I am having trouble printing out triangles recursively involving spaces and asterisks. Apparently stringbuffer or stringbuilder may be necessary to calculate the correct number of spaces and asterisks, but I am having a bit of difficulty. The 2 triangles should look like:
****
***
**
*
and
*
**
***
****
public static String printTriangle(int num)
{
if (num == 0) {
return "";
}
String dots = "";
for (int i = 0; i < num; i++) {
dots = dots + "*";
}
System.out.println(dots);
return printTriangle(num-1) + dots;
}
public static String printTriangle2(int num) {
if (num == 0) {
return "";
}
String dots = printTriangle2(num-1);
dots = dots + ".";
String spaces = "";
for (int i = 0; i < num; i++) {
spaces = spaces + " ";
}
String line = spaces + dots;
System.out.println(line);
return line;
}
This is what I have so far. Any help would be appreciated.
This is the output currently:
****
***
**
*
.
..
...
....
Here's a fairly simple implementation:
static void printTriangle(int n, int len)
{
if(n == len) return;
printRow(n, len);
printTriangle(n+1, len);
printRow(n, len);
}
static void printRow(int n, int len)
{
for(int i=0; i<n; i++) System.out.print(" ");
for(int i=n; i<len; i++) System.out.print("*");
System.out.println();
}
Test:
printTriangle(0, 4);
Output:
****
***
**
*
*
**
***
****
Although I like this:
static void printTriangle(String s)
{
if(!s.contains("*")) return;
System.out.println(s);
printTriangle(" " + s.replaceFirst("\\*", ""));
System.out.println(s);
}
Called with
printTriangle("****");
try the follwing code and call with step=0
printTriangle(4,0)
printTriangle2(4,0)
public static String printTriangle(int num, int step)
{
if (num == 0) {
return "";
}
String ast = "";
for (int i = 0; i < num; i++) {
ast = ast + "*";
}
String sps = "";
for (int i = 0; i < step; i++) {
sps = sps + " ";
}
System.out.println(sps+ast);
return printTriangle(num-1, step+1) ;
}
public static String printTriangle2(int num, int step)
{
if (num == 0) {
return "";
}
String ast = "";
for (int i = 0; i <= step; i++) {
ast = ast + "*";
}
String sps = "";
for (int i = 0; i < num; i++) {
sps = sps + " ";
}
System.out.println(sps+ast);
return printTriangle2(num-1, step+1) ;
}
I am trying to do a summation puzzle, the questions asks to use summation puzzles by enumerating and testing all possible configurations and then it says use it to solve the examples given. The examples given were
pot + pan = bib
dog+cat= pig
boy + girl = baby
I keep getting an error saying left hand side of assignment must be a variable
charSet.charAt(setIndex++) = stringTwo.charAt(loop);
cannot convert from int to bool.
if (exists = 0)
Also in my code where I try to display the output it doesn't run.
import java.util.Scanner;
public class Recursion
{
// Example program
public static String stringOne = new String(new char[10]);
public static String stringTwo = new String(new char[10]);
public static String stringThree = new String(new char[11]);
public static String charSet = new String(new char[11]);
public static int numberOne;
public static int numberTwo;
public static int numberThree;
public static int maxCharCount;
public static int[] numberSet = new int[10];
public static void checkForEquality()
{
numberOne = numberTwo = numberThree = 0;
int loop;
int subloop;
for (loop = 0; loop < stringOne.length(); loop++)
{
for (subloop = 0; subloop < maxCharCount; subloop++)
{
if (stringOne.charAt(loop) == charSet.charAt(subloop))
{
if (loop == 0 && numberSet[subloop] == 0)
return;
//generate the number
numberOne = (numberOne * 10) + numberSet[subloop];
}
}
}
for (loop = 0; loop < stringOne.length(); loop++)
{
for (subloop = 0; subloop < stringTwo.length(); subloop++)
{
if (stringTwo.charAt(loop) == charSet.charAt(subloop))
{
if (loop == 0 && numberSet[subloop] == 0)
return;
//generate the numeber
numberTwo = (numberTwo * 10) + numberSet[subloop];
}
}
}
for (loop = 0; loop < stringThree.length(); loop++)
{
for (subloop = 0; subloop < maxCharCount; subloop++)
{
if (stringThree.charAt(loop) == charSet.charAt(subloop))
{
if (loop == 0 && numberSet[subloop] == 0)
return;
//generate the number
numberThree = (numberThree * 10) + numberSet[subloop];
}
}
}
if (numberOne + numberTwo == numberThree)
{
//display the output
System.out.print(" Summation Puzzle solved. ");
System.out.print("\n");
System.out.print(stringOne);
System.out.print("<==>");
System.out.print(numberOne);
System.out.print("\n");
System.out.print(stringTwo);
System.out.print("<==>");
System.out.print(numberTwo);
System.out.print("\n");
System.out.print(stringThree);
System.out.print("<==>");
System.out.print(numberThree);
System.out.print("\n");
//loop to show the result
for (loop = 0; loop < maxCharCount; loop++)
{
System.out.print(charSet.charAt(loop));
System.out.print("<==>");
System.out.print(numberSet[loop]);
System.out.print("\n");
}
System.exit(0);
}
}
public static void generateCombinations(int indexCounter, int[] availableSet)
{
int loop;
if (indexCounter != 0)
{
for (loop = 0; loop < 10; loop++)
{
numberSet[indexCounter] = loop;
if (availableSet[loop] == 1)
{
availableSet[loop] = 0;
generateCombinations(indexCounter + 1, availableSet);
availableSet[loop] = 1;
}
}
}
if (indexCounter == maxCharCount)
{
checkForEquality();
}
}
public static void createCharSet()
{
int loop;
int setIndex;
int exists;
int subloop;
setIndex = 0;
for (loop = 0; loop < stringOne.length(); loop++)
{
exists = 0;
for (subloop = 0; subloop < setIndex; subloop++)
{
if (stringOne.charAt(loop) == charSet.charAt(subloop))
{
exists = 1;
}
}
if (exists == 0)
{
charSet = StringFunctions.changeCharacter(charSet, setIndex++, stringOne.charAt(loop));
}
}
for (loop = 0; loop < stringTwo.length(); loop++)
{
exists = 0;
for (subloop = 0; subloop < setIndex; subloop++)
{
if (stringTwo.charAt(loop) == charSet.charAt(subloop))
{
exists = 1;
}
}
if (exists == 0)
{
charSet = StringFunctions.changeCharacter(charSet, setIndex++, stringTwo.charAt(loop));
}
}
for (loop = 0; loop < stringThree.length(); loop++)
{
exists = 0;
for (subloop = 0; subloop < setIndex; subloop++)
{
if (stringThree.charAt(loop) == charSet.charAt(subloop))
{
exists = 1;
}
}
if (exists == 0)
{
charSet = StringFunctions.changeCharacter(charSet, setIndex++, stringThree.charAt(loop));
}
}
maxCharCount = setIndex;
}
public static void calculateSummation()
{
int loop;
if (maxCharCount > 10)
{
System.out.print("Please check the input again");
return;
}
else
{
int[] avaliableSet = new int[10];
for (loop = 0; loop < 10; loop++)
{
avaliableSet[loop] = 1;
}
generateCombinations(0, avaliableSet);
}
}
public static void main(String[]args)
{
Scanner scan = new Scanner(System.in);
System.out.print(" Enter the first String :");
stringOne = scan.next();
System.out.print(" Enter the second String :");
stringTwo = scan.next();
System.out.print(" Enter the thirsd String :");
stringThree = scan.next();
createCharSet();
System.out.print(" The character set formed from the given string = ");
System.out.print(charSet);
calculateSummation();
checkForEquality();
}
}
A lot of your problems are stemming from the syntax errors in the code you've written. For example:
line 74: if (stringThree.charAt(loop) == charSet.charAt(subloop) != null)
charSet.charAt(subloop) != null is an invalid comparison since the != operator cannot be used for booleans when comparing to null. If you're trying to determine if the characters return from .charAt(var) exist, use parentheses to make independent comparisons of each object.charAt(var) to null.
line 183: charSet = tangible.StringFunctions.changeCharacter(charSet, setIndex++, stringOne.charAt(loop));
tangible is ironically not tangible, as the variable does not exist locally or has not been defined globally.
charSet.charAt(setIndex++) = stringTwo.charAt(loop);
charSet.charAt(setIndex++) is a method that returns a character. This does not mean you can set the character at the specified index like it's a variable.
line 227: if (exists = 0)
You must use == when conducting comparisons in a conditional.
line 269: Scanner scan = new Scanner(System.in);
The Scanner class was not imported and thus cannot be used.
line 283: charSet.charAt(maxCharCount) = '\0';
Again, you can't use .charAt(var) to set the character at that index like it's a variable.
All of these problems can be self-determined by using a proper IDE, such as Eclipse.
Edit: Try to spend a little more time with pencil and paper working out the logic of your program before writing the code to represent your algorithm. This way you have focus and can write more comprehensive, commented, cleaner code. Here is a bit of a guide to help condense your existing project.
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));
}
}
Here is my code.
Method counter counts the number of times each alphabet occurs in the string.
public class Hard{
public static void counter (String s) {
for (int n = 0; n < s.length() ; n++) {
int count = 0 ,bool = 1;
if (n > 0) {
for (int L = n-1 ; L >= 0 ; L--) {
if (s.charAt(n) == s.charAt(L)) {bool = 0; }
}
}
for (int f = 0; f < s.length() ; f++ ) {
if (bool == 0) { break ; }
if (s.charAt(n) == s.charAt(f)) {count++;}
}
if (count > 0 ) {
System.out.println(s.charAt(n)+" appears "+count+" times.");
}
}
}
public static void main (String[] args) {
counter("bbaddadxzxwfgb$.fgfdf");
}
}
Assuming that you're using Java and assuming A and a are counted as the same letter.
public static int[] counter (String s) {
int [] countArr = new int[26];
for(int i=0; i<s.length(); i++) {
char charAtI = s.charAt(i);
if(Character.isLetter(charAtI)) {
countArr[Character.isUpperCase(charAtI) ? charAtI - 'A' : charAtI - 'a']++;
}
}
return countArr;
}
public static void main (String[] args) {
int [] countArr = counter("asif and abid.");
for(int i = 0; i<countArr.length; i++) {
if(countArr[i] > 0) {
System.out.println(MessageFormat.format("{0} appears {1} times", (char)(i + 'a'), countArr[i]));
}
}
}
The key to the optimization lies in the fact that it performs one pass with no nested loops and once you have all the information you require, then worry about how to present it to the user.
public static HashMap<Character, Integer> counter(String input) {
HashMap<Character, Integer> chCount = new HashMap<>();
for (int i = 0 i < input.length(); i++) {
char c = input.charAt(i);
if (chCount.containsKey(c)) {
int count = chCount.get(c);
chCount.put(c, count + 1);
} else {
chCount.put(c, 1);
}
}
return chCount;
}
You can use hashmaps as well. It will work for other characters like '#', etc and you can treat 'A' and 'a' differently, because the key value will be different.
HashMap tutorials Point
You can count chars like this, using java 8
public static void main(String[] args){
String string="what is this?";
List<Character> chars=string.chars().distinct().mapToObj(i->Character.valueOf((char)i)).filter(p->Character.isAlphabetic(p)).collect(Collectors.toList());
chars.forEach(c->{
long count=string.chars().mapToObj(i->Character.valueOf((char)i)).filter(p->p==c).count();
System.out.println(c+" appear "+count+" times");
});
}
My getCount() method should be returning 2 in this case but it is returning 7. I think the reason why it's counting incorrectly is because it's looping through the for 7 times because that's the length of the string. However, I simply just want to scan the string for the pattern and increment patternCount by 1 each time the pattern occurs in the string I'm scanning. Here's my code:
package a2;
public class DNAStrandAdept {
private String strand;
private String pattern;
private String passedStrand;
private int ACount;
private int CCount;
private int GCount;
private int TCount;
private int patternCount = 0;
public static void main(String[] args) {
DNAStrandAdept test = new DNAStrandAdept("AGGTTGG");
System.out.println("A count: " + test.getACount());
System.out.println("C count: " + test.getCCount());
System.out.println("G count: " + test.getGCount());
System.out.println("T count: " + test.getTCount());
System.out.println("Strand: " + test.getStrandString());
System.out.println("Strand length: " + test.getLength());
System.out.println("Pattern Count: " + test.getCount("GG"));
}
public DNAStrandAdept(String strand) {
passedStrand = strand;
if (passedStrand.contains("a") || passedStrand.contains("c")
|| passedStrand.contains("g") || passedStrand.contains("t")) {
throw new RuntimeException("Illegal DNA strand");
} else if (passedStrand.contains("1") || passedStrand.contains("2")
|| passedStrand.contains("3") || passedStrand.contains("4")
|| passedStrand.contains("5") || passedStrand.contains("6")
|| passedStrand.contains("7") || passedStrand.contains("8")
|| passedStrand.contains("9") || passedStrand.contains("0")) {
throw new RuntimeException("Illegal DNA Strand");
} else if (passedStrand.contains(",") || passedStrand.contains(".")
|| passedStrand.contains("?") || passedStrand.contains("/")
|| passedStrand.contains("<") || passedStrand.contains(">")) {
throw new RuntimeException("Illegal DNA Strand");
}
}
public int getACount() {
for (int i = 0; i < passedStrand.length(); i++) {
if (passedStrand.charAt(i) == 'A') {
ACount++;
}
}
return ACount;
}
public int getCCount() {
for (int i = 0; i < passedStrand.length(); i++) {
if (passedStrand.charAt(i) == 'C') {
CCount++;
}
}
return CCount;
}
public int getGCount() {
for (int i = 0; i < passedStrand.length(); i++) {
if (passedStrand.charAt(i) == 'G') {
GCount++;
}
}
return GCount;
}
public int getTCount() {
for (int i = 0; i < passedStrand.length(); i++) {
if (passedStrand.charAt(i) == 'T') {
TCount++;
}
}
return TCount;
}
public String getStrandString() {
return passedStrand;
}
public int getLength() {
return passedStrand.length();
}
public int getCount(String pattern) {
for (int i = 0; i < passedStrand.length(); i++) {
if (passedStrand.contains(pattern)) {
patternCount++;
}
}
return patternCount;
}
public int findPattern(String pattern, int startIndex) {
return 0;
}
}
Here is my output:
A count: 1
C count: 0
G count: 4
T count: 2
Strand: AGGTTGG
Strand length: 7
Pattern Count: 7
Notice your for loop:
for (int i = 0; i < passedStrand.length(); i++) {
if (passedStrand.contains(pattern)) {
patternCount++;
}
}
If the pattern is there in passedStrand, it will always be true. It doesn't really depend upon any part of the loop. Since the loop is running passedStrand.length() number of times, that condition will be checked that many times. And every time, since it is true, the patternCount is incremented. And hence the final value of patternCount will be passedStrand.length();.
What you rather want to do is, starting at every index, check if next pattern.length() number of characters, make up a string equal to pattern. If yes, then increment patternCount. So, you would need to make use of substring method here:
int patternLen = pattern.length();
for (int i = 0; i < passedStrand.length() - patternLen + 1; i++) {
if (passedStrand.substring(i, i + patternLen).equals(pattern)) {
patternCount++;
}
}
Also notice that, the loop will not really run till the end of passedStrand string. You just need to run till the index, from where there is a possibility of complete occurrence of pattern string.
This method creates extra String objects inside the for loop, due to substring invocation. You can avoid this by using String#indexOf method. You just keep on finding the next index of the pattern in the passedStrand, till you get the index as -1, where it ends.
int startIndex = passedStrand.indexOf(pattern);
while (startIndex != -1) {
patternCount++;
startIndex = passedStrand.indexOf(pattern, startIndex + pattern.length());
}
If efficiency is not a big concern, then regex is really sweet. See how:
public int getCount(String pattern) {
int patternCount = 0;
Matcher matcher = Pattern.compile(pattern).matcher(passedStrand);
while (matcher.find()) {
patternCount++;
}
return patternCount;
}