So I had tried this online challenge but got runtime error.Please help.I am new to programming. I have attached the problem statement and my solution.
The Challenge
Using the Java language, have the function KaprekarsConstant(num) take an integer of four digits (with at least two being distinct) and perform the following routine on said number:
Arrange the digits in descending order and in ascending order.
Subtract the smaller number from the bigger number, padding the difference with zeroes if necessary to maintain a four-digit number.
Then repeat step 1 and 2 using the four-digit difference.
Stop when the difference of the two, permuted numbers equals 6174.
Return the number of times that you had to perform steps 1 and 2 before arriving at a difference with the value of 6174.
nb: performing the routine on 6174 will always give you 6174 (7641 - 1467 = 6174).
For example: if num is 3524 your program should return 3: (pass 1) 5432 - 2345 = 3087, (pass 2) 8730 - 0378 = 8352, (pass 3) 8532 - 2358 = 6174.
Here is my solution:
import java.util.*;
import java.io.*;
class Main {
public static int KaprekarsConstant(int num) {
int diff = 0, count = 0;
while (diff != 6174) {
String s1 = String.valueOf(num);
int[] ch1 = new int[s1.length()];
for (int i = 0; i < ch1.length; i++) {
ch1[i] = s1.charAt(i);
}
Arrays.sort(ch1);
String s2 = String.valueOf(ch1);
String s3 = "";
for (int j = s2.length() - 1; j >= 0; j++) {
s3 += s2.charAt(j);
}
int a = Integer.parseInt(s2);
int b = Integer.parseInt(s3);
if (a > b) {
diff = a - b;
} else if (b > a) {
diff = b - a;
} else {
System.out.println("goal cant be reached");
break;
}
count++;
num = diff;
}
return num;
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.print(KaprekarsConstant(s.nextLine()));
}
}
Try this code:
/**
* KaprekarConstant
*/
import java.util.InputMismatchException;
import java.util.Scanner;
import java.util.Arrays;
import java.util.Collections;
public class KaprekarConstant {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int inputNumber;
char[] digits;
System.out.print("Enter four digit number: ");
try {
inputNumber = scan.nextInt();
if (checkLength(inputNumber))
throw new Exception();
Kaprekar(inputNumber);
} catch (InputMismatchException e) {
System.out.println("Not a number");
} catch (Exception ex) {
System.out.println("Number must have four digits");
} finally {
scan.close();
}
}
public static void Kaprekar(int target) {
int maximum = 0, minimum = 0, result = 0;
Integer[] digits = new Integer[4];
if (target == 6174)
return;
int i = 0;
while (i < 4) {
digits[i] = target % 10;
target /= 10;
i++;
}
Arrays.sort(digits);
minimum = toInt(digits);
Arrays.sort(digits, Collections.reverseOrder());
maximum = toInt(digits);
result = maximum - minimum;
System.out.println(String.format("%d - %d = %d", maximum, minimum, result));
Kaprekar(result);
}
public static boolean checkLength(int number) {
if (String.valueOf(number).length() < 5 && String.valueOf(number).length() > 3)
return false;
return true;
}
public static int toInt(Integer[] digits) {
int number = 0;
for (int digit : digits) {
number *= 10;
number += digit;
}
return number;
}
}
I think this is best method to find the karpekar constant.
There were some basic syntax & logical errors in your code. I have made appropriate changes in it. See if you understand it.
import java.util.*;
import java.io.*;
public class Main {
public static int KaprekarsConstant(int num) {
int diff = 0, count = 0;
while (diff != 6174) {
String s1 = String.valueOf(num);
char[] ch1 = new char[s1.length()];
for (int i = 0; i < ch1.length; i++) {
ch1[i] = s1.charAt(i);
}
Arrays.sort(ch1);
String s2 = new String(ch1);
String s3 = "";
for (int j = s2.length() - 1; j >= 0; j--) {
s3 += s2.charAt(j);
}
int a = Integer.parseInt(s2);
int b = Integer.parseInt(s3);
if (a > b) {
diff = a - b;
} else if (b > a) {
diff = b - a;
} else {
System.out.println("goal cant be reached");
break;
}
count++;
num = diff;
}
return count;
}
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.print(KaprekarsConstant(s.nextInt()));
}
}
This solution has less lines of code:
import static java.lang.System.out;
import java.util.Arrays;
/**
* #see https://en.wikipedia.org/wiki/6174_%28number%29
*/
public class KaprekarConstant {
public static void main(String[] args) {
assert(count(3524) == 3);
assert(count(3087) == 2);
assert(count(8352) == 1);
assert(count(6174) == 1);
out.println("All passed.");
}
public static int count(int start) {
int ct = 0;
do {
start = calc(start);
ct++;
} while (start != 6174);
return ct;
}
static int calc(int n) {
String n1s = String.format("%04d", n);
char[] chs = n1s.toCharArray();
Arrays.sort(chs);
n1s = new String(chs);
String n2s = new StringBuilder(new String(n1s)).reverse().toString();
int n1 = Integer.parseInt(n1s);
int n2 = Integer.parseInt(n2s);
return Math.max(n1, n2) - Math.min(n1, n2);
}
}
Here is a Python implementation:
def kc_count(start_int):
def kc_calc(n):
ns1 = ''.join(sorted("%04d" % n))
ns2 = ns1[::-1]
return max(int(ns1), int(ns2)) - min(int(ns1), int(ns2))
ct = 0;
while True:
start_int = kc_calc(start_int)
ct += 1
if start_int == 6174:
break
return ct
assert(kc_count(3524) == 3)
assert(kc_count(3087) == 2)
assert(kc_count(8352) == 1)
assert(kc_count(6174) == 1)
I want to check if a string has 8 or more characters, and if it has 1 capital letter and 1 number.
This is my code:
import java.util.Scanner;
public class PasswordTest
{
public static void main(String[] args)
{
Scanner keyb = new Scanner(System.in);
System.out.printf("Enter a password to be checked: \n");
String passwordInput = keyb.next();
int numberCharaters = passwordInput.length();
int numberCount = 1;
for (int i = 1; i <= numberCharaters; i++)
{
for(char c = '0'; c <= '9'; c++)
{
if (passwordInput.charAt(i) == c)
{
numberCount++;
}
}
}
int numberNumbers = numberCount - 1;
int captialCount = 1;
for (int i = 1; i <= numberCharaters; i++)
{
for(char c = 'A'; c <= 'Z'; c++)
{
if (passwordInput.charAt(i) == c)
{
captialCount++;
}
}
}
int numberCaptials = captialCount - 1;
if (numberCharaters >= 8 && numberNumbers >= 1 && numberCaptials >= 1)
{
String strongEnough = "Password is strong enough.";
System.out.println(strongEnough);
}
else
{
String strongEnough = "Password is not strong enough.";
System.out.println(strongEnough);
}
}
}
and this is the error message I'm getting
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 5
at java.lang.String.charAt(String.java:658)
at passwordtest.main(passwordtest.java:23)
My input was: Test1
What am I doing wrong? I've been trying to figure out where the java.lang.StringIndexOutOfBoundsException: comes from.
Your program has some mistakes.
You should have initialized numberCount to 0 to avoid having to subtract later, and avoid having to create another variable. Also the error you're getting is because of the <=, since there is no string element = to the length of the string, i.e index is from 0 to length - 1.
int numberCount = 0;
for (int i = 0; i < numberCharaters; i++)
{
for(char c = '0'; c <= '9'; c++)
{
if (passwordInput.charAt(i) == c)
{
numberCount++;
}
}
}
Then, for the capital count you made the same mistake, I would also take into account the advice in the comments to improve the loops.
You are making a minor mistake.
Length = Actual number of elements(characters).
Index = starts from 0 and ends at length-1.
Your code should be look like this:
import java.util.Scanner;
public class PasswordTest
{
public static void main(String[] args)
{
Scanner keyb = new Scanner(System.in);
System.out.printf("Enter a password to be checked: \n");
String passwordInput = keyb.next();
int numberCharaters = passwordInput.length();
int numberCount = 0;
for (int i = 0; i <= numberCharaters-1; i++)
{
for(char c = '0'; c <= '9'; c++)
{
if (passwordInput.charAt(i) == c)
{
numberCount++;
}
}
}
int numberNumbers = numberCount - 0;
int captialCount = 0;
for (int i = 1; i <= numberCharaters; i++)
{
for(char c = 'A'; c <= 'Z'; c++)
{
if (passwordInput.charAt(i) == c)
{
captialCount++;
}
}
}
int numberCaptials = captialCount - 0;
if (numberCharaters >= 8 && numberNumbers >= 1 &&
numberCaptials >= 1)
{
String strongEnough = "Password is strong enough.";
System.out.println(strongEnough);
}
else
{
String strongEnough = "Password is not strong enough.";
System.out.println(strongEnough);
}
}
}
import java.util.*;
public class password
{
public static void main()
{
Scanner sc=new Scanner(System.in);
System.out.print("enter the password : ");
String s=sc.nextLine();
int l=s.length();
int k=0,k1=0;
if(l>=8)
{
for(int i=0;i<l;i++)
{
if(Character.isLetter(s.charAt(i)))
{
if(Character.isUpperCase(s.charAt(i)))
{
k++;
}
}
else
{
if(Character.isDigit(s.charAt(i)))
{
k1++;
}
}
}
if(k>0&&k1>0)
{
System.out.println("Password is strong enough");
}
else
{
System.out.println("Password shoud contain atleast capital letter and one number");
}
}
else
{
System.out.println("Password shoud contain atleast capital letter,one number and shoud have length of 8 or more");
}
}
}
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));
}
}