Develop Reference

Finding first non repeating character in a string in relation to Big O

I solved a task concerning finding the first non-repeating character. For example, given the input "apple" the answer would be "a", the first character that isn't repeated. Even though "e" is not repeated it's not the first character. Another example: "lalas" answer is "s".
public static char firstNonRepeatingCharacter(String input) {
boolean unique;
int count = input.length();
char[] chars = input.toCharArray();
for (int i = 0; i < input.length(); i++) {
unique = true;
for (int j = 0; j < input.length(); j++) {
count--;
char c = chars[i];
if (i != j && c == chars[j]) {
unique = false;
break;
}
}
if (unique) {
return input.charAt(i);
}
}
return (0);
}
I want to simplify this code due to the nested loop having O(n2) complexity. I have been looking at the code trying to figure out if i could make it any faster but nothing comes to mind.

Another way is to find the first and last indexOf the character. If both are same then it is unique.
public static char firstNonRepeatingCharacter(String input) {
for(char c:input.toCharArray())
if(input.indexOf(c) == input.lastIndexOf(c))
return c;
return (0);
}
EDIT:
Or with Java 8+
return (char) input.chars()
.filter(c -> input.indexOf(c) == input.lastIndexOf(c))
.findFirst().orElse(0);

O(n) is better.
Use an intermedian structure to handle the number of repetitions.
public static char firstNonRepeatingCharacter(String input) {
boolean unique;
int count = input.length();
char[] chars = input.toCharArray();
for (int i = 0; i < input.length(); i++) {
unique = true;
for (int j = 0; j < input.length(); j++) {
count--;
char c = chars[i];
if (i != j && c == chars[j]) {
unique = false;
break;
}
}
if (unique) {
return input.charAt(i);
}
}
return (0);
}
public static char firstNonRepeatingCharacterMyVersion(String input) {
Map<String,Integer> map = new HashMap();
// first iteration put in a map the number of times a char appears. Linear O(n)=n
for (char c : input.toCharArray()) {
String character = String.valueOf(c);
if(map.containsKey(character)){
map.put(character,map.get(character) + 1);
} else {
map.put(character,1);
}
}
// Second iteration look for first element with one element.
for (char c : input.toCharArray()) {
String character = String.valueOf(c);
if(map.get(character) == 1){
return c;
}
}
return (0);
}
public static void main(String... args){
System.out.println(firstNonRepeatingCharacter("potatoaonionp"));
System.out.println(firstNonRepeatingCharacterMyVersion("potatoaonionp"));
}

See this solution. Similar to the above #Lucbel. Basically, using a LinkedList. We store all non repeating. However, we will use more space. But running time is O(n).
import java.util.LinkedList;
import java.util.List;
public class FirstNone {
public static void main(String[] args) {
System.out.println(firstNonRepeatingCharacter("apple"));
System.out.println(firstNonRepeatingCharacter("potatoaonionp"));
System.out.println(firstNonRepeatingCharacter("tksilicon"));
}
public static char firstNonRepeatingCharacter(String input) {
List<Character> charsInput = new LinkedList<>();
char[] chars = input.toCharArray();
for (int i = 0; i < input.length(); i++) {
if (charsInput.size() == 0) {
charsInput.add(chars[i]);
} else {
if (!charsInput.contains(chars[i])) {
charsInput.add(chars[i]);
} else if (charsInput.contains(chars[i])) {
charsInput.remove(Character.valueOf(chars[i]));
}
}
}
if (charsInput.size() > 0) {
return charsInput.get(0);
}
return (0);
}
}

private static int Solution(String s) {
// to check is values has been considered once
Set<String> set=new HashSet<String>();
// main loop
for (int i = 0; i < s.length(); i++) {
String temp = String.valueOf(s.charAt(i));
//rest of the values
String sub=s.substring(i+1);
if (set.add(temp) && !sub.contains(temp)) {
return i;
}
}
return -1;
}

PALIN- The next Palindrome - a SPOJ problem

I have opened an account for Ridit, one of 7-years-old students learning Java at SPOJ. The first task i gave to him was PALIN -The Next Palindrome. Here is the link to this problem- PALIN- The next Palindrome- SPOJAfter i explained it to him, he was able to solve it mostly except removing the leading zeros, which i did. Following is his solution of the problem -
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO Auto-generated method stub
try {
Scanner in = new Scanner(System.in);
int t = Integer.parseInt(in.nextLine());
String[] numbersInString = new String[t];
for (int i = 0; i <t; i++) {
String str = in.nextLine();
numbersInString[i] = removeLeadingZeros(str);
}
for (int i = 0 ; i<t; i++) {
int K = Integer.parseInt(numbersInString[i]);
int answer = findTheNextPalindrome(K);
System.out.println(answer);
}
}catch(Exception e) {
return;
}
}
static boolean isPalindrome(int x) {
String str = Integer.toString(x);
int length = str.length();
StringBuffer strBuff = new StringBuffer();
for(int i = length - 1;i>=0;i--) {
char ch = str.charAt(i);
strBuff.append(ch);
}
String str1 = strBuff.toString();
if(str.equals(str1)) {
return true;
}
return false;
}
static int findTheNextPalindrome(int K) {
for(int i = K+1;i<9999999; i++) {
if(isPalindrome(i) == true) {
return i;
}
}
return -1;
}
static String removeLeadingZeros(String str) {
String retString = str;
if(str.charAt(0) != '0') {
return retString;
}
return removeLeadingZeros(str.substring(1));
}
}
It is giving correct answer in Eclipse on his computer, but it is failing in SPOJ. If someone helps this little boy in his first submission, it will definitely make him very happy. I couldn't find any problem with this solution... Thank you in advance...

This might be helpful
import java.io.IOException;
import java.util.Scanner;
public class ThenNextPallindrom2 {
public static void main(String[] args) throws IOException {
// TODO Auto-generated method stub
int t = 0;
Scanner sc = new Scanner(System.in);
if(sc.hasNextInt()) {
t = sc.nextInt();
}
sc.nextLine();
int[] arr, arr2;
while(t > 0) {
t--;
String s = sc.nextLine();
arr = getStringToNumArray(s);
if(all9(arr)) {
arr2 = new int[arr.length + 1];
arr2[0] = 1;
for(int i=0;i<arr.length;i++) {
arr2[i+1] = 0;
}
arr2[arr2.length -1] = 1;
arr = arr2;
} else{
int mid = arr.length/ 2;
int left = mid-1;
int right = arr.length % 2 == 1 ? mid + 1 : mid;
boolean left_small = false;
while(left >= 0 && arr[left] == arr[right]) {
left--;
right++;
}
if(left < 0 || arr[left] < arr[right]) left_small = true;
if(!left_small) {
while(left >= 0) {
arr[right++] = arr[left--];
}
} else {
mid = arr.length/ 2;
left = mid-1;
int carry = 1;
if(arr.length % 2 == 0) {
right = mid;
} else {
arr[mid] += carry;
carry = arr[mid]/10;
arr[mid] %= 10;
right = mid + 1;
}
while(left >= 0) {
arr[left] += carry;
carry = arr[left] / 10;
arr[left] %= 10;
arr[right++] = arr[left--];
}
}
}
printArray(arr);
}
}
public static boolean all9(int[] arr) {
for(int i=0;i<arr.length;i++) {
if(arr[i] != 9)return false;
}
return true;
}
public static void printArray(int[] arr) {
for(int i=0;i<arr.length;i++) {
System.out.print(arr[i]);
}
System.out.println();
}
public static int[] getStringToNumArray(String s) {
int[] arr = new int[s.length()];
for(int i=0; i<s.length();i++) {
arr[i] = Integer.parseInt(String.valueOf(s.charAt(i)));
}
return arr;
}
}

What is wrong with this Vowel question? Out of index

I'm doing a homework task that is:
Find a unique vowel in the string that is preceded by a consonant, and this consonant is preceded by a vowel.
Example: "eeaaAOEacafu"
Result is: u
What i already did:
Main.class
public class Principal {
public static void main(String[] args) {
// TODO Auto-generated method stub
Stream str = new Stream();
str.setText("eeaaAOEacafu");
System.out.println(str.returnChar(str.getVowel()));
}
Stream.class
public class Stream {
String text;
char vowel;
public String getText() {
return texto;
}
public void setText(String text) {
this.text = text;
}
public char getVowel() {
return vowel;
}
public void setVowel(char vowel) {
this.vowel = vowel;
}
public boolean isVowel(String str) {
str = str.toLowerCase();
for(int i=0; i<str.length(); i++) {
char c = str.charAt(i);
if(c=='a' || c=='e' || c=='i' || c=='o'|| c=='u') {
return true;
} else {
return false;
}
}
return false;
}
public char returnChar(String str) {
char last;
char next;
char result = '0';
int j=1;
for(int i=0; i<str.length(); i++) {
last = str.charAt(i-1);
next = str.charAt(i+1);
j++;
if(!vogal(str.charAt(i))) {
if(vogal(last) && vogal(next)) {
result = next;
}
}
}
this.setVowel(result);
return result;
} }
This returns: String index out of range: -1
This j=1, was to fix this -1 out of range. It fix but i got new one: out of range 11 because of the next.
The thing is: I have to use pure java and no API.
Can you guys help me?

use regular expressions for the test and locating the character
[aeiouAEIOU][bcdfghjklmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ]([aeiouAEIOU])

Use a String as a cheap map to keep track of which vowels you've already seen. Also, keep a count of how many consecutive consonants you've encountered. Then, when you hit a vowel that you haven't seen before preceded by a single consonant you've found your answer.
public static void main(String[] args)
{
String s = "eeaaAOEacafu".toLowerCase();
int consCount = 0;
String seenVowels = "";
for(int i=0; i<s.length(); i++)
{
char c = s.charAt(i);
if("aeiou".indexOf(c) >= 0)
{
if(seenVowels.indexOf(c) == -1)
{
if(consCount == 1)
{
System.out.println("Result: " + c);
break;
}
seenVowels += c;
}
consCount = 0;
}
else consCount++;
}
}
Output:
Result: u
The above works if we take 'unique' to mean that we haven't seen the vowel before. If the vowel has to be unique within the input string then things are a little more complicated. Now we have to keep track of each vowel that meets the original criteria, but remove the solution if we subsequently encounter another instance of the same vowel.
Here's some code to illustrate:
public static void main(String[] args)
{
String s = "afuxekozue".toLowerCase();
int consCount = 0;
String seenVowels = "";
String answer = "";
for(int i=0; i<s.length(); i++)
{
char c = s.charAt(i);
if("aeiou".indexOf(c) >= 0)
{
if(seenVowels.indexOf(c) == -1)
{
if(consCount == 1)
{
answer += c;
}
seenVowels += c;
}
else if(answer.indexOf(c) >= 0)
{
answer = answer.replaceAll(String.valueOf(c), "");;
}
consCount = 0;
}
else consCount++;
}
if(answer.length() > 0)
System.out.println("Result: " + answer.charAt(0));
}
Output:
Result: o

Time Complexity of the below Solution? Its already been asked, but i still didnt understand the exponential part! Kindly explain

The following code prints all strings of length k where the characters are in sorted order. It does this by generating all strings of length k and then checking if each is sorted. What is its runtime?
public static int numChars = 26;
public static void printSortedStrings(int remaining) {
printSortedStrings(remaining, "");
}
public static void printSortedStrings(int remaining, String prefix) {
if (remaining == 0) {
if (isInOrder(prefix)) {
System.out.println(prefix); // Printing the string
}
} else {
for (int i = 0; i < numChars; i++) {
char c = ithLetter(i);
printSortedStrings(remaining - 1, prefix + c);
}
}
}
public static boolean isInOrder(String s) {
for (int i = 1; i < s.length(); i++) {
int prev = ithLetter(s.charAt(i - 1));
int curr = ithLetter(s.charAt(i));
if (prev > curr) {
return false;
}
}
return true;
}
public static char ithLetter(int i) {
return (char) (((int) 'a') + i);
}
public static void main(String[] args) {
printSortedStrings(2);
}

How to find the most common character in a String

I have a quick question. How would I find the most common character in a string in Java. I know logically how to do it, but I am not sure if my syntax in correct:
public class HelloWorld {
public static void main(String[] args){
String votes = "ABBAB";
char[] StoringArray = votes.toCharArray();
int numOFB = 0;
int numOFA = 0;
if (StoringArray.contains("A")) {
numOFA++;
} else if (StoringArray.contains("B")) {
numOFAB++;
}
if (numOFA = numOFB) {
System.out.println("Tie");
} else if (numOFA > B) {
System.out.println("A");
} else {
System.out.println("B");
}
}
}
Could anyone help me with how to correctly do this in Java?

You can not compare char Array with string, below logic should work and give you what you need:
public static void main(String[] args){
String votes = "ABBAB";
char[] storingArray = votes.toCharArray();
int numOFB = 0;
int numOFA = 0;
for(char c : storingArray) {
if(c == 'A') {
numOFA++;
}
if(c == 'B') {
numOFB++;
}
}
if (numOFA == numOFB) {
System.out.println("Tie");
} else if (numOFA > numOFB) {
System.out.println("A");
} else {
System.out.println("B");
}
}
There are couple of mistakes in your code:
You can not use if (numOFA = numOFB) it is not valid expression. You should use == to compare
You can not compare char Array with contains method. It should be used on String object

As the comments said; it looks like you're counting the number of A's or B's, not the longest substring. Are you only analyzing a String composed of A's and B's?
Also, you're using = to check for equality when you should be using ==. I would recommend using an IDE like Eclipse which would show you when you're doing this.
Edit: also, you're not looping through the array. You're just checking if the String contains an A or a B and adding 1 if it does. You need to loop through the entire array.

Actually, I was working with it, and I found this is the nicest way to do it:
String votes = "ABBAB";
char[] StoringArray = votes.toCharArray();
int B = 0;
int A = 0;
for (int i = 0; i < StoringArray.length; i ++) {
if (StoringArray[i] == 'A') {
A++;
} else if (StoringArray[i] == 'B') {
B++;
}
}
if (A == B) {
System.out.println("Tie");
} else if (A > B) {
System.out.println("A");
} else {
System.out.println("B");
}

I would give you a more abstract solution:
public class Counter{
private char c;
private int count;
Counter(char c, int count){
this.c=c;
this.count=count;
}
public char getC() {
return c;
}
public void setC(char c) {
this.c = c;
}
public int getCount() {
return count;
}
public void addOcurrence() {
this.count++;
}
#Override
public boolean equals(Object obj) {
if(obj!=null)
if(((Counter)obj).getC()== this.c)
return true;
return false;
}
}
public static void main(String[] args){
String votes = "whateveryouwanttoputhereorcanbefromaparameter";
char[] storingArray = votes.toCharArray();
List<Counter> listCounter = new ArrayList<Counter>();
for(char aChar : storingArray){
Counter compareCounter = new Counter(aChar,1);
if(listCounter.contains(compareCounter)){
listCounter.get(listCounter.indexOf(compareCounter)).addOcurrence();
}else{
listCounter.add(compareCounter);
}
}
Counter max = listCounter.get(0);
for( Counter c : listCounter){
if(c.getCount() > max.getCount()){
max = c;
}
}
System.out.println("the character with more ocurrence is: "+max.getC());
}