate the number of ways to express a number using certain powers, remember that result. If you need to calculate it again later, just look up the answer instead of recalculating it.
  6. This 'remembering' strategy, along with carefully choosing whether or not to use a power at each step, avoids exploring redundant possibilities and leads to a fast solution.

Code Implementation

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)))

Big(O) Analysis

Time Complexity
O(n^(n/x))The algorithm uses recursion with memoization to explore different combinations of powers that sum up to n. The depth of the recursion is related to the largest power less than or equal to n, which depends on x (the power to which numbers are raised). Memoization drastically reduces redundant calculations, but in the worst case, it might explore a significant portion of the possible combinations. Because the number of unique combinations to explore is related to the size of n to the power of n/x, we have O(n^(n/x)) operations. Specifically, there are n possibilities for the maximum power to use, and approximately n/x possibilities for other powers to use until the sum is achieved or n is exhausted. So the time complexity is O(n^(n/x)).
Space Complexity
O(N*log(N))The space complexity is dominated by the memoization used to store the results of intermediate calculations. The memoization table stores the number of ways to express a number (up to N) using powers up to a certain limit. The number of powers less than or equal to N^1/X (where X is the power) is approximately log(N). Therefore, the memoization table can have at most N rows and approximately log(N) columns, leading to a space complexity of O(N*log(N)). The recursion stack also contributes to the space complexity, having a maximum depth of approximately log(N), however, the memoization dominates.

Edge Cases

CaseHow to Handle
n is zeroIf 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 zeroIf 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 oneIf x is one, the number of ways is 1 if n is 1, otherwise it's 0.
n is oneThe only way to represent it is x^1, so return 1.
x is negativeIf 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 powersUse long data type or check for overflow before calculating the power to prevent incorrect results.
x raised to n is greater than the target numberPrune 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 recursionConvert the recursive solution to an iterative dynamic programming solution to avoid stack overflow errors.