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Ways to Express an Integer as Sum of Powers

Number: 2882

Difficulty: Medium

Paid? No

Companies: N/A

Problem Description

Given two positive integers n and x, count the number of ways n can be expressed as the sum of the xth powers of unique positive integers. In other words, find the number of sets of unique integers [n1, n2, ..., nk] such that n = n1^x + n2^x + ... + nk^x. Since the result can be very large, return it modulo 10^9+7.

Key Insights

  • Only consider positive integers whose xth power is ≤ n.
  • Use recursion or backtracking to explore combinations, enforcing uniqueness by always progressing to higher numbers.
  • Use memoization or DP to avoid recalculating results for the same (n, current) state.
  • The modulo constraint (10^9+7) ensures results remain within integer bounds.

Space and Time Complexity

Time Complexity: O(n * m) where m is the number of candidate integers such that candidate^x ≤ n. Space Complexity: O(n) for the recursion call stack and memoization storage.

Solution

The problem can be solved using a recursive Depth First Search (DFS) approach. Start with the smallest candidate (1) and recursively try to reduce n by subtracting candidate^x. To maintain uniqueness and avoid counting duplicates, always consider the next candidate (i.e., candidates greater than the current one). Additionally, use memoization with the state (remaining n, current candidate) to avoid redundant computation. Finally, result values are taken modulo 10^9+7.

Code Solutions

MOD = 10**9 + 7

def waysToExpress(n, x):
    # memoization dictionary with key (remaining_sum, current_candidate)
    memo = {}
    
    def dfs(remaining, current):
        # If exactly met the target, we found a valid solution
        if remaining == 0:
            return 1
        # If remaining sum goes negative, it's not a valid path
        if remaining < 0:
            return 0
        # Use memoization to avoid duplicate work
        if (remaining, current) in memo:
            return memo[(remaining, current)]
        
        count = 0
        # Try all candidates starting from current that satisfy candidate^x <= remaining
        candidate = current
        while candidate ** x <= remaining:
            power_val = candidate ** x
            # Recursively reduce remaining sum and move to the next candidate
            count = (count + dfs(remaining - power_val, candidate + 1)) % MOD
            candidate += 1
        
        memo[(remaining, current)] = count
        return count
    
    return dfs(n, 1)

# Example usage:
print(waysToExpress(10, 2))  # Expected output: 1
print(waysToExpress(4, 1))   # Expected output: 2
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