Power Sums: Count the Ways!

 Ways to Express an Integer as Sum of Powers 

Today’s problem: Find the number of ways to express n as the sum of the xth power of unique integers. At first, it might seem like a simple number-crunching task, but when you look deeper, it’s a great opportunity to learn about backtracking and dynamic programming.

Let’s explore how to solve this problem with two different approaches: brute force and optimized dynamic programming.

The Best Data Structure for Solving It (And Why!)

When tackling this problem, the key is to keep track of the unique sums.

  • Brute force (backtracking) is straightforward but can be slow since it checks all combinations.

  • Dynamic programming (DP), on the other hand, allows us to break the problem into smaller subproblems and solve them efficiently without repeating work.

Both methods have their place, but we’ll start with brute force to build the intuition and then move to the optimized DP solution.

Different Approaches – Brute Force to Optimized Solutions

1. Brute Force (Backtracking)

This approach uses backtracking to explore all subsets of powers of integers and check if they sum up to n. It’s simple but inefficient because it explores all possibilities without any memoization or pruning.

class Solution {
    public int numOfWays(int n, int x) {
        return backtrack(n, x, 1, 0);
    }

    // Helper function for backtracking
    private int backtrack(int n, int x, int start, int sum) {
        if (sum == n) {
            return 1; // Found a valid combination
        }
        if (sum > n) {
            return 0; // Exceeded the target sum, no valid combination
        }

        int count = 0;
        for (int i = start; Math.pow(i, x) <= n - sum; i++) {
            int power = (int) Math.pow(i, x);
            count += backtrack(n, x, i + 1, sum + power); // Include the current number
        }

        return count;
    }
}

Explanation:

  • The backtrack function tries to add powers of integers starting from 1 and recursively checks if we can form the sum n.

  • The base case checks if the current sum equals n (valid combination), and if it exceeds n, it returns 0.

  • We iterate through integers starting from the current index to ensure the numbers are unique (no duplicates).

2. Optimized Solution (Dynamic Programming)
Now, let's improve the brute force approach using dynamic programming (DP). The idea is to use a 1D DP array to store the number of ways to express each sum from 0 to n. We avoid redundant calculations by building up the solution step by step.

class Solution {
    public int numOfWays(int n, int x) {
        final int MOD = 1000000007;
        int max = (int) Math.pow(n, 1.0 / x);  // Largest integer to consider
        int[] dp = new int[n + 1];
        dp[0] = 1; // Base case: one way to make 0

        // Iterate over all integers whose xth power is <= n
        for (int i = 1; i <= max; i++) {
            int power = (int) Math.pow(i, x);  // i^x
            for (int j = n; j >= power; j--) {  // Iterate backwards to avoid double counting
                dp[j] = (dp[j] + dp[j - power]) % MOD;
            }
        }

        return dp[n];
    }
}

Explanation:

  • We initialize dp[0] = 1, as there’s exactly one way to sum to 0—by choosing no numbers.

  • For each integer i, we compute i^x and check if it can be used to form a valid sum up to n.

  • The DP array dp[j] keeps track of the number of ways to form the sum j using the powers of integers up to n.

⚠️ Common Mistakes & Pitfalls to Avoid

  • Double Counting: In the brute force method, make sure to increment the index properly so that we don't use the same integer more than once.

  • Modulo Operation: Don’t forget to apply the modulo 10^9 + 7 to avoid overflow, especially in the DP solution.

  • Off-by-One Errors: In both approaches, ensure that you correctly update the sum and avoid including integers multiple times.

📊 Time and Space Complexity Analysis

  • Brute Force:

    • Time Complexity: Exponential time complexity O(2^k) because we're checking all subsets. This is not efficient for larger values of n but works for small inputs.

    • Space Complexity: O(k) for the recursion stack, where k is the number of recursive calls (dependent on the size of the tree formed by recursive calls).

  • Dynamic Programming:

    • Time Complexity: O(n * k), where n is the target number and k is the number of integers whose xth power is ≤ n.

    • Space Complexity: O(n) due to the DP array used to store the results.

🌍 Real-World Applications

This type of problem shows up in combinatorics and dynamic programming optimization tasks. Some potential real-world uses:

  • Resource Allocation: Distributing limited resources in the most efficient way.

  • Game Design: Finding combinations of moves or actions that sum up to a specific value in games.

  • Cryptography: Using powers to generate hashes or codes based on certain criteria.


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

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