# Java Challenge - Number Permutation

This community-built FAQ covers the “Number Permutation” code challenge in Java. You can find that challenge here, or pick any challenge you like from our list.

## Top Discussions on the Java challenge Number Permutation

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I tried a brute force approach instead of recursion.
I generated all possible permutations [for the number and everything less] (including the permutations with zeros) and then I counted only the ones that worked.
To generate the permutations, I used the binary representation of a long integer as a permutation.
It is horribly inefficient: it requires 2 ^ (z + 4) - 1 permutations to be tested, given number z.
Do not use this algorithm; its really bad.

``````public class NumberPermutation {

public static boolean checkPermutationBinary(long x, int expectedBinarySum) {
int maxNumberOf0s = 4;
// is one less than the maximum number of numbers in permutation
int maxConsecutive1s = 9;  // max number we can use in permutation
//int i = 0; // position in the binary representation
long b = 1; // used to test if each binary digit is a 1
boolean previousWas0 = false;
int numberOf0s = 0;
int numberOf1s = 0;
int numberOfConsecutive1s = 0;
if ((b & x) != b) { // starts with 0
return false;
}

while (b <= x) {
if ((b & x) == b) { // have a 1
numberOf1s++;
numberOfConsecutive1s++;
if (numberOf1s > expectedBinarySum) {
return false;
}
if (numberOfConsecutive1s > maxConsecutive1s) {
return false;
}
previousWas0 = false;
}
else { // hava a 0
numberOf0s++;
numberOfConsecutive1s = 0;
if (previousWas0 || (numberOf0s > maxNumberOf0s)) {
return false;
}
previousWas0 = true;
}
b = b << 1; // go to next binary digit
//i += 1;
}
/*  // for displaying the permutation
if (numberOf1s == expectedBinarySum) {
b = 1;
int aggregator = 0;
while(b <= x) {
if ((b & x) == b) {
aggregator++;
}
else {
System.out.print(aggregator);
System.out.print(" ");
aggregator = 0;
}
b = b << 1;
}
System.out.println(aggregator);
}
*/
return (numberOf1s == expectedBinarySum);
}

public static int makeNumber(int z) {
if (z > 45) {
return 0;
}
long maxToCheck = Math.round(Math.pow(2, z + 4)) - 1;
int count = 0;
for (long n = 1; n <= maxToCheck; n++) {
if (checkPermutationBinary(n, z)) {
count++;
}
}
return count;
//return (int)Math.round(Math.pow(2, z - 1)); // only for z < 6
}

public static void main(String[] args) {
System.out.println(makeNumber(6));
}
}
``````

I did the challenge with a branching recursion (which I did by having recursion in a loop).
I made lots of little helper functions. And one big recursive function.

I actually generated all the relevant permutations as arrays (which wasn’t really necessary) and created a method to display them.
I did some object-oriented stuff: using an object to store information that didn’t change, and to store the `ArrayList` of all the possible permutations.
(Note that these permutations are actually permutations with repetition, not the classic permutations usually seen in a Probability class).

My code is quite long.

my code
``````import java.util.Arrays;
import java.util.ArrayList;

public class NumberPermutation {

public int requiredTotal; // number z
public ArrayList<int[]> list; // to store the permutations
public int maxRecursionDepth = 5; // permuation of 5 numbers at most.

public NumberPermutation() {
this.requiredTotal = 0;
this.list = new ArrayList<int[]>();
}
public NumberPermutation(int total) {
this.requiredTotal = total;
this.list = new ArrayList<int[]>();
}
public NumberPermutation(int total, ArrayList<int[]> listOfArrays) {
this.requiredTotal = total;
this.list = listOfArrays;
}
public NumberPermutation(int total, int initialCapacity) {
this.requiredTotal = total;
this.list = new ArrayList<int[]>(initialCapacity);
}

public void print() { // to display the permutations
for (int[] permutation : this.list) {
for (int x: permutation) {
System.out.print(x);
}
System.out.print(" ");
}
System.out.println();
}

private static int min(int a, int b) { return (a < b) ? a : b; }
private static int max(int a, int b) { return (a > b) ? a : b; }

private static int sum(int[] array) {
if (array == null) {
return 0;
}
int length = array.length;
int total = 0;
for (int i = 0; i < length; i++) {
total += array[i];
}
}

public static int[] combine(int[] array, int end) {
// makes new array by appending an integer at the end
if (array == null) {
return new int[]{end};
}
int length = array.length;
int[] combined = new int[length + 1];
for (int i = 0; i < length; i++) {
combined[i] = array[i];
}
combined[length] = end;
//length++;
return combined;
}

public int getPerms(int[] soFar, int minToUse, int maxToUse, int depth) {
/* soFar is array to store the numbers being considered
for the permutation so far
*/
int totalSoFar = sum(soFar);
// base cases:
if (depth > this.maxRecursionDepth) {
return 0;
}
if (totalSoFar > this.requiredTotal) {
return 0;
}
else if (totalSoFar == this.requiredTotal) {
return 1;
}
// recursive case:
else if (depth < this.maxRecursionDepth) {
int count = 0;
// recursion in a loop to do branching recursion:
for (int i = minToUse; i <= maxToUse; i++) {
if ((totalSoFar + i) == this.requiredTotal) {
count += 1;
break;
}
else if ((totalSoFar + i) < this.requiredTotal) {
int[] combined = combine(soFar, i);
int newMax = min(this.requiredTotal - totalSoFar, maxToUse);
// main recursion:
count += getPerms(combined, minToUse, newMax, depth + 1);
}
}
return count;
}
return 0;
}

public static int makeNumber(int z) {
if (z > 45 || z < 1) {
return 0;
}
NumberPermutation nPerm = new NumberPermutation(z);
/* nPerm.getPerms(soFar, minToUse, maxToUse, depth)
arguments:
starting with no permuations yet (so use null).
put min possible number in permutation as 1, or 9 - (45 - z),
whichever is higher.
max possible number in permutation is 9 or z,
whichever is lower.