Add Power Sum (fixes #2634) (#2645)

Backtracking/PowerSum.java

@@ -0,0 +1,52 @@
+package Backtracking;
+
+import java.util.Scanner;
+/*
+ * Problem Statement :
+ * Find the number of ways that a given integer, N , can be expressed as the sum of the Xth powers of unique, natural numbers.
+ * For example, if N=100 and X=3, we have to find all combinations of unique cubes adding up to 100. The only solution is 1^3+2^3+3^3+4^3.
+ * Therefore output will be 1.
+*/
+public class PowerSum {
+
+	public static void main(String[] args) {
+		Scanner sc = new Scanner(System.in);
+		System.out.println("Enter the number and the power");
+		int N = sc.nextInt();
+		int X = sc.nextInt();
+		PowerSum ps = new PowerSum();
+		int count = ps.powSum(N,X);
+		//printing the answer.
+		System.out.println("Number of combinations of different natural number's raised to "+X+" having sum "+N+" are : ");
+		System.out.println(count);
+		sc.close();
+	}
+	private int count = 0,sum=0;
+    public int powSum(int N, int X) {
+        Sum(N,X,1);
+        return count;
+    }
+    //here i is the natural number which will be raised by X and added in sum.
+    public void Sum(int N, int X,int i) {
+    	//if sum is equal to N that is one of our answer and count is increased.
+        if(sum == N) {
+        	count++;
+        	return;
+        }
+        //we will be adding next natural number raised to X only if on adding it in sum the result is less than N.
+        else if(sum+power(i,X)<=N) {
+        	sum+=power(i,X);
+            Sum(N,X,i+1);
+            //backtracking and removing the number added last since no possible combination is there with it.
+            sum-=power(i,X);
+        }
+        if(power(i,X)<N) {
+        	//calling the sum function with next natural number after backtracking if when it is raised to X is still less than X.
+        	Sum(N,X,i+1);
+        }
+    }
+    //creating a separate power function so that it can be used again and again when required. 
+    private int power(int a , int b ){
+        return (int)Math.pow(a,b);
+    }
+}