Develop Reference

Levenshtein distance in java outputting the wrong number

For my university assignment in java I have been asked to provide "extra analytics functions" I decided to use Levenshtein distance but I have an issue where the number outputted to the console is one less than the actual answer. So the distance between "cat" and "hat" should be 1 but it's displaying as 0
public class Levenshtein {
public Levenshtein(String first, String second) {
char [] s = first.toCharArray();
char [] t = second .toCharArray();
int Subcost = 0;
int[][] array = new int[first.length()][second.length()];
for (int i = 0; i < array[0].length; i++)
{
array[0][i] = i;
}
for (int j = 0; j < array.length; j++)
{
array [j][0]= j;
}
for (int i = 1; i < second.length(); i++)
{
for (int j = 1; j < first.length(); j++)
{
if (s[j] == t [i])
{
Subcost = 0;
}
else
{
Subcost = 1;
}
array [j][i] = Math.min(array [j-1][i] +1,
Math.min(array [j][i-1] +1,
array [j-1][i-1] + Subcost) );
}
}
UI.output("The Levenshtein distance is -> " + array[first.length()-1][second.length()-1]);
}
}

Apparently you're using the following algorithm:
https://en.wikipedia.org/wiki/Levenshtein_distance#Iterative_with_full_matrix
I think you were not too accurate with indices. I'm not sure where exactly the problem is, but here is a working version:
public int calculateLevenshteinDistance(String first, String second) {
char[] s = first.toCharArray();
char[] t = second.toCharArray();
int substitutionCost = 0;
int m = first.length();
int n = second.length();
int[][] array = new int[m + 1][n + 1];
for (int i = 1; i <= m; i++) {
array[i][0] = i;
}
for (int j = 1; j <= n; j++) {
array[0][j] = j;
}
for (int j = 1; j <= n; j++) {
for (int i = 1; i <= m; i++) {
if (s[i - 1] == t[j - 1]) {
substitutionCost = 0;
} else {
substitutionCost = 1;
}
int deletion = array[i - 1][j] + 1;
int insertion = array[i][j - 1] + 1;
int substitution = array[i - 1][j - 1] + substitutionCost;
int cost = Math.min(
deletion,
Math.min(
insertion,
substitution));
array[i][j] = cost;
}
}
return array[m][n];
}

How to use LSD String Sort without having to enter a fixed length?

sorry for my bad english. I'm styding LSD String Sorts algorithm and I have a question related to it. Here my code. I want input W not fixed, for example:
String[] a = {"38A", "3TW723", "2IYEA938", "3CI34780720"};
public static void sort(String[] a, int w) { // Sort a[] on leading W characters.
int R = 256;
int N = a.length;
//For each of the character from right to left
for (int d = w - 1; d >= 0; d--) {
//1. count the frequencies
int[] count = new int[R + 1];
for (int i = 0; i < N; i++) {
count[a[i].charAt(d) + 1]++;
}
//2. Transform counts to indices
for (int r = 0; r < R; r++) {
count[r + 1] += count[r];
}
//3. Distribute
String aux[] = new String[N];
for (int i = 0; i < N; i++) {
aux[count[a[i].charAt(d)]] = a[i];
count[a[i].charAt(d)]++;
}
//4. Copyback
System.arraycopy(aux, 0, a, 0, N);
}
}

To develop an implementation of LSD string sort that works for variable-length strings, we need to do many works on the base of your code. We need to find the longest length of string in string[] a, so when d >= a[i].length(), we return 0, which means we add extra 0 to make every string in the same length. This is my code.
// Develop an implementation of LSD string sort
// that works for variable-length strings.
public class LSDForVariableLengthStrings {
// do not instantiate
private LSDForVariableLengthStrings() { }
// find longest length string in string[] a.
public static int findLongestLength(String[] a) {
int longest = 0;
for (int i = 0; i < a.length; ++i) {
if (a[i].length() > longest) {
longest = a[i].length();
}
}
return longest;
}
// if d >= 0 && d < a[i].length(), return a[i].charAt(d);
// else , return 0, which means least value to sort.
public static int findCharAtInString(int i, int d, String[] a) {
if (d < 0 || d >= a[i].length()) {
return 0;
}
return a[i].charAt(d);
}
// Rearranges the array of variable-length strings.
public static void sort(String[] a) {
int n = a.length;
int R = 256; // extended ASCII alphabet size.
String[] aux = new String[n];
int w = findLongestLength(a); // w is the length of longest string in a.
for (int d = w - 1; d >= 0; d--) {
// sort by key-indexed counting on dth character
// compute frequency counts
int[] count = new int[R + 1];
for (int i = 0; i < n; ++i) {
int c = findCharAtInString(i, d, a);
count[c + 1]++;
}
// compute cumulates
for (int r = 0; r < R; ++r) {
count[r + 1] += count[r];
}
// move data
for (int i = 0; i < n; ++i) {
int c = findCharAtInString(i, d, a);
aux[count[c]++] = a[i];
}
// copy back
for (int i = 0; i < n; ++i) {
a[i] = aux[i];
}
}
}
public static void main(String[] args) {
String[] a = {"38A", "3TW723", "2IYEA938", "3CI34780720"};
int n = a.length;
// sort the strings
sort(a);
// prints results
for (int i = 0; i < n; ++i) {
System.out.println(a[i]);
}
}
}

LSD sorts only fixed-length strings. Use MSD instead

Why does my Biginteger.multiply() shows NullPointerException?

I tried to initialize factorials from 1 to 1000 to an biginteger array and calculating the sum of the digits. Why this code showing java.lang.NullPointerException? I think everything was initialized correctly.
class Main {
public static void main(String[] args) {
BigInteger[] b = new BigInteger[1010];
int[] ara = new int[1010];
BigInteger c;
b[0] = BigInteger.ONE;
b[1] = BigInteger.ONE;
ara[0] = ara[1] = 1;
String s;
int l, sum;
for (int i = 2; i <= 1001; i++) {
c = b[i - 1];
b[i] = b[i].multiply(c);
s = b[i].toString();
l = s.length();
sum = 0;
for (int j = 0; j < l; j++) {
sum += Character.getNumericValue(s.charAt(j));
}
ara[i] = sum;
}

Problem :
b[i] = b[i].multiply(c);
And look at your b array which you initialised
b[0] = BigInteger.ONE;
b[1] = BigInteger.ONE;
And now look at the for loop
for (int i = 2; i <= 1001; i++) {
c = b[i - 1];
b[i] = b[i].multiply(c);
You have only 0,1 indexes. It will throw NPE for index 2.
You are trying loop on 1001 elements and there are only 2 elements inside your array. Fill the b array with zeros first.
Solution :
Change your for loop as below and keep everything same. It works.
for (int i = 2; i <= 1001; i++) {
b[i] = BigInteger.ONE;
c = b[i - 1];
b[i] = b[i].multiply(c);
s = b[i].toString();

The algorithm for factorial is to take some value n and multiply that n by n - 1 until the value 1 is arrived at. Your algorithm doesn't appear to do that (it generates ones). I think you wanted something like
int len = 1010;
BigInteger[] b = new BigInteger[len];
int[] ara = new int[len];
for (int i = 0; i < len; i++) {
// calculate factorial.
b[i] = BigInteger.valueOf(i + 1);
for (int j = i; j > 1; j--) {
b[i] = b[i].multiply(BigInteger.valueOf(j));
}
// now sum digits.
for (char ch : b[i].toString().toCharArray()) {
ara[i] += Character.getNumericValue(ch);
}
}

Suffix array O(NlogN) implementation

I'm looking into the specific O(NlogN) implementation of suffix array found at this link : https://sites.google.com/site/indy256/algo/suffix_array
I'm able to understand the core concepts but understanding the implementation in its entirety is a problem.
public static int[] suffixArray(CharSequence S) {
int n = S.length();
Integer[] order = new Integer[n];
for (int i = 0; i < n; i++)
order[i] = n - 1 - i;
// stable sort of characters
Arrays.sort(order, (a, b) -> Character.compare(S.charAt(a), S.charAt(b)));
int[] sa = new int[n];
int[] classes = new int[n];
for (int i = 0; i < n; i++) {
sa[i] = order[i];
classes[i] = S.charAt(i);
}
// sa[i] - suffix on i'th position after sorting by first len characters
// classes[i] - equivalence class of the i'th suffix after sorting by first len characters
for (int len = 1; len < n; len *= 2) {
int[] c = classes.clone();
for (int i = 0; i < n; i++) {
// condition sa[i - 1] + len < n simulates 0-symbol at the end of the string
// a separate class is created for each suffix followed by simulated 0-symbol
classes[sa[i]] = i > 0 && c[sa[i - 1]] == c[sa[i]] && sa[i - 1] + len < n && c[sa[i - 1] + len / 2] == c[sa[i] + len / 2] ? classes[sa[i - 1]] : i;
}
// Suffixes are already sorted by first len characters
// Now sort suffixes by first len * 2 characters
int[] cnt = new int[n];
for (int i = 0; i < n; i++)
cnt[i] = i;
int[] s = sa.clone();
for (int i = 0; i < n; i++) {
// s[i] - order of suffixes sorted by first len characters
// (s[i] - len) - order of suffixes sorted only by second len characters
int s1 = s[i] - len;
// sort only suffixes of length > len, others are already sorted
if (s1 >= 0)
sa[cnt[classes[s1]]++] = s1;
}
}
return sa;
}
I'm wondering about the use of cnt[] array and places it is useful.
Any pointers would be helpful.
Thanks.

How to generate combinations obtained by permuting 2 positions in Java

I have this problem, I need to generate from a given permutation not all combinations, but just those obtained after permuting 2 positions and without repetition. It's called the region of the a given permutation, for example given 1234 I want to generate :
2134
3214
4231
1324
1432
1243
the size of the region of any given permutation is , n(n-1)/2 , in this case it's 6 combinations .
Now, I have this programme , he does a little too much then what I want, he generates all 24 possible combinations :
public class PossibleCombinations {
public static void main(String[] args) {
Scanner s=new Scanner(System.in);
System.out.println("Entrer a mumber");
int n=s.nextInt();
int[] currentab = new int[n];
// fill in the table 1 TO N
for (int i = 1; i <= n; i++) {
currentab[i - 1] = i;
}
int total = 0;
for (;;) {
total++;
boolean[] used = new boolean[n + 1];
Arrays.fill(used, true);
for (int i = 0; i < n; i++) {
System.out.print(currentab[i] + " ");
}
System.out.println();
used[currentab[n - 1]] = false;
int pos = -1;
for (int i = n - 2; i >= 0; i--) {
used[currentab[i]] = false;
if (currentab[i] < currentab[i + 1]) {
pos = i;
break;
}
}
if (pos == -1) {
break;
}
for (int i = currentab[pos] + 1; i <= n; i++) {
if (!used[i]) {
currentab[pos] = i;
used[i] = true;
break;
}
}
for (int i = 1; i <= n; i++) {
if (!used[i]) {
currentab[++pos] = i;
}
}
}
System.out.println(total);
}
}
the Question is how can I fix this programme to turn it into a programme that generates only the combinations wanted .

How about something simple like
public static void printSwapTwo(int n) {
int count = 0;
StringBuilder sb = new StringBuilder();
for(int i = 0; i < n - 1;i++)
for(int j = i + 1; j < n; j++) {
// gives all the pairs of i and j without repeats
sb.setLength(0);
for(int k = 1; k <= n; k++) sb.append(k);
char tmp = sb.charAt(i);
sb.setCharAt(i, sb.charAt(j));
sb.setCharAt(j, tmp);
System.out.println(sb);
count++;
}
System.out.println("total=" + count+" and should be " + n * (n - 1) / 2);
}