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Improve power sum algorithm (#5652)
* Update directory * Improve PowerSum algorithm implementation and documentation This commit enhances the PowerSum class in the backtracking package. The changes focus on improving code quality, readability, and documentation. Key improvements include: 1. Enhanced code structure and efficiency: - Removed class-level variables for better thread safety - Optimized the recursive approach to avoid unnecessary calculations - Simplified the overall logic for easier understanding 2. Improved readability: - Used more descriptive variable names (e.g., 'targetSum' instead of 'n', 'power' instead of 'x') - Enhanced method structure with a private recursive helper method 3. Better documentation: - Added comprehensive JavaDoc comments explaining the algorithm's purpose and implementation - Clarified the meaning of parameters, especially relating them to the original problem statement (N and X) - Improved inline comments for better code understanding 4. Adhered to Java best practices: - Improved encapsulation by making the recursive method private - Used Math.pow() directly instead of a custom power method 5. Maintained core functionality: - The algorithm still solves the same problem as before, but with improved code quality * updated PowerSum * Refactor PowerSum algorithm implementation and documentation * Refactor PowerSum algorithm implementation and documentation * Refactor code formatting and remove unnecessary line in PowerSum.java * Refactor code formatting and add newline at end of file in .clang-format --------- Co-authored-by: manishraj27 <manishraj27@users.noreply.github.com> Co-authored-by: Bama Charan Chhandogi <b.c.chhandogi@gmail.com>
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@ -1,45 +1,51 @@
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package com.thealgorithms.backtracking;
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/*
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* Problem Statement :
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* Find the number of ways that a given integer, N , can be expressed as the sum of the Xth powers
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* of unique, natural numbers. For example, if N=100 and X=3, we have to find all combinations of
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* unique cubes adding up to 100. The only solution is 1^3+2^3+3^3+4^3. Therefore output will be 1.
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/**
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* Problem Statement:
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* Find the number of ways that a given integer, N, can be expressed as the sum of the Xth powers
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* of unique, natural numbers.
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* For example, if N=100 and X=3, we have to find all combinations of unique cubes adding up to 100.
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* The only solution is 1^3 + 2^3 + 3^3 + 4^3. Therefore, the output will be 1.
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*
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* N is represented by the parameter 'targetSum' in the code.
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* X is represented by the parameter 'power' in the code.
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*/
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public class PowerSum {
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private int count = 0;
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private int sum = 0;
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public int powSum(int n, int x) {
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sum(n, x, 1);
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return count;
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/**
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* Calculates the number of ways to express the target sum as a sum of Xth powers of unique natural numbers.
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*
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* @param targetSum The target sum to achieve (N in the problem statement)
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* @param power The power to raise natural numbers to (X in the problem statement)
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* @return The number of ways to express the target sum
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*/
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public int powSum(int targetSum, int power) {
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// Special case: when both targetSum and power are zero
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if (targetSum == 0 && power == 0) {
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return 1; // by convention, one way to sum to zero: use nothing
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}
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return sumRecursive(targetSum, power, 1, 0);
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}
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// here i is the natural number which will be raised by X and added in sum.
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public void sum(int n, int x, int i) {
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// if sum is equal to N that is one of our answer and count is increased.
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if (sum == n) {
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count++;
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return;
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} // we will be adding next natural number raised to X only if on adding it in sum the
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// result is less than N.
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else if (sum + power(i, x) <= n) {
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sum += power(i, x);
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sum(n, x, i + 1);
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// backtracking and removing the number added last since no possible combination is
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// there with it.
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sum -= power(i, x);
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}
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if (power(i, x) < n) {
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// calling the sum function with next natural number after backtracking if when it is
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// raised to X is still less than X.
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sum(n, x, i + 1);
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/**
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* Recursively calculates the number of ways to express the remaining sum as a sum of Xth powers.
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*
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* @param remainingSum The remaining sum to achieve
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* @param power The power to raise natural numbers to (X in the problem statement)
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* @param currentNumber The current natural number being considered
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* @param currentSum The current sum of powered numbers
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* @return The number of valid combinations
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*/
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private int sumRecursive(int remainingSum, int power, int currentNumber, int currentSum) {
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int newSum = currentSum + (int) Math.pow(currentNumber, power);
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if (newSum == remainingSum) {
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return 1;
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}
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if (newSum > remainingSum) {
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return 0;
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}
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// creating a separate power function so that it can be used again and again when required.
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private int power(int a, int b) {
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return (int) Math.pow(a, b);
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return sumRecursive(remainingSum, power, currentNumber + 1, newSum) + sumRecursive(remainingSum, power, currentNumber + 1, currentSum);
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}
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}
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