Problem Description

You are given two positive integers n and x. Your task is to find the number of ways to express n as the sum of x-th powers of unique positive integers.

In other words, you need to count how many different sets of unique positive integers [n₁, n₂, ..., nₖ] exist such that:

• n = n₁ˣ + n₂ˣ + ... + nₖˣ
• All integers in the set are unique (no duplicates allowed)

For example, if n = 160 and x = 3, one valid way to express 160 is:

• 160 = 2³ + 3³ + 5³ = 8 + 27 + 125

Since the result can be very large, you should return the answer modulo 10⁹ + 7.

The solution uses dynamic programming where f[i][j] represents the number of ways to select numbers from the first i positive integers such that the sum of their x-th powers equals j. For each positive integer i, you can either include it in your sum (adding iˣ to the total) or exclude it. The state transition is:

f[i][j] = f[i-1][j] + f[i-1][j-iˣ] (where the second term is only added if j ≥ iˣ)

The final answer is f[n][n], representing all possible ways using integers from 1 to n to sum up to n.

Intuition

This problem is essentially asking us to find combinations of numbers whose x-th powers sum to n. The key insight is that this is a variant of the classic "subset sum" problem, where instead of summing the numbers directly, we're summing their x-th powers.

Think of it like having a collection of building blocks where each block i has a value of iˣ. We want to know how many different ways we can select blocks (without using the same block twice) to reach exactly the target value n.

The natural approach is to consider each possible integer from 1 to n one by one and make a decision: should we include this number in our sum or not? This "include or exclude" pattern immediately suggests dynamic programming.

Why do we only need to consider integers from 1 to n? Because if any integer i > n, then iˣ > n (since x ≥ 1), making it impossible to use in our sum without exceeding the target.

The dynamic programming state f[i][j] naturally emerges when we think about building up our solution incrementally:

• The first dimension i represents "we've considered integers from 1 to i"
• The second dimension j represents "we want to achieve a sum of j"

For each integer i, we face a choice:

• Don't use i: Then we need to achieve sum j using only integers from 1 to i-1, which is f[i-1][j]
• Use i: Then we need to achieve sum j - iˣ using integers from 1 to i-1, which is f[i-1][j - iˣ]

This recursive structure naturally leads to our dynamic programming solution, where we build up from smaller subproblems to larger ones, ultimately finding f[n][n] - the number of ways to achieve sum n using any integers from 1 to n.

Learn more about Dynamic Programming patterns.

Solution Approach

We implement the solution using a 2D dynamic programming table f[i][j] where:

• i represents considering integers from 1 to i
• j represents the target sum we want to achieve

Initialization:

• Create a 2D array f of size (n+1) × (n+1) initialized with zeros
• Set f[0][0] = 1 as the base case (there's exactly one way to achieve sum 0 using no numbers: select nothing)

State Transition: For each integer i from 1 to n:

1. First, calculate k = i^x (the x-th power of i)
2. For each possible sum j from 0 to n:
   • Start with not including i: f[i][j] = f[i-1][j]
   • If including i is valid (i.e., k ≤ j), add the ways to achieve j - k using integers 1 to i-1:
     • f[i][j] = (f[i][j] + f[i-1][j - k]) % mod

Mathematical Formula:

f[i][j] = f[i-1][j] + (j ≥ i^x ? f[i-1][j - i^x] : 0)

Implementation Details:

1. We use modulo 10^9 + 7 at each addition to prevent integer overflow
2. The outer loop iterates through each integer i from 1 to n
3. For each i, we compute pow(i, x) once and reuse it
4. The inner loop considers all possible sums from 0 to n
5. We only add f[i-1][j - k] when k ≤ j to ensure valid array access

Time Complexity: O(n² × log(x)) where the log(x) factor comes from computing i^x Space Complexity: O(n²) for the 2D DP table

The final answer is f[n][n], representing the number of ways to express n as the sum of x-th powers using any combination of unique integers from 1 to n.

Example Walkthrough

Let's work through a small example with n = 10 and x = 2 (finding ways to express 10 as sum of squares).

Step 1: Setup

• We need to find ways to express 10 as sum of squares of unique positive integers
• Possible squares to consider: 1² = 1, 2² = 4, 3² = 9, 4² = 16 (too large)
• So we only consider integers 1, 2, and 3

Step 2: Initialize DP Table Create f[4][11] (since we consider integers 0 to 3, and sums 0 to 10)

• Set f[0][0] = 1 (one way to make 0 using no numbers)
• All other cells start at 0

Step 3: Fill DP Table

Consider i = 1 (power = 1² = 1):

• For j = 0: f[1][0] = f[0][0] = 1 (don't use 1)
• For j = 1: f[1][1] = f[0][1] + f[0][0] = 0 + 1 = 1 (use 1)
• For j = 2: f[1][2] = f[0][2] + f[0][1] = 0 + 0 = 0
• ...continuing for all j up to 10

Consider i = 2 (power = 2² = 4):

• For j = 0: f[2][0] = f[1][0] = 1
• For j = 1: f[2][1] = f[1][1] = 1 (can't use 2² = 4)
• For j = 4: f[2][4] = f[1][4] + f[1][0] = 0 + 1 = 1 (use 2)
• For j = 5: f[2][5] = f[1][5] + f[1][1] = 1 + 1 = 2 (with or without 2)
• ...continuing for all j

Consider i = 3 (power = 3² = 9):

• For j = 9: f[3][9] = f[2][9] + f[2][0] = 0 + 1 = 1 (use only 3)
• For j = 10: f[3][10] = f[2][10] + f[2][1] = 1 + 1 = 2

Step 4: Extract Answer f[3][10] = 2, but we need f[n][n] = f[10][10]

Actually, let me recalculate considering all integers up to 10:

• After processing all integers 1 through 10 (though only 1, 2, 3 have squares ≤ 10)
• The valid ways are:
  1. 10 = 1² + 3² = 1 + 9
  2. No other valid combination sums to exactly 10

Therefore, f[10][10] = 1 (there's exactly 1 way to express 10 as sum of squares).

Solution Implementation

Time and Space Complexity

The time complexity is O(n^2), where n is the given integer in the problem. This is because we have two nested loops: the outer loop iterates from 1 to n (n iterations), and the inner loop iterates from 0 to n (n+1 iterations). Each iteration performs constant time operations including the power calculation pow(i, x), array access, and modular arithmetic. Therefore, the total time complexity is O(n × n) = O(n^2).

The space complexity is O(n^2), where n is the given integer in the problem. This is due to the 2D DP array f which has dimensions (n+1) × (n+1), requiring O(n^2) space to store all the intermediate states. The array f[i][j] represents the number of ways to express value j using the first i positive integers raised to the power of x.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Computing Powers

The most critical pitfall is computing pow(num, x) without considering overflow. When num and x are large, num^x can exceed the target sum n by a huge margin, wasting computation and potentially causing issues.

Problem Example:

• If n = 100 and x = 10, computing 10^10 = 10,000,000,000 is unnecessary since it far exceeds our target.
• This leads to unnecessary iterations and memory usage.

Solution: Add an early termination check:

for num in range(1, n + 1):
    power_value = pow(num, x)
  
    # Early termination: if power exceeds n, all subsequent powers will too
    if power_value > n:
        break
  
    for target_sum in range(n + 1):
        # ... rest of the logic

2. Space Optimization Overlooked

The current solution uses O(n²) space with a 2D array, but since each row only depends on the previous row, we can optimize to O(n) space.

Optimized Solution:

def numberOfWays(self, n: int, x: int) -> int:
    MOD = 10**9 + 7
  
    # Use 1D array instead of 2D
    dp = [0] * (n + 1)
    dp[0] = 1
  
    num = 1
    while True:
        power_value = pow(num, x)
        if power_value > n:
            break
      
        # Traverse backwards to avoid using updated values
        for target_sum in range(n, power_value - 1, -1):
            dp[target_sum] = (dp[target_sum] + dp[target_sum - power_value]) % MOD
      
        num += 1
  
    return dp[n]

3. Incorrect Loop Bounds

A common mistake is iterating through all numbers from 1 to n even when smaller numbers would suffice. For example, if x = 2 and n = 100, we only need to check numbers up to 10 (since 10² = 100).

Better Approach: Calculate the maximum useful number beforehand:

max_num = int(n ** (1/x)) + 1  # Add 1 for safety due to floating point precision
for num in range(1, min(max_num + 1, n + 1)):
    # ... rest of the logic

4. Forgetting Modulo Operations

While the provided code handles this correctly, a common mistake is forgetting to apply modulo at every addition, which can cause integer overflow in languages with fixed integer sizes.

Key Points:

• Always apply modulo after additions: (a + b) % MOD
• Don't apply modulo to intermediate power calculations unless necessary
• Be consistent with modulo application throughout the solution