I have the below String variable
String s = "abc,xyz,lmn,ijk";
I want to extract only the portion of the String (i.e - 'lmn')
And, Should not use in-built functions like, SubString(), Split(), IndexOf(). But I can use charArray()
And this question was asked in my interview.
I tried the below code,
But not sure how to proceed. Can any one please provide your thoughts?
String s = "abc,xyz,lmn,ijk";
int counter = 0;
char[] ch = s.toCharArray();
for (int i = 0; i < ch.length; i++) {
if (ch[i] == ',') {
counter++;
}
}
Here's one way:
public static void main(String[] args) {
String s = "abc,xyz,lmn,ijk";
char[] ch = s.toCharArray();
int counter = 0;
int place = 2;
for (int i = 0; i < ch.length-2; i++) {
if(ch[i] == ',') {
counter++;
}
if(counter == place && ch[i] != ',') {
System.out.print(ch[i]);
}
}
}
It prints everything after the second comma, but before the third one.
public static void main(String[] args) {
// TODO Auto-generated method stub
String s = "abc,xyz,lmn,ijk";
StringBuffer sb=new StringBuffer();
char[] ch = s.toCharArray();
for (int i = 0; i < ch.length; i++) {
if(','==(ch[i]))
{
if (sb.toString().equals("lmn")) {
System.out.println(sb.toString());
}
else
{
int length=sb.length();
sb.delete(0, length);
}
}
else
{
sb.append(ch[i]);
}
}
}
I would do it this way.
String s = "abc,xyz,lmn,ijk";
String x = "c,x"; // String to found
String r = "";
boolean coincidence = false;
int a=0; // Initial index if of the first character in x is found
int b=0; // Last index If it was possible to search for the last character of x
int c=0; // Index "iterator" on String x
char[] ch = s.toCharArray();
for (int i = 0; i < ch.length; i++) {
if(c == x.length())
break;
else{
if(ch[i] == x.charAt(c) && !coincidence){
a = i; b = i; c++;
coincidence = true;
}
else if(ch[i] == x.charAt(c) && coincidence){
b++; c++;
}else{
coincidence = false;
a = 0; b = 0; c = 0;
}
}
}
System.out.println("String: " + s);
System.out.println("String to find: " + x);
System.out.println("Was found? " + ((coincidence)? "Yes" : "No"));
if(coincidence){
System.out.println("Intervals indexes in String: ["+a + "," + b +"]");
// String extration
for (int i = a; i <= b; i++)
r += s.charAt(i);
System.out.println("String extracted: " + r);
}
I m writing a method to find the first non repeating character in a string. I saw this method in a previous stackoverflow question
public static char findFirstNonRepChar(String input){
char currentChar = '\0';
int len = input.length();
for(int i=0;i<len;i++){
currentChar = input.charAt(i);
if((i!=0) && (currentChar!=input.charAt(i-1)) && (i==input.lastIndexOf(currentChar))){
return currentChar;
}
}
return currentChar;
}
I came up with a solution using a hashtable where I have two for loops (not nested) where I interate through the string in one loop writing down each occurance of a letter (for example in apple, a would have 1, p would have 2, etc.) then in the second loop I interate through the hashtable to see which one has a count of 1 first. What is the benefit to the above method over what I came up with? I am new to Java does having two loops (not nested) hinder time complexity. Both these algorithms should have O(n) right? Is there another faster, less space complexity algorithm for this question than these two solutions?
public class FirstNonRepeatCharFromString {
public static void main(String[] args) {
String s = "java";
for(Character ch:s.toCharArray()) {
if(s.indexOf(ch) == s.lastIndexOf(ch)) {
System.out.println("First non repeat character = " + ch);
break;
}
}
}
}
As you asked if your code is from O(n) or not, I think it's not, because in the for loop, you are calling lastIndexOf and it's worst case is O(n). So it is from O(n^2).
About your second question: having two loops which are not nested, also makes it from O(n).
If assuming non unicode characters in your input String, and Uppercase or Lowercase characters are assumed to be different, the following would do it with o(n) and supports all ASCII codes from 0 to 255:
public static Character getFirstNotRepeatedChar(String input) {
byte[] flags = new byte[256]; //all is initialized by 0
for (int i = 0; i < input.length(); i++) { // O(n)
flags[(int)input.charAt(i)]++ ;
}
for (int i = 0; i < input.length(); i++) { // O(n)
if(flags[(int)input.charAt(i)] > 0)
return input.charAt(i);
}
return null;
}
Thanks to Konstantinos Chalkias hint about the overflow, if your input string has more than 127 occurrence of a certain character, you can change the type of flags array from byte[] to int[] or long[] to prevent the overflow of byte type.
Hope it would be helpful.
The algorithm you showed is slow: it looks for each character in the string, it basically means that for each character you spend your time checking the string twice!! Huge time loss.
The best naive O(n) solution basically holds all the characters in order of insertion (so the first can be found) and maps a mutable integer to them. When we're done, analyzing, we go through all the entries and return the first character that was registered and has a count of 1.
There are no restrictions on the characters you can use. And AtomicInteger is available with import java.util.concurrent.atomic.AtomicInteger.
Using Java 8:
public static char findFirstNonRepChar(String string) {
Map<Integer,Long> characters = string.chars().boxed()
.collect(Collectors.groupingBy(Function.identity(), LinkedHashMap::new, Collectors.counting()));
return (char)(int)characters.entrySet().stream()
.filter(e -> e.getValue() == 1L)
.findFirst()
.map(Map.Entry::getKey)
.orElseThrow(() -> new RuntimeException("No unrepeated character"));
}
Non Java 8 equivalent:
public static char findFirstNonRepChar(String string) {
Map<Character, AtomicInteger> characters = new LinkedHashMap<>(); // preserves order of insertion.
for (int i = 0; i < string.length(); i++) {
char c = string.charAt(i);
AtomicInteger n = characters.get(c);
if (n == null) {
n = new AtomicInteger(0);
characters.put(c, n);
}
n.incrementAndGet();
}
for (Map.Entry<Character, AtomicInteger> entry: characters.entries()) {
if (entry.getValue().get() == 1) {
return entry.getKey();
}
}
throw new RuntimeException("No unrepeated character");
}
import java.util.LinkedHashMap;
import java.util.Map;
public class getFirstNonRep {
public static char get(String s) throws Exception {
if (s.length() == 0) {
System.out.println("Fail");
System.exit(0);
} else {
Map<Character, Integer> m = new LinkedHashMap<Character, Integer>();
for (int i = 0; i < s.length(); i++) {
if (m.containsKey(s.charAt(i))) {
m.put(s.charAt(i), m.get(s.charAt(i)) + 1);
} else {
m.put(s.charAt(i), 1);
}
}
for (Map.Entry<Character, Integer> hm : m.entrySet()) {
if (hm.getValue() == 1) {
return hm.getKey();
}
}
}
return 0;
}
public static void main(String[] args) throws Exception {
System.out.print(get("Youssef Zaky"));
}
}
This solution takes less space and less time, since we iterate the string only one time.
Works for any type of characters.
String charHolder; // Holds
String testString = "8uiuiti080t8xt8t";
char testChar = ' ';
int count = 0;
for (int i=0; i <= testString.length()-1; i++) {
testChar = testString.charAt(i);
for (int j=0; j < testString.length()-1; j++) {
if (testChar == testString.charAt(j)) {
count++;
}
}
if (count == 1) { break; };
count = 0;
}
System.out.println("The first not repeating character is " + testChar);
I accumulated all possible methods with string length 25'500 symbols:
private static String getFirstUniqueChar(String line) {
String result1 = null, result2 = null, result3 = null, result4 = null, result5 = null;
int length = line.length();
long start = System.currentTimeMillis();
Map<Character, Integer> chars = new LinkedHashMap<Character, Integer>();
char[] charArray1 = line.toCharArray();
for (int i = 0; i < length; i++) {
char currentChar = charArray1[i];
chars.put(currentChar, chars.containsKey(currentChar) ? chars.get(currentChar) + 1 : 1);
}
for (Map.Entry<Character, Integer> entry : chars.entrySet()) {
if (entry.getValue() == 1) {
result1 = entry.getKey().toString();
break;
}
}
long end = System.currentTimeMillis();
System.out.println("1st test:\n result: " + result1 + "\n time: " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < length; i++) {
String current = Character.toString(line.charAt(i));
String left = line.substring(0, i);
if (!left.contains(current)) {
String right = line.substring(i + 1);
if (!right.contains(current)) {
result2 = current;
break;
}
}
}
end = System.currentTimeMillis();
System.out.println("2nd test:\n result: " + result2 + "\n time: " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < length; i++) {
char currentChar = line.charAt(i);
if (line.indexOf(currentChar) == line.lastIndexOf(currentChar)) {
result3 = Character.toString(currentChar);
break;
}
}
end = System.currentTimeMillis();
System.out.println("3rd test:\n result: " + result3 + "\n time: " + (end - start));
start = System.currentTimeMillis();
char[] charArray4 = line.toCharArray();
for (int i = 0; i < length; i++) {
char currentChar = charArray4[i];
int count = 0;
for (int j = 0; j < length; j++) {
if (currentChar == charArray4[j] && i != j) {
count++;
break;
}
}
if (count == 0) {
result4 = Character.toString(currentChar);
break;
}
}
end = System.currentTimeMillis();
System.out.println("4th test:\n result: " + result4 + "\n time: " + (end - start));
start = System.currentTimeMillis();
for (int i = 0; i < length; i++) {
char currentChar = line.charAt(i);
int count = 0;
for (int j = 0; j < length; j++) {
if (currentChar == line.charAt(j) && i != j) {
count++;
break;
}
}
if (count == 0) {
result5 = Character.toString(currentChar);
break;
}
}
end = System.currentTimeMillis();
System.out.println("5th test:\n result: " + result5 + "\n time: " + (end - start));
return result1;
}
And time results (5 times):
1st test:
result: g
time: 13, 12, 12, 12, 14
2nd test:
result: g
time: 55, 56, 59, 70, 59
3rd test:
result: g
time: 2, 3, 2, 2, 3
4th test:
result: g
time: 3, 3, 2, 3, 3
5th test:
result: g
time: 6, 5, 5, 5, 6
public static char NonReapitingCharacter(String str) {
Set<Character> s = new HashSet();
char ch = '\u0000';
for (char c : str.toCharArray()) {
if (s.add(c)) {
if (c == ch) {
break;
} else {
ch = c;
}
}
}
return ch;
}
Okay I misread the question initially so here's a new solution. I believe is this O(n). The contains(Object) of HashSet is O(1), so we can take advantage of that and avoid a second loop. Essentially if we've never seen a specific char before, we add it to the validChars as a potential candidate to be returned. The second we see it again however, we add it to the trash can of invalidChars. This prevents that char from being added again. At the end of the loop (you have to loop at least once no matter what you do), you'll have a validChars hashset with n amount of elements. If none are there, then it will return null from the Character class. This has a distinct advantage as the char class has no good way to return a 'bad' result so to speak.
public static Character findNonRepeatingChar(String x)
{
HashSet<Character> validChars = new HashSet<>();
HashSet<Character> invalidChars = new HashSet<>();
char[] array = x.toCharArray();
for (char c : array)
{
if (validChars.contains(c))
{
validChars.remove(c);
invalidChars.add(c);
}
else if (!validChars.contains(c) && !invalidChars.contains(c))
{
validChars.add(c);
}
}
return (!validChars.isEmpty() ? validChars.iterator().next() : null);
}
If you are only interested for characters in the range a-z (lowercase as OP requested in comments), you can use this method that requires a minimum extra storage of two bits per character Vs a HashMap approach.
/*
* It works for lowercase a-z
* you can scale it to add more characters
* eg use 128 Vs 26 for ASCII or 256 for extended ASCII
*/
public static char getFirstNotRepeatedChar(String input) {
boolean[] charsExist = new boolean[26];
boolean[] charsNonUnique = new boolean[26];
for (int i = 0; i < input.length(); i++) {
int index = 'z' - input.charAt(i);
if (!charsExist[index]) {
charsExist[index] = true;
} else {
charsNonUnique[index] = true;
}
}
for (int i = 0; i < input.length(); i++) {
if (!charsNonUnique['z' - input.charAt(i)])
return input.charAt(i);
}
return '?'; //example return of no character found
}
In case of two loops (not nested) the time complexity would be O(n).
The second solution mentioned in the question can be implemented as:
We can use string characters as keys to a map and maintain their count. Following is the algorithm.
1.Scan the string from left to right and construct the count map.
2.Again, scan the string from left to right and check for count of each character from the map, if you find an element who’s count is 1, return it.
package com.java.teasers.samples;
import java.util.Map;
import java.util.HashMap;
public class NonRepeatCharacter {
public static void main(String[] args) {
String yourString = "Hi this is javateasers";//change it with your string
Map<Character, Integer> characterMap = new HashMap<Character, Integer>();
//Step 1 of the Algorithm
for (int i = 0; i < yourString.length(); i++) {
Character character = yourString.charAt(i);
//check if character is already present
if(null != characterMap.get(character)){
//in case it is already there increment the count by 1.
characterMap.put(character, characterMap.get(character) + 1);
}
//in case it is for the first time. Put 1 to the count
else
characterMap.put(character, 1);
}
//Step 2 of the Algorithm
for (int i = 0; i < yourString.length(); i++) {
Character character = yourString.charAt(i);
int count = characterMap.get(character);
if(count == 1){
System.out.println("character is:" + character);
break;
}
}
}
}
public char firstNonRepeatedChar(String input) {
char out = 0;
int length = input.length();
for (int i = 0; i < length; i++) {
String sub1 = input.substring(0, i);
String sub2 = input.substring(i + 1);
if (!(sub1.contains(input.charAt(i) + "") || sub2.contains(input
.charAt(i) + ""))) {
out = input.charAt(i);
break;
}
}
return out;
}
Since LinkedHashMap keeps the order of insertion
package com.company;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Scanner;
public class Main {
public static void main(String[] argh) {
Scanner sc = new Scanner(System.in);
String l = sc.nextLine();
System.out.println(firstCharNoRepeated(l));
}
private static String firstCharNoRepeated(String l) {
Map<String, Integer> chars = new LinkedHashMap();
for(int i=0; i < l.length(); i++) {
String c = String.valueOf(l.charAt(i));
if(!chars.containsKey(c)){
chars.put(c, i);
} else {
chars.remove(c);
}
}
return chars.keySet().iterator().next();
}
}
Few lines of code, works for me.
public class FirstNonRepeatingCharacter {
final static String string = "cascade";
public static void main(String[] args) {
char[] charArr = string.toCharArray();
for (int i = 0; charArr.length > i; i++) {
int count = 0;
for (int j = 0; charArr.length > j; j++) {
if (charArr[i] == charArr[j]) {
count++;
}
}
if (count == 1){
System.out.println("First Non Repeating Character is: " + charArr[i]);
break;
}
}
}
}
Constraint for this solution:
O(n) time complexity. My solution is O(2n), follow Time Complexity analysis,O(2n) => O(n)
import java.util.HashMap;
public class FindFirstNonDuplicateCharacter {
public static void main(String args[]) {
System.out.println(findFirstNonDuplicateCharacter("abacbcefd"));
}
private static char findFirstNonDuplicateCharacter(String s) {
HashMap<Character, Integer> chDupCount = new HashMap<Character, Integer>();
char[] charArr = s.toCharArray();
for (char ch: charArr) { //first loop, make the tables and counted duplication by key O(n)
if (!chDupCount.containsKey(ch)) {
chDupCount.put(ch,1);
continue;
}
int dupCount = chDupCount.get(ch)+1;
chDupCount.replace(ch, dupCount);
}
char res = '-';
for(char ch: charArr) { //second loop, get the first duplicate by count number, O(2n)
// System.out.println("key: " + ch+", value: " + chDupCount.get(ch));
if (chDupCount.get(ch) == 1) {
res = ch;
break;
}
}
return res;
}
}
Hope it help
char firstNotRepeatingCharacter(String s) {
for(int i=0; i< s.length(); i++){
if(i == s.lastIndexOf(s.charAt(i)) && i == s.indexOf(s.charAt(i))){
return s.charAt(i);
}
}
return '_';
}
String a = "sampapl";
char ar[] = a.toCharArray();
int dya[] = new int[256];
for (int i = 0; i < dya.length; i++) {
dya[i] = -1;
}
for (int i = 0; i < ar.length; i++) {
if (dya[ar[i]] != -1) {
System.out.println(ar[i]);
break;
} else {
dya[ar[i]] = ar[i];
}
}
This is solution in python:
input_str = "interesting"
#input_str = "aabbcc"
#input_str = "aaaapaabbcccq"
def firstNonRepeating(param):
counts = {}
for i in range(0, len(param)):
# Store count and index repectively
if param[i] in counts:
counts[param[i]][0] += 1
else:
counts[param[i]] = [1, i]
result_index = len(param) - 1
for x in counts:
if counts[x][0] == 1 and result_index > counts[x][1]:
result_index = counts[x][1]
return result_index
result_index = firstNonRepeating(input_str)
if result_index == len(input_str)-1:
print("no such character found")
else:
print("first non repeating charater found: " + input_str[result_index])
Output:
first non repeating charater found: r
import java.util.*;
public class Main {
public static void main(String[] args) {
String str1 = "gibblegabbler";
System.out.println("The given string is: " + str1);
for (int i = 0; i < str1.length(); i++) {
boolean unique = true;
for (int j = 0; j < str1.length(); j++) {
if (i != j && str1.charAt(i) == str1.charAt(j)) {
unique = false;
break;
}
}
if (unique) {
System.out.println("The first non repeated character in String is: " + str1.charAt(i));
break;
}
}
}
}
public class GFG {
public static void main(String[] args) {
String s = "mmjjjjmmn";
for (char c : s.toCharArray()) {
if (s.indexOf(c) == s.lastIndexOf(c)) {
System.out.println("First non repeated is:" + c);
break;
}
}
}
output = n
Non Repeated Character String in Java
public class NonRepeatedCharacter {
public static void main(String[] args) {
String s = "ffeeddbbaaclck";
for (int i = 0; i < s.length(); i++) {
boolean unique = true;
for (int j = 0; j < s.length(); j++) {
if (i != j && s.charAt(i) == s.charAt(j)) {
unique = false;
break;
}
}
if (unique) {
System.out.println("First non repeated characted in String \""
+ s + "\" is:" + s.charAt(i));
break;
}
}
}
}
Output:
First non repeated characted in String "ffeeddbbaaclck" is:l
For More Details
In this coding i use length of string to find the first non repeating letter.
package com.string.assingment3;
import java.util.Scanner;
public class FirstNonRepetedChar {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter a String : ");
String str = in.next();
char[] ch = str.toCharArray();
int length = ch.length;
int x = length;
for(int i=0;i<length;i++) {
x = length-i;
for(int j=i+1;j<length;j++) {
if(ch[i]!=ch[j]) {
x--;
}//if
}//inner for
if(x==1) {
System.out.println(ch[i]);
break;
}
else {
continue;
}
}//outer for
}
}// develope by NDM
In Kotlin
fun firstNonRepeating(string: String): Char?{
//Get a copy of the string
var copy = string
//Slice string into chars then convert them to string
string.map { it.toString() }.forEach {
//Replace first occurrance of that character and check if it still has it
if (copy.replaceFirst(it,"").contains(it))
//If it has the given character remove it
copy = copy.replace(it,"")
}
//Return null if there is no non-repeating character
if (copy.isEmpty())
return null
//Get the first character from what left of that string
return copy.first()
}
https://pl.kotl.in/KzL-veYNZ
public static void firstNonRepeatFirstChar(String str) {
System.out.println("The given string is: " + str);
for (int i = 0; i < str.length(); i++) {
boolean unique = true;
for (int j = 0; j < str.length(); j++) {
if (i != j && str.charAt(i) == str.charAt(j)) {
unique = false;
break;
}
}
if (unique) {
System.out.println("The first non repeated character in String is: " + str.charAt(i));
break;
}
}
}
Using Set with single for loop
public static Character firstNonRepeatedCharacter(String str) {
Character result = null;
if (str != null) {
Set<Character> set = new HashSet<>();
for (char c : str.toCharArray()) {
if (set.add(c) && result == null) {
result = c;
} else if (result != null && c == result) {
result = null;
}
}
}
return result;
}
You can achieve this in single traversal of String using LinkedHashSet as follows:
public static Character getFirstNonRepeatingCharacter(String str) {
Set<Character> result = new LinkedHashSet<>(256);
for (int i = 0; i< str.length(); ++i) {
if(!result.add(str.charAt(i))) {
result.remove(str.charAt(i));
}
}
if(result.iterator().hasNext()) {
return result.iterator().next();
}
return null;
}
For Java;
char firstNotRepeatingCharacter(String s) {
HashSet<String> hs = new HashSet<>();
StringBuilder sb =new StringBuilder(s);
for (int i = 0; i<s.length(); i++){
char c = sb.charAt(i);
if(s.indexOf(c) == i && s.indexOf(c, i+1) == -1 ) {
return c;
}
}
return '_';
}
public class FirstNonRepeatingChar {
public static void main(String[] args) {
String s = "hello world i am here";
s.chars().boxed()
.collect(Collectors.groupingBy(Function.identity(), LinkedHashMap::new, Collectors.counting()))
.entrySet().stream().filter(e -> e.getValue() == 1).findFirst().ifPresent(e->System.out.println(e.getKey()));
}
}
package looping.concepts;
import java.util.Scanner;
public class Line {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter name: ");
String a = sc.nextLine();
int i = 0;
int j = 0;
for (i = 0; i < a.length(); i++) {
char ch = a.charAt(i);
int counter = 0;
// boolean repeat = false;
for (j = 0; j < a.length(); j++) {
if (ch == a.charAt(j)) {
counter++;
}
}
if (counter == 1) {
System.out.print(ch);
}
else
{
System.out.print("There is no non repeated character");
break;
}
}
}
}
import java.util.Scanner;
public class NonRepaeated1
{
public static void main(String args[])
{
String str;
char non_repeat=0;
int len,i,j,count=0;
Scanner s = new Scanner(System.in);
str = s.nextLine();
len = str.length();
for(i=0;i<len;i++)
{
non_repeat=str.charAt(i);
count=1;
for(j=0;j<len;j++)
{
if(i!=j)
{
if(str.charAt(i) == str.charAt(j))
{
count=0;
break;
}
}
}
if(count==1)
break;
}
if(count == 1)
System.out.print("The non repeated character is : " + non_repeat);
}
}
package com.test.util;
public class StringNoRepeat {
public static void main(String args[]) {
String st = "234123nljnsdfsdf41l";
String strOrig=st;
int i=0;
int j=0;
String st1="";
Character ch=' ';
boolean fnd=false;
for (i=0;i<strOrig.length(); i++) {
ch=strOrig.charAt(i);
st1 = ch.toString();
if (i==0)
st = strOrig.substring(1,strOrig.length());
else if (i == strOrig.length()-1)
st=strOrig.substring(0, strOrig.length()-1);
else
st=strOrig.substring(0, i)+strOrig.substring(i+1,strOrig.length());
if (st.indexOf(st1) == -1) {
fnd=true;
j=i;
break;
}
}
if (!fnd)
System.out.println("The first no non repeated character");
else
System.out.println("The first non repeated character is " +strOrig.charAt(j));
}
}
I know this question was asked many times, but I didn't find any of the answers helpful in my case. I have a method that receives a String. I want to check if any of the characters in the string are repeated. If so the method will return an empty String. If not it will return the String back.
The method is looking for any repeated character in the String.
private String visit(String word) {
int count = 0;
if(word == ""){
return "<empty>";
}
//alphabet is an array that holds all characters that could be used in the String
for(int i = 0; i < alphabet.length; i++){
for(int j = 0; j < word.length(); j++){
if(alphabet[i] == word.charAt(j)){
count++;
}
if(count == 2){
return "";
}
}
count = 0;
}
return word;
}
Ok, I publish my solution to this:
package main;
import java.util.Arrays;
public class main {
public static void main(String[] args) {
System.out.println(hasDups("abc"));
System.out.println(hasDups("abcb"));
}
public static String hasDups(String arg) {
String[] ar = arg.split("");
Arrays.sort(ar);
boolean noDups = true;
for (int i = 1; i < ar.length && noDups; i++) {
if (ar[i].equals(ar[i-1])) noDups = false;
}
if (noDups) return arg; else return "";
}
}
This might not be the best way of doing what you want, but you can use two for loops to check each character against all the other characters to see if it is repeated.
public static String hasRepeated(String word) {
if (word.isEmpty()) return "<empty>";
char[] charArray = word.toCharArray();
for (int i = 0; i < charArray.length; i++) {
for (int j = 0; j < charArray.length; j++) {
if (i == j) {
} else if (Character.toString(charArray[i]).
equalsIgnoreCase(Character.toString(charArray[j]))) {
return "";
}
}
}
return word;
}
Note: This code assumes that the case of the character doesn't matter, it just checks if it is repeated.
/**
* Returns the index of the first character repeated, or -1 if no repeats
*/
public static int firstRepeated( String s ) {
if ( s != null ) {
int n = s.length();
for (int i = 0; i < (n - 1); i++) {
int indx = s.indexOf( s.charAt( i ), i + 1 );
if ( indx > 0 ) {
return i;
}
}
}
return -1;
}
This works!!
public static String checkDuplicate(String str)
{
int count = 0;
char[] charArray = str.toCharArray();
for(int i=0;i<charArray.length;i++)
{
count = 0;
for(int j=0;j<charArray.length;j++)
{
if(charArray[i]==charArray[j])
{
count++;
if(count==2) break;
}
}
if(count==2) break;
}
if(count==2)
return "";
else
return str;
}
}
I need to write a static method that takes a String as a parameter and returns a new String obtained by replacing every instance of repeated adjacent letters with a single instance of that letter without using regular expressions. For example if I enter "maaaakkee" as a String, it returns "make".
I already tried the following code, but it doesn't seem to display the last character.
Here's my code:
import java.util.Scanner;
public class undouble {
public static void main(String [] args){
Scanner console = new Scanner(System.in);
System.out.println("enter String: ");
String str = console.nextLine();
System.out.println(removeSpaces(str));
}
public static String removeSpaces(String str){
String ourString="";
int j = 0;
for (int i=0; i<str.length()-1 ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
return ourString;
}
}
You could use regular expressions for that.
For instance:
String input = "ddooooonnneeeeee";
System.out.println(input.replaceAll("(.)\\1{1,}", "$1"));
Output:
done
Pattern explanation:
"(.)\\1{1,}" means any character (added to group 1) followed by itself at least once
"$1" references contents of group 1
maybe:
for (int i=1; i<str.length() ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
The problem is with your condition. You say compare i and i+1 in each iteration and in last iteration you have both i and j pointing to same location so it will never print the last character. Try this unleass you want to use regex to achive this:
EDIT:
public void removeSpaces(String str){
String ourString="";
for (int i=0; i<str.length()-1 ; i++){
if(i==0){
ourString = ""+str.charAt(i);
}else{
if(str.charAt(i-1) != str.charAt(i)){
ourString = ourString +str.charAt(i);
}
}
}
System.out.println(ourString);
}
if you cannot use replace or replaceAll, here is an alternative. O(2n), O(N) for stockage and O(N) for creating the string. It removes all repeated chars in the string put them in a stringbuilder.
input : abcdef , output : abcdef
input : aabbcdeef, output : cdf
private static String remove_repeated_char(String str)
{
StringBuilder result = new StringBuilder();
HashMap<Character, Integer> items = new HashMap<>();
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == null)
items.put(current, 1);
else
items.put(current, ocurrence + 1);
}
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == 1)
result.append(current);
}
return result.toString();
}
import java.util.*;
public class string2 {
public static void main(String[] args) {
//removes repeat character from array
Scanner sc=new Scanner(System.in);
StringBuffer sf=new StringBuffer();
System.out.println("enter a string");
sf.append(sc.nextLine());
System.out.println("string="+sf);
int i=0;
while( i<sf.length())
{
int j=1+i;
while(j<sf.length())
{
if(sf.charAt(i)==sf.charAt(j))
{
sf.deleteCharAt(j);
}
else
{
j=j+1;
}
}
i=i+1;
}
System.out.println("string="+sf);
}
}
Input AABBBccDDD, Output BD
Input ABBCDDA, Outout C
private String reducedString(String s){
char[] arr = s.toCharArray();
String newString = "";
Map<Character,Integer> map = new HashMap<Character,Integer>();
map.put(arr[0],1);
for(int index=1;index<s.length();index++)
{
Character key = arr[index];
int value;
if(map.get(key) ==null)
{
value =0;
}
else
{
value = map.get(key);
}
value = value+1;
map.put(key,value);
}
Set<Character> keyset = map.keySet();
for(Character c: keyset)
{
int value = map.get(c);
if(value%2 !=0)
{
newString+=c;
}
}
newString = newString.equals("")?"Empty String":newString;
return newString;
}
public class RemoveDuplicateCharecterInString {
static String input = new String("abbbbbbbbbbbbbbbbccccd");
static String output = "";
public static void main(String[] args)
{
// TODO Auto-generated method stub
for (int i = 0; i < input.length(); i++) {
char temp = input.charAt(i);
boolean check = false;
for (int j = 0; j < output.length(); j++) {
if (output.charAt(j) == input.charAt(i)) {
check = true;
}
}
if (!check) {
output = output + input.charAt(i);
}
}
System.out.println(" " + output);
}
}
Answer : abcd
public class RepeatedChar {
public static void main(String[] args) {
String rS = "maaaakkee";
String outCome= rS.charAt(0)+"";
int count =0;
char [] cA =rS.toCharArray();
for(int i =0; i+1<cA.length; ++i) {
if(rS.charAt(i) != rS.charAt(i+1)) {
outCome += rS.charAt(i+1);
}
}
System.out.println(outCome);
}
}
TO WRITE JAVA PROGRAM TO REMOVE REPEATED CHARACTERS:
package replace;
public class removingrepeatedcharacters
{
public static void main(String...args){
int i,j=0,count=0;
String str="noordeen";
String str2="noordeen";
char[] ch=str.toCharArray();
for(i=0;i<=5;i++)
{
count=0;
for(j=0;j<str2.length();j++)
{
if(ch[i]==str2.charAt(j))
{
count++;
System.out.println("at the index "+j +"position "+ch[i]+ "+ count is"+count);
if(count>=2){
str=str2;
str2=str.replaceFirst(Character.toString(ch[j]),Character.toString(' '));
}
System.out.println("after replacing " +str2);
}
}
}
}
}
String outstr = "";
String outstring = "";
for(int i = 0; i < str.length() - 1; i++) {
if(str.charAt(i) != str.charAt(i + 1)) {
outstr = outstr + str.charAt(i);
}
outstring = outstr + str.charAt(i);
}
System.out.println(outstring);
public static void remove_duplicates(String str){
String outstr="";
String outstring="";
for(int i=0;i<str.length()-1;i++) {
if(str.charAt(i)!=str.charAt(i+1)) {
outstr=outstr+str.charAt(i);
}
outstring=outstr+str.charAt(i);
}
System.out.println(outstring);
}
More fun with java 7:
System.out.println("11223344445555".replaceAll("(?<nums>.+)\\k<nums>+","${nums}"));
No more cryptic numbers in regexes.
public static String removeDuplicates(String str) {
String str2 = "" + str.charAt(0);
for (int i = 1; i < str.length(); i++) {
if (str.charAt(i - 1) == str.charAt(i) && i != 0) {
continue;
}
str2 = str2 + str.charAt(i);
}
return str2;
}