I'm looking for a way to compare two strings to each other character by character.
It should show whether there are other characters which don't appear in both strings.
Does someone have a solution for?
for(int j=0; j < min; j++) {
s1 = w1.substring(j,j+1);
s2 = w2.substring(j,j+1);
if (!s1.equalsIgnoreCase(s2) ){
counter++;
}
}
This only looks sequential for differences. But I want to find out whether there are differences between those two strings at all.
So **abc** and **cab** should count as a hit
I think what you want is to show that 2 strings have the same letters and the same amount of each letter. Use 2 hashsets, where the key is the character and the value is the number of occurrence in the string. you'll have one hashset for each string, then loop through the string add the characters to the set and compare to see if the sets are equal.
Turn them into charArrays and add the differences to an empty string as a for loop scans through them, for instance
String a = "abc";
char[] aa = a.toCharArray();
String b = "cba";
char[] bb = b.toCharArray();
String dif;
public void differ() {
for(int i = 0; i < aa.length - 1; i++) {
if(!aa[i].equals(bb[i])) {
dif += aa[i];
}
}
}
I believe running differ() would return "ac" because they both have b in the middle. Is this what you wanted?
Related
I am currently solving one of Leetcode's questions on strings. I need to check that all of my string literals are letters.
I have the attached code below:
for (int i = 0; i < s.length(); i++) {
char character = s.charAt(i);
if (!Character.isLetter(character))
return s;
lastIndex[character - 'a'] = i;
}
What is the time complexity of checking that a string contains all letters?
I think it's O(n) though since the check will be done for all the characters in the string.
Thanks.
In a given two words, is it possible to use regex to find multiple strings matching character as well index.
For example:
String1 = cat
String2 = carrot
the first 2 characters and indexes are matching (ca). t does not count because it is not in the same index.
I've tried for loop. However it appears to be not working and not very efficient.
for (int i = 0; i < string1.length(); i++){
for (int j = 0; j < string2.length(); j++){
char ch1 = string1.charAt(i);
char ch2 = string2.charAt(j);
if (ch1 == ch2) {
count char++;
}
}
What you are probably looking for is the longest common prefix.
See Find longest common prefix?
Regex is for pattern matching. It is a solution to a different problem.
For loop can still work for this job to find positions where each string has same char and number of times this occurs:
ArrayList<Integer> places = new ArrayList<Integer>();
for (int i = 0; i < Math.min(string1.length(), string2.length()); i++) {
a = string1.charAt(i);
b = string2.charAt(i);
if (a == b) {
count++;
places.add(i); //To say at which indices the 2 strings have the same chars
}
}
I guess you want to count the number of characters repeated at the same positions in two words. (Not same prefix)
In words cat carrot, you want to get 2 since c and a are in the same position, but t is not.
in words carrot cabra, you will get 3, since c, a and r (4th) are the same in the same position.
You only need to iterate one time over the two strings at the same time:
String string1 = "car";
String string2 = "carrot";
int minLength = Math.min( string1.length(), string2.length() );
int count = 0;
for (int i = 0; i < minLength; i++){
char ch1 = string1.charAt(i);
char ch2 = string2.charAt(i);
if (ch1 == ch2) {
count++;
}
}
We use minLength since we only need to check until the length of the smallest word.
We use string1.charAt(i) and string2.charAt(i), with same index i, since we want to check characters in the same position.
This is my assignment. I am not allowed to use if statements.
Write a program NumStrings.java that receives two strings through the command line as
input and then prints out the number of times the second string occurs as a substring in the
first.
My bad code:
public class Test {
public static void main(String[] args) {
String a = "HelloHelloHelloHello";
String b = "Hello";
int times = 0;
for(int i=0; i <= a.length()-5; i++){
for (int z=4; z<=(a.length()-1) && a.compareTo(b)==0; z++){
times = times +1;
}
}
System.out.print(times);
}
}
Here is the correct way to do it, using subString() (documentation here: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int)):
String a = "HelloHelloHelloHello";
String b = "Hello";
int times = 0;
for (int i = 0; i <= a.length() - b.length(); i++) {
String substring = a.subString(i, i + b.length());
if (substring.equals(b)) {
times = times + 1;
}
}
System.out.println(times);
And here is a way to do it without if statements... Which I don't recommend. But if you have to do it that way, this will work.
String a = "HelloHelloHelloHello";
String b = "Hello";
int times = 0;
for (int i = 0; i <= a.length() - b.length(); i++) {
String substring = a.substring(i, i + b.length());
for (int j = 0; substring.equals(b) && j < 1; j++) {
times = times + 1;
}
}
System.out.println(times);
Look at it this way: you don't have to count how often you find the second string in the first String, because you always have to check if you found it or not. So, to avoid all sorts of conditions or if statements, consider using firstString.split(secondString).
split(someString) will return you an array of remaining substrings once you "split" the base string everytime it finds your substring:
String first = "bababa";
String second = "a";
String[] substrings = first.split(second);
now substrings will look like this: ["b", "b", b"] because every a has been removed and the rest put in separate Strings.
Next you have to check the size of the array and you'll see how often your first String was split.
int count = substrings.length; // 3
However, this is not the end of it because we still have the following case:
String first = "bababaa";
With the above solution you would get an array of size 3: ["b", "b", "b"]. The last occurrence of a will only be removed without leaving any substring behind (not even an empty one '').
So you can take advantage of another (slightly different) split():
first.split(second, limit);
Where limit is the maximum number of occurrences the method tries to find. So how often can you find your second string in the first one? As many letters the first string has: int limit = first.length
first.split(second, first.length); // will produce [b, b, b, , ]
Can you see what happens? there are two empty strings at the end where there where two a. You get an array of substrings for everything that is found before or after the occurrence of the second String.
Naturally, when you split the string ba you would get ["b", ] so 2 substrings. But you don't care about the b just the "commas" in the middle (for every a a ,).
first.split(second, first.length).length -1; // that's how many commas there are, and thats how many second strings there are
EDIT
(thanks #saka1029 !) So, the "split" method still misses something when first="aaa" and second="aa" because it counts only 1 not 2 occurrences.
To correct that I thought of looping through the whole first string and checking only for the very first occurrence, and then removing the first letter and continuing (since OP already accepted another answer, I just post my code):
String first = "ZZZ";
String second = "ZZ";
int counter = 0; // counts the number of occurrences
int n = first.length(); // needs to be fixed, because we'll change the length of the first string in the loop
for(int i = 0; i < n; i++){ // check the first string letter by letter
String[] split = first.split(second, 2); // we want one substring and the rest (2 in total) like: ['', 'Z'] when we cut off the first 'ZZ'
counter += split.length - 1; // now add the number of occurrences (which will be either 0 or 1 in our case)
first = first.substring(1); // cut off the first letter of the first string and continue
}
System.out.println("counter = " + counter); // now we should get 3 for 'ZZZ'
I have a four letter string of letters, such as ASDF and I want to find all 3 (three) letter combinations using these letters and the three letter combination does not need to form a real word.Ex.
AAA AAS AAD AAF
ADA ADS ADD ADF
...............
SSA SSD SSF SSS
I am relatively new to Java, having just learned how to use the String class and using loops and conditional statements. The only way that I know how to do this is by a massive and very tedious set of for loops and if statements that account for every possibility that could arise. This would look like:
public static void main(String[] args)
{
String combo = "";
for(int counter = 1; counter <= 16; counter++){
combo = "A";
if(counter == 1){
combo = combo + "AA";
}
// This would continue on for all the possibilities starting with "A" and
// then move on to "S" as the lead character
}
}
I know that this is one of the worst ways to go about this problem, but I am really stuck as to how to do it another way. It would be easier if I had 3 letters and made the 3 letter combos, as then I could just get each letter from the array and just rearrange them, but as I'm only using 3 of the 4 letters it is more difficult. Any advice on how to get this done in an more efficient manner?
Use a recursive function.
Like this (not tested, don't have a Java compiler on my laptop).
Performance could probably be boosted by using StringBuilder.
static void printAllPossibilities(String charSet, int length) {
printAllPossibilities_(charSet, length, "");
}
static void printAllPossibilities_(String charSet, int length, String temp) {
if (length == 0) {
System.out.println(temp);
return;
}
for (int i = 0; i < charSet.length(); i++)
printAllPossibilities_(charSet, length - 1, temp + charSet.charAt(i));
}
Usage:
printAllPossibilities("ASDF", 4); // Print all 4-letter combinations out of "ASDF"
printAllPossibilities("bar", 2); // Print all 2-letter combinations out of "bar"
For the general case (all combinations of N chars out of M), the solution by #Qntm is exactly what you need... but, use a StringBuilder as he said, and just change the last character instead of constructing strings like 'temp + charSet.charAt(i)'.
If you need exactly 3 chars out of N, it's easier to just do 3 nested loops:
for (int char1 = 0; char1 < charSet.length(); char1++) {
for (int char2 = 0; char2 < charSet.length(); char2++) {
for (int char3 = 0; char3 < charSet.length(); char3++) {
System.out.println(""+charSet.charAt(char1)+charSet.charAt(char2)+charSet.charAt(char3));
}
}
}
This question already has answers here:
How to map character to numeric position in java?
(7 answers)
Closed 5 years ago.
How to get numeric position of alphabets in java ?
Suppose through command prompt i have entered abc then as a output i need to get 123 how can i get the numeric position of alphabets in java?
Thanks in advance.
String str = "abcdef";
char[] ch = str.toCharArray();
for(char c : ch){
int temp = (int)c;
int temp_integer = 96; //for lower case
if(temp<=122 & temp>=97)
System.out.print(temp-temp_integer);
}
Output:
123456
#Shiki for Capital/UpperCase letters use the following code:
String str = "DEFGHI";
char[] ch = str.toCharArray();
for(char c : ch){
int temp = (int)c;
int temp_integer = 64; //for upper case
if(temp<=90 & temp>=65)
System.out.print(temp-temp_integer);
}
Output:
456789
Another way to do this problem besides using ASCII conversions is the following:
String input = "abc".toLowerCase();
final static String alphabet = "abcdefghijklmnopqrstuvwxyz";
for(int i=0; i < input.length(); i++){
System.out.print(alphabet.indexOf(input.charAt(i))+1);
}
Convert each character to its ASCII code, subtract the ASCII code for "a" and add 1. I'm deliberately leaving the code as an exercise.
This sounds like homework. If so, please tag it as such.
Also, this won't deal with upper case letters, since you didn't state any requirement to handle them, but if you need to then just lowercase the string before you start.
Oh, and this will only deal with the latin "a" through "z" characters without any accents, etc.
char letter;
for(int i=0; i<text.length(); i++)
{
letter = text.charAt(i);
if(letter>='A' && letter<='Z')
System.out.println((int)letter - 'A'+1);
if(letter>='a' && letter<= 'z')
System.out.println((int)letter - 'a'+1);
}
This depends on the alphabet but for the english one, try this:
String input = "abc".toLowerCase(); //note the to lower case in order to treat a and A the same way
for( int i = 0; i < input.length(); ++i) {
int position = input.charAt(i) - 'a' + 1;
}
First you need to write a loop to iterate over the characters in the string. Take a look at the String class which has methods to give you its length and to find the charAt at each index.
For each character, you need to work out its numeric position. Take a look at this question to see how this could be done.
just logic I can suggest take two arrays.
one is Char array
and another is int array.
convert ur input string to char array,get the position of char from char and int array.
dont expect source code here
String word = "blah blah";
for(int i =0;i<word.length;++i)
{
if(Character.isLowerCase(word.charAt(i)){
System.out.print((int)word.charAt(i) - (int)'a'+1);
}
else{
System.out.print((int)word.charAt(i)-(int)'A' +1);
}
}