The task is to write the logic of checking the magnitude of the coincidence of the player's attempt with the hidden word.
More formally, let there be a string S — a hidden word and a string Q — a player's attempt.
Both strings have the same length N. For each position 1 ≤ i ≤ N of string Q, we need to calculate the type of match in this position with string S.
If Q[i] = S[i], then at position i the match type should be equal to "correct".
If Q[i]≠S[i], but there is another position 1 ≤ j≤ N such that Q[i] = S[j], then in position i the match type must be equal to "present".
Each letter of the string S can be used in no more than one match of
the type "correct" or "present".
Priority is always given to the "correct" type.
Of all possible use cases in the "present" type, the program selects
the leftmost position in the Q string.
In other positions, the match type must be equal to "absent".
Input format:
The first line contains the string S (1≤ S ≤ 10^6) — the hidden word.
The second line contains the string Q ( Q = S) — the player's attempt.
It is guaranteed that the strings S and Q contain only uppercase Latin letters.
Example:
input:
COVER
CLEAR
output:
correct
absent
present
absent
correct
My program does this very slowly, how can I speed it up?
import java.util.*;
public class Task1 {
private final static String cor = "correct";
private final static String abs = "absent";
private final static String pre = "present";
static String[] stringArr;
static java.util.Map<Integer, Integer> map = new HashMap<>();
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String a = sc.nextLine();
String b = sc.nextLine();
stringArr = new String[a.length()];
int length1 = a.length();
char[] arr1 = a.toCharArray();
char[] arr2 = b.toCharArray();
for (int i = 0; i < length1; i++) {
if (arr2[i] == arr1[i]) {
stringArr[i] = cor;
map.put(i, i);
}
}
for (int i = 0; i < length1; i++) {
if (arr2[i] != arr1[i]) {
while (stringArr[i] == null) {
boolean finded = false;
for (int j = 0; j < arr1.length; j++) {
if (arr2[i] == arr1[j] && !map.containsKey(j)) {
stringArr[i] = pre;
finded = true;
map.put(j, j);
break;
}
}
if (!finded) stringArr[i] = abs;
}
}
}
for (String s : stringArr) {
System.out.println(s);
}
}
}
You can use an int array (or a HashMap<Char, Integer>) count to store those "presented" characters.
We need two for loops. The first loop will set all "correct" positions (if arr1[i] == arr2[i], then stringArr[i] = cor); and for those unmatched positions (arr1[i] != arr2[i]), we count the number of occurrences for those characters in the hidden word (count[arr1[i] - 'A']++).
Then in the second for loop, we check those unmatched positions with the help of count.
If count[arr2[i] - 'A'] > 0, it means that the hidden word contains the character arr2[i], so we can set stringArr[i] = pre and decrement count[arr2[i] - 'A'] (since each letter can be used once)
Otherwise, there's no matching letter in the hidden word, stringArr[i] should be set to abs.
import java.util.Scanner;
public class Task1 {
private final static String cor = "correct";
private final static String abs = "absent";
private final static String pre = "present";
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String a = sc.nextLine();
String b = sc.nextLine();
String[] stringArr = new String[a.length()];
int length1 = a.length();
char[] arr1 = a.toCharArray();
char[] arr2 = b.toCharArray();
int[] count = new int[26];
for (int i = 0; i < length1; i++) {
if (arr2[i] == arr1[i]) {
stringArr[i] = cor;
}
else {
count[arr1[i] - 'A']++;
}
}
for (int i = 0; i < length1; i++) {
if (arr1[i] != arr2[i]) {
if (count[arr2[i] - 'A'] > 0) {
stringArr[i] = pre;
count[arr2[i] - 'A']--;
}
else {
stringArr[i] = abs;
}
}
}
for (String s : stringArr) {
System.out.println(s);
}
}
}
This can be done using hashmaps and taking advantage of the o(1) insert/delete time.
I'm assuming that each word is exactly the same length. Consequently, to get the answer, we need to check each letter in each word. If they have a length of n, then that means at a minimum it will take o(2n) == o(n) to find the answer.
[1] Then, if you're able to use extra space, I would just use a hashmap as an index, where the key is a character in secret, and the value is the set of indices that letter occurs in secret. So it would look something like this:
//cover
c:<0>
o:<1>
etx...
[2] Then check each letter in guess against the index.
If there is no key for a letter, the put absent
If there is a key, then look in the set of indices to see if there is a matching index. Since I used a hashset, that's another o(1) lookup
So sum total, you walk through secret and guess once each which is o(n) runtime. That, plus o(1) for each index lookup produces o(n) runtime.
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
public class Task1 {
public static void main(String[] args)
{
ArrayList<String> res = solve("COVER", "CLEAR");
System.out.println(res);
}
public static ArrayList<String> solve(String secret, String guess){
ArrayList<String> result = new ArrayList<>();
HashMap<Character, HashSet<Integer>> index = new HashMap<>();
//shortcut
if(secret.equals(guess)){
for(int i = 0; i < secret.length(); i++) result.add("correct");
return result;
}
//make the index of char:<indices> for the chars in "secret"
for(int i = 0; i < secret.length(); i++){
char c = secret.charAt(i);
HashSet<Integer> bucket = index.get(c);
if(bucket == null){
bucket = new HashSet<>();
index.put(c, bucket);
}
bucket.add(i);
}
//check each char in "guess" against the index and tally the result
for(int i = 0; i < guess.length(); i++){
char c = guess.charAt(i);
HashSet<Integer> bucket = index.get(c);
if(bucket == null) result.add("absent");
else if(bucket.contains(i)) result.add("correct");
else result.add("present");
}
return result;
}
}
I tried to code a program to detect an anagram with 2 Strings given.
My approach is to convert both strings to char Arrays and then sort them before comparing them.
I know I could use the sort() function but I don't want to use any imports for training purposes.
The problem is i want my programm to ignore blanks while scanning for an anagram.
in the current version the ouput is like this:
(triangle, relating) ---> true
(tri angle, relating) ---> false
while it should be both true.
i would be thankful for any help!
heres my code, (please ignore my comments):
public static boolean anagramCheck(String a, String b) {
boolean r = true;
// In Char Arrays umwandeln /
char[] Ca = a.toCharArray();
char[] Cb = b.toCharArray();
// Laengen Abfrage
int L1 = Ca.length;
int L2 = Cb.length;
// Erste For-Schleife
for (int i = 0; i < L1; i++) {
for (int j = i + 1; j < L1; j++) {
if (Ca[j] < Ca[i]) {
char temp = Ca[i];
Ca[i] = Ca[j];
Ca[j] = temp;
}
}
}
// Zweite For-schleife
for (int i = 0; i < L2; i++) {
for (int j = i + 1; j < L2; j++) {
if (Cb[j] < Cb[i]) {
char temp = Cb[i];
Cb[i] = Cb[j];
Cb[j] = temp;
}
}
}
// Char Arrays zu Strings
String S1 = String.valueOf(Ca);
String S2 = String.valueOf(Cb);
// Vergleich und Ausgabe
if (S1.compareTo(S2) == 0) {
return r;
}
else {
r = false;
return r;
}
}
}
The String.replace(String, String) is the non-regexp replace method.
So remove all spaces:
String S1 = String.valueOf(Ca).replace(" ", "");
String S2 = String.valueOf(Cb).replace(" ", "");
It would be nicer to do this on a and b.
public static boolean isAnagram(String one, String two) {
char[] letters = new char[26];
for (int i = 0; i < one.length(); i++) {
char ch = Character.toLowerCase(one.charAt(i));
if (Character.isLetter(ch))
letters[ch - 'a']++;
}
for (int i = 0; i < two.length(); i++) {
char ch = Character.toLowerCase(two.charAt(i));
if (Character.isLetter(ch))
letters[ch - 'a']--;
}
for (int i = 0; i < letters.length; i++)
if (letters[i] != 0)
return false;
return true;
}
Generally, less code is better (if it’s readable). And learning a language means learning the built in libraries.
Here’s a method to return a String of sorted chars:
public static String sortChars(String str) {
return str.replace(" ", "").chars().sorted()
.mapToObj(c -> (char)c + "")
.collect(Collectors.joining(""));
}
With this method, your main method becomes:
public static boolean anagramCheck(String a, String b) {
return sortedChars(a).equals(sortedChars(b));
}
Refactoring like this, using well-named methods makes your code easier to understand, test, debug and maintain.
It’s worth noting that you don’t actually need a sorted String… a sorted array would serve equally well, and requires less code:
public static int[] sortChars(String str) {
return str.replace(" ", "").chars().sorted().toArray();
}
public static boolean anagramCheck(String a, String b) {
return Arrays.equal(sortedChars(a), sortedChars(b));
}
A frequency map could be created with the characters from String one incrementing and the characters from String two decrementing, then the resulting map should contain only 0 as values.
To skip non-letters, Character::isLetter can be used.
public static boolean isAnagram(String a, String b) {
Map<Character, Integer> frequencies = new HashMap<>();
for (int i = 0, na = a.length(), nb = b.length(), n = Math.max(na, nb); i < n; i++) {
if (i < na && Character.isLetter(a.charAt(i)))
frequencies.merge(Character.toLowerCase(a.charAt(i)), 1, Integer::sum);
if (i < nb && Character.isLetter(b.charAt(i)))
frequencies.merge(Character.toLowerCase(b.charAt(i)), -1, Integer::sum);
}
return frequencies.values().stream().allMatch(x -> x == 0);
}
I am solving this question as an assignment of the school. But the two of my test cases are coming out wrong when I submit the code? I don't know what went wrong. I have checked various other test cases and corner cases and it all coming out right.
Here is my code:
public static boolean isPermutation(String input1, String input2) {
if(input1.length() != input2.length())
{
return false;
}
int index1 =0;
int index2 =0;
int count=0;
while(index2<input2.length())
{
while(index1<input1.length())
{
if( input1.charAt(index1)==input2.charAt(index2) )
{
index1=0;
count++;
break;
}
index1++;
}
index2++;
}
if(count==input1.length())
{
return true;
}
return false;
}
SAMPLE INPUT
abcde
baedc
output
true
SAMPLE INPUT
abc
cbd
output
false
A simpler solution would be to sort the characters in both strings and compare those character arrays.
String.toCharArray() returns an array of characters from a String
Arrays.sort(char \[\]) to sort a character array
Arrays.equals(char \[\], char \[\]) to compare the arrays
Example
public static void main(String[] args) {
System.out.println(isPermutation("hello", "olleh"));
System.out.println(isPermutation("hell", "leh"));
System.out.println(isPermutation("world", "wdolr"));
}
private static boolean isPermutation(String a, String b) {
char [] aArray = a.toCharArray();
char [] bArray = b.toCharArray();
Arrays.sort(aArray);
Arrays.sort(bArray);
return Arrays.equals(aArray, bArray);
}
A more long-winded solution without sorting would to be check every character in A is also in B
private static boolean isPermutation(String a, String b) {
char[] aArray = a.toCharArray();
char[] bArray = b.toCharArray();
if (a.length() != b.length()) {
return false;
}
int found = 0;
for (int i = 0; i < aArray.length; i++) {
char eachA = aArray[i];
// check each character in A is found in B
for (int k = 0; k < bArray.length; k++) {
if (eachA == bArray[k]) {
found++;
bArray[k] = '\uFFFF'; // clear so we don't find again
break;
}
}
}
return found == a.length();
}
You have two ways to proceed
Sort both strings and then compare both strings
Count the characters in string and then match.
Follow the tutorial here
In case you String is ASCII you may use the next approach:
Create 256 elements int array
Increment element of corresponding character whenever it's found in string1
Decrement element of corresponding character whenever it's found in string2
If all elements are 0, then string2 is permutation of string1
Overall complexity of this approach is O(n). The only drawback is space allocation for charCount array:
public static boolean isPermutation(String s1, String s2) {
if (s1.length() != s2.length()) {
return false;
}
int[] charCount = new int[256];
for (int i = 0; i < s1.length(); i++) {
charCount[s1.charAt(i)]++;
charCount[s2.charAt(i)]--;
}
for (int i = 0; i < charCount.length; i++) {
if (charCount[i] != 0) {
return false;
}
}
return true;
}
If your strings can hold non-ASCII values, the same approach could be implemented using HashMap<String, Integer> as character count storage
I have a recursive method to solve the permutations problem. I think that this code will seem to be tough but if you will try to understand it you will see the beauty of this code. Recursion is always hard to understand but good to use! This method returns all the permutations of the entered String 's' and keeps storing them in the array 'arr[]'. The value of 't' initially is blank "" .
import java.io.*;
class permute_compare2str
{
static String[] arr= new String [1200];
static int p=0;
void permutation(String s,String t)
{
if (s.length()==0)
{
arr[p++]=t;
return;
}
for(int i=0;i<s.length();i++)
permutation(s.substring(0,i)+s.substring(i+1),t+s.charAt(i));
}
public static void main(String kr[])throws IOException
{
int flag = 0;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the first String:");
String str1 = br.readLine();
System.out.println("Enter the second String:");
String str2 = br.readLine();
new permute_compare2str().permutation(str1,"");
for(int i = 0; i < p; ++i)
{
if(arr[i].equals(str2))
{
flag = 1;
break;
}
}
if(flag == 1)
System.out.println("True");
else
{
System.out.println("False");
return;
}
}
}
One limitation that I can see is that the length of the array is fixed and so will not be able to return values for a large String value 's'. Please alter the same as per the requirements. There are other solution to this problem as well.
I have shared this code because you can actually use this to get the permutations of a string printed directly without the array as well.
HERE:
void permutations(String s,String t)
{
if (s.length()==0)
{
System.out.print(t+" ");
return;
}
for(int i=0;i<s.length();i++)
permutations(s.substring(0,i)+s.substring(i+1),t+s.charAt(i));
}
Value of 's' is the string whose permutations is needed and value of 't' is again empty "".
it will take O(n log n) cuz i sort each string and i used more space to store each one
public static boolean checkPermutation(String a,String b){
return sortString(a).equals(sortString(b));
}
private static String sortString(String test) {
char[] tempChar = test.toCharArray();
Arrays.sort(tempChar);
return new String(tempChar);
}
String checkPermutation(String a,String b){
char[] aArr = a.toCharArray();
char[] bArr = b.toCharArray();
Arrays.sort(aArr);
Arrays.sort(bArr);
if(aArr.length != bArr.length){
return "NO";
}
int p = 0, q = 0;
while(p < aArr.length){
if(aArr[p] != bArr[q]){
return "NO";
}
p++;
q++;
}
return "YES";
}
public static boolean isStringArePermutate(String s1, String s2){
if(s1==null || s2==null) {
return false;
}
int len1 = s1.length();
int len2 =s2.length();
if(len1!=len2){
return false;
}
for(int i =0;i<len1;i++){
if(!s1.contains(String.valueOf(s2.charAt(i)))) {
return false;
}
s1=s1.replaceFirst(String.valueOf(s2.charAt(i)), "");
}
if(s1.equals("")) {
return true;
}
return false;
}
How to determine if two strings are permutations of each other
Sort the two strings's characters.
Compare the results to see if they're identical.
Edit:
The above method is reasonably efficient - O(n*log(n)) and, as others have shown, very easy to implement using the standard Java API. Even more efficient (but also more work) would be counting and comparing the occurrence of each character, using the char value as index into an array of counts.
I do not thing there is an efficient way to do it recursively. An inefficient way (O(n^2), worse if implemented straightforwardly) is this:
If both strings consist of one identical character, return true
Otherwise:
remove one character from the first string
Look through second string for occurrence of this character
If not present, return false
Otherwise, remove said character and apply algorithm recursively to the remainders of both strings.
To put #Michael Borgwardt's words in to code:
public boolean checkAnagram(String str1, String str2) {
if (str1.length() != str2.length())
return false;
char[] a = str1.toCharArray();
char[] b = str2.toCharArray();
Arrays.sort(a);
Arrays.sort(b);
return Arrays.equals(a, b);
}
Create a Hashmap with the characters of the first string as keys and the number of occurances as value; then go through the second string and for each character, look up the hash table and decrement the number if it is greater than zero. If you don't find an entry or if it is already 0, the strings are not a permutation of each other. Obviously, the string must have the same length.
Linear Time solution in HashMap. Traverse and put first String in HashMap, keep the count of each character. Traverse second String and if it is already in the hashmap decrease the count by 1. At the end if all character were in the string the value in hashmap will be 0 for each character.
public class StringPermutationofEachOther {
public static void main(String[] args)
{
String s1= "abc";
String s2 ="bbb";
System.out.println(perm(s1,s2));
}
public static boolean perm(String s1, String s2)
{ HashMap<Character, Integer> map = new HashMap<Character, Integer>();
int count =1;
if(s1.length()!=s2.length())
{
return false;
}
for(Character c: s1.toCharArray())
{
if(!map.containsKey(c))
map.put(c, count);
else
map.put(c, count+1);
}
for(Character c: s2.toCharArray())
{
if(!map.containsKey(c))
return false;
else
map.put(c, count-1);
}
for(Character c: map.keySet())
{
if(map.get(c)!=0)
return false;
}
return true;
}
}
You can try to use XOR, if one string is a permeation of the other, they should have essentially identical chars. The only difference is just the order of chars. Therefore using XOR trick can help you get rid of the order and focus only on the chars.
public static boolean isPermutation(String s1, String s2){
if (s1.length() != s2.length()) return false;
int checker = 0;
for(int i = 0; i < s1.length();i++ ){
checker ^= s1.charAt(i) ^ s2.charAt(i);
}
return checker == 0;
}
Sort the 2 strings by characters and compare if they're the same (O(n log n) time, O(n) space), or
Tally the character frequency of the 2 strings and compare if they're the same (O(n) time, O(n) space).
You might take a look at String.toCharArray and Arrays.sort
First you check the lengths (n), if they are not same, they cannot be permutations of each other. Now create two HashMap<Character, Integer>. Iterate over each string and put the number of times each character occur in the string. E.g. if the string is aaaaa, the map will have just one element with key a and value 5. Now check if the two maps are identical. This is an O(n) algorithm.
EDIT with code snippet :
boolean checkPermutation(String str1, String str2) {
char[] chars1 = str1.toCharArray();
char[] chars2 = str2.toCharArray();
Map<Character, Integer> map1 = new HashMap<Character, Integer>();
Map<Character, Integer> map2 = new HashMap<Character, Integer>();
for (char c : chars1) {
int occ = 1;
if (map1.containsKey(c) {
occ = map1.get(c);
occ++;
}
map1.put(c, occ);
}
// now do the same for chars2 and map2
if (map1.size() != map2.size()) {
return false;
}
for (char c : map1.keySet()) {
if (!map2.containsKey(c) || map1.get(c) != map2.get(c)) {
return false;
}
}
return true;
}
I'm working on a Java library that should simplify your task. You can re-implement this algorithm using only two method calls:
boolean arePermutationsOfSameString(String s1, String s2) {
s1 = $(s1).sort().join();
s2 = $(s2).sort().join();
return s1.equals(s2);
}
testcase
#Test
public void stringPermutationCheck() {
// true cases
assertThat(arePermutationsOfSameString("abc", "acb"), is(true));
assertThat(arePermutationsOfSameString("bac", "bca"), is(true));
assertThat(arePermutationsOfSameString("cab", "cba"), is(true));
// false cases
assertThat(arePermutationsOfSameString("cab", "acba"), is(false));
assertThat(arePermutationsOfSameString("cab", "acbb"), is(false));
// corner cases
assertThat(arePermutationsOfSameString("", ""), is(true));
assertThat(arePermutationsOfSameString("", null), is(true));
assertThat(arePermutationsOfSameString(null, ""), is(true));
assertThat(arePermutationsOfSameString(null, null), is(true));
}
PS
In the case you can clone the souces at bitbucket.
I did this, and it works well and quickly:
public static boolean isPermutation(String str1, String str2)
{
char[] x = str1.toCharArray();
char[] y = str2.toCharArray();
Arrays.sort(x);
Arrays.sort(y);
if(Arrays.equals(x, y))
return true;
return false;
}
public boolean isPermutationOfOther(String str, String other){
if(str == null || other == null)
return false;
if(str.length() != other.length())
return false;
Map<Character, Integer> characterCount = new HashMap<Character, Integer>();
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
int count = 1;
if(characterCount.containsKey(c)){
int k = characterCount.get(c);
count = count+k;
}
characterCount.put(c, count);
}
for (int i = 0; i < other.length(); i++) {
char c = other.charAt(i);
if(!characterCount.containsKey(c)){
return false;
}
int count = characterCount.get(c);
if(count == 1){
characterCount.remove(c);
}else{
characterCount.put(c, --count);
}
}
return characterCount.isEmpty();
}
Variation on other approaches but this one uses 2 int arrays to track the chars, no sorting, and you only need to do 1 for loop over the strings. The for loop I do over the int arrays to test the permutation is a constant, hence not part of N.
Memory is constant.
O(N) run time.
// run time N, no sorting, assume 256 ASCII char set
public static boolean isPermutation(String v1, String v2) {
int length1 = v1.length();
int length2 = v2.length();
if (length1 != length2)
return false;
int s1[] = new int[256];
int s2[] = new int[256];
for (int i = 0; i < length1; ++i) {
int charValue1 = v1.charAt(i);
int charValue2 = v2.charAt(i);
++s1[charValue1];
++s2[charValue2];
}
for (int i = 0; i < s1.length; ++i) {
if (s1[i] != s2[i])
return false;
}
return true;
}
}
Unit Tests
#Test
public void testIsPermutation_Not() {
assertFalse(Question3.isPermutation("abc", "bbb"));
}
#Test
public void testIsPermutation_Yes() {
assertTrue(Question3.isPermutation("abc", "cba"));
assertTrue(Question3.isPermutation("abcabcabcabc", "cbacbacbacba"));
}
If we do the initial length check, to determine if the two strings are of the same length or not,
public boolean checkIfPermutation(String str1, String str2) {
if(str1.length()!=str2.length()){
return false;
}
int str1_sum = getSumOfChars(str1);
int str2_sum = getSumOfChars(str2);
if (str1_sum == str2_sum) {
return true;
}
return false;
}
public int getSumOfChars(String str){
int sum = 0;
for (int i = 0; i < str.length(); i++) {
sum += str.charAt(i);
}
return sum;
}
The obligatory Guava one-liner:
boolean isAnagram(String s1, String s2) {
return ImmutableMultiset.copyOf(Chars.asList(s1.toCharArray())).equals(ImmutableMultiset.copyOf(Chars.asList(s2.toCharArray())));
}
(Just for fun. I don't recommend submitting this for your assignment.)
As you requested, here's a complete solution using recursion.
Now all you have to do is:
Figure out what language this is
Translate it to Java.
Good luck :-)
proc isAnagram(s1, s2);
return {s1, s2} = {''} or
(s2 /= '' and
(exists c = s1(i) |
s2(1) = c and
isAnagram(s1(1..i-1) + s1(i+1..), s2(2..))));
end isAnagram;
I did it using C#
bool Arepermutations(string string1, string string2)
{
char[] str1 = string1.ToCharArray();
char[] str2 = string2.ToCharArray();
if (str1.Length !=str2.Length)
return false;
Array.Sort(str1);
Array.Sort(str2);
if (str1.Where((t, i) => t!= str2[i]).Any())
{
return false;
}
return true;
}
public boolean permitation(String s,String t){
if(s.length() != t.length()){
return false;
}
int[] letters = new int[256];//Assumes that the character set is ASCII
char[] s_array = s.toCharArray();
for(char c:s_array){ /////count number of each char in s
letters[c]++;
}
for(int i=0;i<t.length();i++){
int c = (int)t.charAt(i);
if(--letters[c]<0){
return false;
}
}
return true;
}
import java.io.*;
public class permute
{
public static String sort(String s)
{
char[] str = s.toCharArray();
java.util.Arrays.sort(str);
return new String(str);
}
public static boolean permutation(String s,String t)
{
if(s.length()!=t.length())
{
return false;
}
return sort(s).equals(sort(t));
}
public static void main(String[] args) throws IOException
{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String string = null;
boolean x=true;
System.out.println("Input String:");
string = bf.readLine();
System.out.println("Input another String:");
String result = bf.readLine();
String resultant = sort(string);
if(x==permutation(result,resultant))
{
System.out.println("String"+" "+"("+result+")"+"is a permutation of String"+" "+"("+string+")");
}
else
{
System.out.println("Sorry No anagram found");
}
}
}
public class TwoStrgPermutation {
public int checkForUnique(String s1, String s2)
{
int[] array1 = new int[256];
int[] array2 = new int[256];
array1 = arrayStringCounter(array1,s1);
array2 = arrayStringCounter(array2,s2);
if(Arrays.equals(array1, array2))
return 0;
else
return 1;
}
public int[] arrayStringCounter(int[] array,String s)
{
int val;
for(int i=0;i<s.length();i++)
{
val = (int)s.charAt(i);
array[val]++;
}
return array;
}
public static void main(String[] args) {
// TODO Auto-generated method stub
InputStreamReader in = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(in);
TwoStrgPermutation obj = new TwoStrgPermutation();
try {
String string1 = br.readLine();
String string2 = br.readLine();
int len1 = string1.length();
int len2 = string2.length();
if(len1==len2)
{
int result = obj.checkForUnique(string1,string2);
if (result == 0){
System.out.println("yes it is a permutation");
}
else if (result >0)
{
System.out.println("no it is not");
}
}
else
{
System.out.println("no it is not");
}
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
>>>def isPermutation = lambda x, y: set([i for i in x]) == set([j for j in x])
>>>isPermutation("rasp", "spar")
>>>True
This is an O(N) solution, in which N is the length of the shorter string.
It has a disadvantage, which is that only ASCII characters are acceptable.
If we want better applicability, we may substitute a hash table for the int charNums[].
But it also means C wouldn't be a good choice, coz' there is no standard hash table implementation for C.
int is_permutation(char *s1, char *s2)
{
if ((NULL == s1) ||
(NULL == s2)) {
return false;
}
int static const
_capacity = 256; // Assumption
int
charNums[_capacity] = {0};
char
*cRef1 = s1,
*cRef2 = s2;
while ('\0' != *cRef1 && '\0' != *cRef2) {
charNums[*cRef1] += 1;
charNums[*cRef2] -= 1;
cRef1++;
cRef2++;
}
if ('\0' != *cRef1 || '\0' != *cRef2) {
return false;
}
for (int i = 0; i < _capacity; i++) {
if (0 != charNums[i]) {
return false;
}
}
return true;
}
Create two methods:
1. First method takes a string and returns a sorted string:
public String sort(String str) {
char char_set[] = str.toCharArray();
Arrays.sort(char_set);
return new String(char_set);
}
2. Second method takes two strings and return a boolean:
`public boolean sort(String x, String y) {
if (x.length() != y.length()) {
System.out.println("false");
return false;
}
System.out.println(sort(x).equals(sort(y)));
return sort(x).equals(sort(y));
}`
It can be done by using a Dictionary in C#. A basic implementation is like :
private static bool IsPermutation(string str1, string str2)
{
if (str1.Length != str2.Length)
return false;
var dictionary = new Dictionary<char, int>();
foreach(var x in str1)
{
if (dictionary.ContainsKey(x))
dictionary[x] += 1;
else
dictionary.Add(x, 1);
}
foreach (var y in str2)
{
if (dictionary.ContainsKey(y))
{
if (dictionary[y] > 0)
dictionary[y] -= 1;
else
return false;
}
else
return false;
}
foreach(var el in dictionary)
{
if (el.Value != 0)
return false;
}
return true;
}
Time Complexity is O(n), linear solution.
public static boolean isPermutation(String s1, String s2) {
if (s1.length() != s2.length()) {
return false;
}else if(s1.length()==0 ){
return true;
}
else if(s1.length()==1){
return s1.equals(s2);
}
char[] s = s1.toCharArray();
char[] t = s2.toCharArray();
for (int i = 0; i < s.length; i++) {
for (int j = 0; j < t.length; j++) {
if (s.length == s1.length() && (i == 0 && j == t.length - 1) && s[i] != t[j]) {
return false;
}
if (s[i] == t[j]) {
String ss = new String(s);
String tt = new String(t);
s = (ss.substring(0, i) + ss.substring(i + 1, s.length)).toCharArray();
t = (tt.substring(0, j) + tt.substring(j + 1, t.length)).toCharArray();
System.out.println(new String(s));
System.out.println(new String(t));
i = 0;
j = 0;
}
}
}
return s[0]==t[0] ;
}
This solution works for any charset. With an O(n) complexity.
The output for: isPermutation("The Big Bang Theory", "B B T Tehgiangyroeh")
he Big Bang Theory
B B Tehgiangyroeh
e Big Bang Theory
B B Tegiangyroeh
Big Bang Theory
B B Tgiangyroeh
Big Bang Theory
BB Tgiangyroeh
ig Bang Theory
B Tgiangyroeh
g Bang Theory
B Tgangyroeh
Bang Theory
B Tangyroeh
Bang Theory
B Tangyroeh
Bng Theory
B Tngyroeh
Bg Theory
B Tgyroeh
B Theory
B Tyroeh
BTheory
BTyroeh
Bheory
Byroeh
Beory
Byroe
Bory
Byro
Bry
Byr
By
By
B
B
true
Best Way to do this is by sorting the two strings first and then compare them. It's not the most efficient way but it's clean and is bound to the runtime of the sorting routine been used.
boolean arePermutation(String s1, String s2) {
if(s1.lenght() != s2.lenght()) {
return false;
}
return mySort(s1).equals(mySort(s2));
}
String mySort(String s) {
Char letters[] = s.toCharArray();
Arrays.sort(letters);
return new String(letters);
}
String str= "abcd";
String str1 ="dcazb";
int count=0;
char s[]= str.toCharArray();
char s1[]= str1.toCharArray();
for(char c:s)
{
count = count+(int)c ;
}
for(char c1:s1)
{
count=count-(int)c1;
}
if(count==0)
System.out.println("String are Anagram");
else
System.out.println("String are not Anagram");
}
Here is a simple program I wrote that gives the answer in O(n) for time complexity and O(1) for space complexity. It works by mapping every character to a prime number and then multiplying together all of the characters in the string's prime mappings together. If the two strings are permutations then they should have the same unique characters each with the same number of occurrences.
Here is some sample code that accomplishes this:
// maps keys to a corresponding unique prime
static Map<Integer, Integer> primes = generatePrimes(255); // use 255 for
// ASCII or the
// number of
// possible
// characters
public static boolean permutations(String s1, String s2) {
// both strings must be same length
if (s1.length() != s2.length())
return false;
// the corresponding primes for every char in both strings are multiplied together
int s1Product = 1;
int s2Product = 1;
for (char c : s1.toCharArray())
s1Product *= primes.get((int) c);
for (char c : s2.toCharArray())
s2Product *= primes.get((int) c);
return s1Product == s2Product;
}
private static Map<Integer, Integer> generatePrimes(int n) {
Map<Integer, Integer> primes = new HashMap<Integer, Integer>();
primes.put(0, 2);
for (int i = 2; primes.size() < n; i++) {
boolean divisible = false;
for (int v : primes.values()) {
if (i % v == 0) {
divisible = true;
break;
}
}
if (!divisible) {
primes.put(primes.size(), i);
System.out.println(i + " ");
}
}
return primes;
}
Based on the comment on this solution, https://stackoverflow.com/a/31709645/697935 here's that approach, revised.
private static boolean is_permutation(String s1, String s2) {
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
int count = 1;
if(s1.length()!=s2.length()) {
return false;
}
for(Character c: s1.toCharArray()) {
if(!map.containsKey(c)) {
map.put(c, 1);
}
else {
map.put(c, map.get(c) + 1);
}
}
for(Character c: s2.toCharArray()) {
if(!map.containsKey(c)) {
return false;
}
else {
map.put(c, map.get(c) - 1);
}
}
for(Character c: map.keySet()) {
if(map.get(c) != 0) { return false; }
}
return true;
}
bool is_permutation1(string str1, string str2) {
sort(str1.begin(), str1.end());
sort(str2.begin(), str2.end());
for (int i = 0; i < str1.length(); i++) {
if (str1[i] != str2[i]) {
return false;
}
}
return true;
}
I wanted to give a recursive solution for this problem as I did not find any answer which was recursive. I think that this code seems to be tough but if you'll try to understand it you'll see the beauty of this code. Recursion is sometimes hard to understand but good to use! This method returns all the permutations of the entered String 's' and keeps storing them in the array 'arr[]'. The value of 't' initially is blank "" .
import java.io.*;
class permute_compare2str
{
static String[] arr= new String [1200];
static int p=0;
void permutation(String s,String t)
{
if (s.length()==0)
{
arr[p++]=t;
return;
}
for(int i=0;i<s.length();i++)
permutation(s.substring(0,i)+s.substring(i+1),t+s.charAt(i));
}
public static void main(String kr[])throws IOException
{
int flag = 0;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter the first String:");
String str1 = br.readLine();
System.out.println("Enter the second String:");
String str2 = br.readLine();
new permute_compare2str().permutation(str1,"");
for(int i = 0; i < p; ++i)
{
if(arr[i].equals(str2))
{
flag = 1;
break;
}
}
if(flag == 1)
System.out.println("True");
else
{
System.out.println("False");
return;
}
}
}
One limitation that I can see is that the length of the array is fixed and so will not be able to return values for a large String value 's'. Please alter the same as per the requirements. There are other solutions to this problem as well.
I have shared this code because you can actually use this to get the permutations of a string printed directly without the array as well.
HERE:
void permutations(String s, String t)
{
if (s.length()==0)
{
System.out.print(t+" ");
return;
}
for(int i=0;i<s.length();i++)
permutations(s.substring(0,i)+s.substring(i+1),t+s.charAt(i));
}
Value of 's' is the string whose permutations is needed and value of 't' is again empty "".
Reference: Introduction to Programming in Java