Java compressing Strings - java
I need to create a method that receives a String and also returns a String.
Ex input: AAABBBBCC
Ex output: 3A4B2C
Well, this is quite embarrassing and I couldn't manage to do it on the interview that I had today ( I was applying for a Junior position ), now, trying at home I made something that works statically, I mean, not using a loop which is kind of useless but I don't know if I'm not getting enough hours of sleep or something but I can't figure it out how my for loop should look like. This is the code:
public static String Comprimir(String texto){
StringBuilder objString = new StringBuilder();
int count;
char match;
count = texto.substring(texto.indexOf(texto.charAt(1)), texto.lastIndexOf(texto.charAt(1))).length()+1;
match = texto.charAt(1);
objString.append(count);
objString.append(match);
return objString.toString();
}
Thanks for your help, I'm trying to improve my logic skills.
Loop through the string remembering what you last saw. Every time you see the same letter count. When you see a new letter put what you have counted onto the output and set the new letter as what you have last seen.
String input = "AAABBBBCC";
int count = 1;
char last = input.charAt(0);
StringBuilder output = new StringBuilder();
for(int i = 1; i < input.length(); i++){
if(input.charAt(i) == last){
count++;
}else{
if(count > 1){
output.append(""+count+last);
}else{
output.append(last);
}
count = 1;
last = input.charAt(i);
}
}
if(count > 1){
output.append(""+count+last);
}else{
output.append(last);
}
System.out.println(output.toString());
You can do that using the following steps:
Create a HashMap
For every character, Get the value from the hashmap
-If the value is null, enter 1
-else, replace the value with (value+1)
Iterate over the HashMap and keep concatenating (Value+Key)
use StringBuilder (you did that)
define two variables - previousChar and counter
loop from 0 to str.length() - 1
each time get str.charat(i) and compare it to what's stored in the previousChar variable
if the previous char is the same, increment a counter
if the previous char is not the same, and counter is 1, increment counter
if the previous char is not the same, and counter is >1, append counter + currentChar, reset counter
after the comparison, assign the current char previousChar
cover corner cases like "first char"
Something like that.
The easiest approach:- Time Complexity - O(n)
public static void main(String[] args) {
String str = "AAABBBBCC"; //input String
int length = str.length(); //length of a String
//Created an object of a StringBuilder class
StringBuilder sb = new StringBuilder();
int count=1; //counter for counting number of occurances
for(int i=0; i<length; i++){
//if i reaches at the end then append all and break the loop
if(i==length-1){
sb.append(str.charAt(i)+""+count);
break;
}
//if two successive chars are equal then increase the counter
if(str.charAt(i)==str.charAt(i+1)){
count++;
}
else{
//else append character with its count
sb.append(str.charAt(i)+""+count);
count=1; //reseting the counter to 1
}
}
//String representation of a StringBuilder object
System.out.println(sb.toString());
}
In the count=... line, lastIndexOf will not care about consecutive values, and will just give the last occurence.
For instance, in the string "ABBA", the substring would be the whole string.
Also, taking the length of the substring is equivalent to subtracting the two indexes.
I really think that you need a loop.
Here is an example :
public static String compress(String text) {
String result = "";
int index = 0;
while (index < text.length()) {
char c = text.charAt(index);
int count = count(text, index);
if (count == 1)
result += "" + c;
else
result += "" + count + c;
index += count;
}
return result;
}
public static int count(String text, int index) {
char c = text.charAt(index);
int i = 1;
while (index + i < text.length() && text.charAt(index + i) == c)
i++;
return i;
}
public static void main(String[] args) {
String test = "AAABBCCC";
System.out.println(compress(test));
}
Please try this one. This may help to print the count of characters which we pass on string format through console.
import java.util.*;
public class CountCharacterArray {
private static Scanner inp;
public static void main(String args[]) {
inp = new Scanner(System.in);
String str=inp.nextLine();
List<Character> arrlist = new ArrayList<Character>();
for(int i=0; i<str.length();i++){
arrlist.add(str.charAt(i));
}
for(int i=0; i<str.length();i++){
int freq = Collections.frequency(arrlist, str.charAt(i));
System.out.println("Frequency of "+ str.charAt(i)+ " is: "+freq);
}
}
}
Java's not my main language, hardly ever use it, but I wanted to give it a shot :]
Not even sure if your assignment requires a loop, but here's a regexp approach:
public static String compress_string(String inp) {
String compressed = "";
Pattern pattern = Pattern.compile("([\\w])\\1*");
Matcher matcher = pattern.matcher(inp);
while(matcher.find()) {
String group = matcher.group();
if (group.length() > 1) compressed += group.length() + "";
compressed += group.charAt(0);
}
return compressed;
}
This is just one more way of doing it.
public static String compressor(String raw) {
StringBuilder builder = new StringBuilder();
int counter = 0;
int length = raw.length();
int j = 0;
while (counter < length) {
j = 0;
while (counter + j < length && raw.charAt(counter + j) == raw.charAt(counter)) {
j++;
}
if (j > 1) {
builder.append(j);
}
builder.append(raw.charAt(counter));
counter += j;
}
return builder.toString();
}
The following can be used if you are looking for a basic solution. Iterate through the string with one element and after finding all the element occurrences, remove that character. So that it will not interfere in the next search.
public static void main(String[] args) {
String string = "aaabbbbbaccc";
int counter;
String result="";
int i=0;
while (i<string.length()){
counter=1;
for (int j=i+1;j<string.length();j++){
System.out.println("string length ="+string.length());
if (string.charAt(i) == string.charAt(j)){
counter++;
}
}
result = result+string.charAt(i)+counter;
string = string.replaceAll(String.valueOf(string.charAt(i)), "");
}
System.out.println("result is = "+result);
}
And the output will be :=
result is = a4b5c3
private String Comprimir(String input){
String output="";
Map<Character,Integer> map=new HashMap<Character,Integer>();
for(int i=0;i<input.length();i++){
Character character=input.charAt(i);
if(map.containsKey(character)){
map.put(character, map.get(character)+1);
}else
map.put(character, 1);
}
for (Entry<Character, Integer> entry : map.entrySet()) {
output+=entry.getValue()+""+entry.getKey().charValue();
}
return output;
}
One other simple way using Multiset of guava-
import java.util.Arrays;
import com.google.common.collect.HashMultiset;
import com.google.common.collect.Multiset;
import com.google.common.collect.Multiset.Entry;
public class WordSpit {
public static void main(String[] args) {
String output="";
Multiset<String> wordsMultiset = HashMultiset.create();
String[] words="AAABBBBCC".split("");
wordsMultiset.addAll(Arrays.asList(words));
for (Entry<String> string : wordsMultiset.entrySet()) {
if(!string.getElement().isEmpty())
output+=string.getCount()+""+string.getElement();
}
System.out.println(output);
}
}
consider the below Solution in which the String s1 identifies the unique characters that are available in a given String s (for loop 1), in the second for loop build a string s2 that contains unique character and no of times it is repeated by comparing string s1 with s.
public static void main(String[] args)
{
// TODO Auto-generated method stub
String s = "aaaabbccccdddeee";//given string
String s1 = ""; // string to identify how many unique letters are available in a string
String s2=""; //decompressed string will be appended to this string
int count=0;
for(int i=0;i<s.length();i++) {
if(s1.indexOf(s.charAt(i))<0) {
s1 = s1+s.charAt(i);
}
}
for(int i=0;i<s1.length();i++) {
for(int j=0;j<s.length();j++) {
if(s1.charAt(i)==s.charAt(j)) {
count++;
}
}
s2=s2+s1.charAt(i)+count;
count=0;
}
System.out.println(s2);
}
It may help you.
public class StringCompresser
{
public static void main(String[] args)
{
System.out.println(compress("AAABBBBCC"));
System.out.println(compress("AAABC"));
System.out.println(compress("A"));
System.out.println(compress("ABBDCC"));
System.out.println(compress("AZXYC"));
}
static String compress(String str)
{
StringBuilder stringBuilder = new StringBuilder();
char[] charArray = str.toCharArray();
int count = 1;
char lastChar = 0;
char nextChar = 0;
lastChar = charArray[0];
for (int i = 1; i < charArray.length; i++)
{
nextChar = charArray[i];
if (lastChar == nextChar)
{
count++;
}
else
{
stringBuilder.append(count).append(lastChar);
count = 1;
lastChar = nextChar;
}
}
stringBuilder.append(count).append(lastChar);
String compressed = stringBuilder.toString();
return compressed;
}
}
Output:
3A4B2C
3A1B1C
1A
1A2B1D2C
1A1Z1X1Y1C
The answers which used Map will not work for cases like aabbbccddabc as in that case the output should be a2b3c2d2a1b1c1.
In that case this implementation can be used :
private String compressString(String input) {
String output = "";
char[] arr = input.toCharArray();
Map<Character, Integer> myMap = new LinkedHashMap<>();
for (int i = 0; i < arr.length; i++) {
if (i > 0 && arr[i] != arr[i - 1]) {
output = output + arr[i - 1] + myMap.get(arr[i - 1]);
myMap.put(arr[i - 1], 0);
}
if (myMap.containsKey(arr[i])) {
myMap.put(arr[i], myMap.get(arr[i]) + 1);
} else {
myMap.put(arr[i], 1);
}
}
for (Character c : myMap.keySet()) {
if (myMap.get(c) != 0) {
output = output + c + myMap.get(c);
}
}
return output;
}
O(n) approach
No need for hashing. The idea is to find the first Non-matching character.
The count of each character would be the difference in the indices of both characters.
for a detailed answer: https://stackoverflow.com/a/55898810/7972621
The only catch is that we need to add a dummy letter so that the comparison for
the last character is possible.
private static String compress(String s){
StringBuilder result = new StringBuilder();
int j = 0;
s = s + '#';
for(int i=1; i < s.length(); i++){
if(s.charAt(i) != s.charAt(j)){
result.append(i-j);
result.append(s.charAt(j));
j = i;
}
}
return result.toString();
}
The code below will ask the user for user to input a specific character to count the occurrence .
import java.util.Scanner;
class CountingOccurences {
public static void main(String[] args) {
Scanner inp = new Scanner(System.in);
String str;
char ch;
int count=0;
System.out.println("Enter the string:");
str=inp.nextLine();
System.out.println("Enter th Char to see the occurence\n");
ch=inp.next().charAt(0);
for(int i=0;i<str.length();i++)
{
if(str.charAt(i)==ch)
{
count++;
}
}
System.out.println("The Character is Occuring");
System.out.println(count+"Times");
}
}
public static char[] compressionTester( char[] s){
if(s == null){
throw new IllegalArgumentException();
}
HashMap<Character, Integer> map = new HashMap<>();
for (int i = 0 ; i < s.length ; i++) {
if(!map.containsKey(s[i])){
map.put(s[i], 1);
}
else{
int value = map.get(s[i]);
value++;
map.put(s[i],value);
}
}
String newer="";
for( Character n : map.keySet()){
newer = newer + n + map.get(n);
}
char[] n = newer.toCharArray();
if(s.length > n.length){
return n;
}
else{
return s;
}
}
package com.tell.datetime;
import java.util.Stack;
public class StringCompression {
public static void main(String[] args) {
String input = "abbcccdddd";
System.out.println(compressString(input));
}
public static String compressString(String input) {
if (input == null || input.length() == 0)
return input;
String finalCompressedString = "";
String lastElement="";
char[] charArray = input.toCharArray();
Stack stack = new Stack();
int elementCount = 0;
for (int i = 0; i < charArray.length; i++) {
char currentElement = charArray[i];
if (i == 0) {
stack.push((currentElement+""));
continue;
} else {
if ((currentElement+"").equalsIgnoreCase((String)stack.peek())) {
stack.push(currentElement + "");
if(i==charArray.length-1)
{
while (!stack.isEmpty()) {
lastElement = (String)stack.pop();
elementCount++;
}
finalCompressedString += lastElement + "" + elementCount;
}else
continue;
}
else {
while (!stack.isEmpty()) {
lastElement = (String)stack.pop();
elementCount++;
}
finalCompressedString += lastElement + "" + elementCount;
elementCount=0;
stack.push(currentElement+"");
}
}
}
if (finalCompressedString.length() >= input.length())
return input;
else
return finalCompressedString;
}
}
public class StringCompression {
public static void main(String[] args){
String s = "aabcccccaaazdaaa";
char check = s.charAt(0);
int count = 0;
for(int i=0; i<s.length(); i++){
if(s.charAt(i) == check) {
count++;
if(i==s.length()-1){
System.out.print(s.charAt(i));
System.out.print(count);
}
} else {
System.out.print(s.charAt(i-1));
System.out.print(count);
check = s.charAt(i);
count = 1;
if(i==s.length()-1){
System.out.print(s.charAt(i));
System.out.print(count);
}
}
}
}
// O(N) loop through entire character array
// match current char with next one, if they matches count++
// if don't then just append current char and counter value and then reset counter.
// special case is the last characters, for that just check if count value is > 0, if it's then append the counter value and the last char
private String compress(String str) {
char[] c = str.toCharArray();
String newStr = "";
int count = 1;
for (int i = 0; i < c.length - 1; i++) {
int j = i + 1;
if (c[i] == c[j]) {
count++;
} else {
newStr = newStr + c[i] + count;
count = 1;
}
}
// this is for the last strings...
if (count > 0) {
newStr = newStr + c[c.length - 1] + count;
}
return newStr;
}
public class StringCompression {
public static void main(String... args){
String s="aabbcccaa";
//a2b2c3a2
for(int i=0;i<s.length()-1;i++){
int count=1;
while(i<s.length()-1 && s.charAt(i)==s.charAt(i+1)){
count++;
i++;
}
System.out.print(s.charAt(i));
System.out.print(count);
}
System.out.println(" ");
}
}
This is a leet code problem 443. Most of the answers here uses StringBuilder or a HashMap, the actual problem statement is to solve using the input char array and in place array modification.
public int compress(char[] chars) {
int startIndex = 0;
int lastArrayIndex = 0;
if (chars.length == 1) {
return 1;
}
if (chars.length == 0) {
return 0;
}
for (int j = startIndex + 1; j < chars.length; j++) {
if (chars[startIndex] != chars[j]) {
chars[lastArrayIndex] = chars[startIndex];
lastArrayIndex++;
if ((j - startIndex) > 1) {
for (char c : String.valueOf(j - startIndex).toCharArray()) {
chars[lastArrayIndex] = c;
lastArrayIndex++;
}
}
startIndex = j;
}
if (j == chars.length - 1) {
if (j - startIndex >= 1) {
j = chars.length;
chars[lastArrayIndex] = chars[startIndex];
lastArrayIndex++;
for (char c : String.valueOf(j - startIndex).toCharArray()) {
chars[lastArrayIndex] = c;
lastArrayIndex++;
}
} else {
chars[lastArrayIndex] = chars[startIndex];
lastArrayIndex++;
}
}
}
return lastArrayIndex;
}
}
Related
String compression algorithm in Java
I am looking to implement a method to perform basic string compression in the form of: aabcccccaaa -> a2b1c5a3 I have this program: import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String str = sc.nextLine(); System.out.println(compress(str)); } public static String compress(String str) { char[] chars = str.toCharArray(); int count = 0; String result = ""; for (int i = 0; i < chars.length; i++) { char curr = chars[i]; result += curr; for (int j = i; j < chars.length; j++) { if (chars[j] == curr) { count++; } else { i += count; break; } } result += count; count = 0; } return result; } } But in my tests I am always missing the last character count. I assume this is because the program gets out of the inner for loop before it should, but why is this the case? Thanks a lot
You don't need two for loops for this and can do it in one go like so String str = "aaabbbbccccca"; char[] chars = str.toCharArray(); char currentChar = str.length() > 0 ? chars[0] : ' '; char prevChar = ' '; int count = 1; StringBuilder finalString = new StringBuilder(); if(str.length() > 0) for(int i = 1; i < chars.length; i++) { if(currentChar == chars[i]) { count++; }else{ finalString.append(currentChar + "" + count); prevChar = currentChar; currentChar = chars[i]; count = 1; } } if(str.length() > 0 && prevChar != currentChar) finalString.append(currentChar + "" + count); System.out.println(finalString.toString()); Output is: a3b4c5a1 for aaabbbbccccca
Keep a track of character that you are reading and compare it with next character of the string. If it is different, reset the count. public static void stringCompression (String compression) { String finalCompressedString = ""; char current = '1'; int count = 0; compression = compression + '1'; for (int i = 0; i < compression.length(); i++) { if (compression.charAt(i) == current) { count = count + 1; } else { if (current != '1') finalCompressedString = finalCompressedString + (current + Integer.toString(count)); count = 1; current = compression.charAt(i); } } System.out.println(finalCompressedString); }
My answer for String Compression in java. In this what i have done is and what you should have done is that , Keep a record of the characters that that are coming for a specific number of times, do so by comparing the current character with the next character , and when the current and the next character become unequal reset the value of count and repeat the whole process again for the next different character. Hope it helps! import java.util.*; public class Main { public static void main(String args[]) { Scanner sc = new Scanner(System.in); String str = sc.nextLine(); int count = 0; for (int i=0; i<str.length(); ++i) { int j=i+1; count=1; while (j!=str.length() && str.charAt(i) == str.charAt(j)) { count += 1; j += 1; i += 1; } System.out.print(str.charAt(i)); if (count > 1) { System.out.print(count); } } } }
Java Compress no return?
what my code is supposed to be doing is converting input strings and outputing the compressed versions Ex. Input: "qqqwww" Output: "3q3w". But my code returns nothing. P.S. IO is just an input system. public class Compress { public static String compress(String original){ String s = ""; char s1 = original.charAt(0); int count = 1; for(int i = 0; i < original.length(); i++){ char c = original.charAt(i); if(c == s1){ count++; } else{ s = s + count + s1; //i think the problem is here right??? count = 1; } s1 = c; } return s; } public static void main(String[] args){ String s = IO.readString(); String y = compress(s); System.out.println(y); } }
Your could should look like this : String returnString=""; for (int index = 0; index < original.length();) { char currentChar = original.charAt(index); int counter=1; while(++index < original.length() && currentChar==original.charAt(index)) { counter++; } returnString=returnString+counter+currentChar; } return returnString; } Here we loop thought the string (outer for loop) and check if the adjacent values are same the we keep adding them. (inner while loop)
How to remove repeated letter from words in Java
i have problem writing java code to remove repeated letters from word.This code will remove repeated letter by accepting only one of the letter which is repeating . Suppose, if input is "SUSHIL" then output would be "SUHIL". This java code i write. import java.io.*; import java.util.*; public class Repeat { public static void main(String args[]) { Scanner sc = new Scanner(System.in); String name = sc.nextLine(); char ch1, ch2; int i, j; int l = name.length(); String result = ""; for (i = 0; i < l; i++) { for (j = 1; j < l; j++) { ch1 = name.charAt(i); ch2 = name.charAt(j); if (ch1 != ch2) { result = result + ch1; break; } } } System.out.println("Output:" + result); } }
try this: private static String removeRepeat(String input){ Set<Character> str = new LinkedHashSet<Character>(); for(int n=0;n<input.length();n++){ str.add(input.charAt(n)); } return str.toString(); } good point from the comment, changed to LinkedHashSet. It may be the crap code, but what I mean is don't reinvent the wheel, only if you have to
char ch1,ch2; int l=name.length(); String result=""; for(int i=0;i<l;i++){ if(name.indexOf(name.charAt(i))==i){ result+=name.charAt(i); } } System.out.println(result); input = SUSHSILHI output = SUHIL
You should do the opposite: add the first letter to result, then check if the next letter is already in result: boolean exist=false; result=name.charAt(0); for (i=1; i<l;i++) { exist=false; int j=0; while (exist=false && j<i) { if(name.charAt(i)==charAt(j)) { exist=true; } j++; } if(exist==false){ result=result+name.charAt(i); } } The for checks for all the string name, then the while checks for the characters already in result, if it doesn't already exist, else it doesn't do anything.
Using indexOf() , one for loop should work, like below String name="SUSHIL"; String newName=""; int i=0; int l=name.length(); for(i=0;i<l;i++) { char ch1=name.charAt(i); if(!(newName.indexOf(ch1)>-1)) { newName=newName + ch1; } } System.out.println("Output:"+newName);
String name = "SUSHIL"; char ch1 = 0, ch2; int i, j; int l = name.length(); StringBuilder sb = new StringBuilder(); for (i = 0; i < l; i++) { //this is used to append char to StringBuilder boolean shouldAppend = true; //if we don't check if the length is equal to 0 to start then the below loop will never run and the result would be an empty string so just append the first character to the StringBuilder if (sb.length() == 0) { sb.append(name.charAt(i)); shouldAppend = false; } else { for (j = 0; j < sb.length(); j++) { ch1 = name.charAt(i); ch2 = sb.charAt(j); if (ch1 == ch2) { //StringBuilder contains ch1 so turn shouldAppend to false and break out of this inner loop shouldAppend = false; break; } } } if (shouldAppend) sb.append(ch1); } System.out.println("Output:" + sb.toString());
Try: //Globally List<Character> list = new ArrayList<Character>(); public String remRepeats(String original) { char ch = original.charAt(0); if (original.length() == 1) return original; if (list.contains(ch)) return remRepeats(original.substring(1)); else { list.add(ch); return ch + remRepeats(original.substring(1)); } }
List<Character> characters = new ArrayList<>(); char[] chars = name.toCharArray(); StringBuilder stringBuilder = new StringBuilder(); for(char currChar:chars) { if (!characters.contains(currChar)) { characters.add(currChar); stringBuilder.append(currChar); } } System.out.println(stringBuilder);
How to return longest sequence of chars in a string in java?
Following is what I end up doing but i did not find right answer. Example - If I have the sequence "hellloo" the output will be "lll". Please tell me what is wrong? public class LongestSequenceOfChar { static String testcase1="hellloo"; public static void main(String[] args) { LongestSequenceOfChar test = new LongestSequenceOfChar(); String result = test.longestSequenceOfChar(testcase1); System.out.println(result); } public String longestSequenceOfChar(String str){ String result=""; for(int i=0;i<str.length();i++){ char ch=str.charAt(i); for(int j=i+1;j<str.length();j++){ char ch1=str.charAt(j); if(ch!=ch1){ continue; } result+=ch; } } return result; } }
You should have a counter that counts the number of the longest sequence for now. When you find a longer sequence, you should reset result and update the counter accordingly. However, you can have better solutions: Have an array of size 26 (the size of the English alphabet). Now you iterate on the String and for each char in it you add 1 in the corresponding cell in the helper array. Use a HashMap that has the char as a key and the number it appears as the value. If it's a new char you simply put it with 0 value, if it exists, you increment the existing value. Tip: Use a debugger, it can save your life.
1. Create a HashMap<Character,Integer>.. Integer-->count 2. Start from the beginning of your String.. For each character, check if it is already present in the hashmap a. If Yes, just increment the count b. if No, then add the character as key to the Map and set its count value to 1.
If there are three 'l' you only add two and in the next step are two 'l' and you add one of them. Then the same with the two 'o' where you are adding one. You only have to clear the result string when you step to the next letter and before save the result in another variable, but only if its is longer! public String longestSequenceOfChar(String str){ String interimresult=""; String result=""; //final result for(int i=0;i<str.length();i++){ char ch=str.charAt(i); interimresult += ch; //add the letter once for(int j=i+1;j<str.length();j++){ char ch1=str.charAt(j); if(ch!=ch1){ break; } interimresult +=ch; } if(interimresult.length()>result.length())//store the result if it is longer result = interimresult; interimresult = ""; //clear to continue with the next letter } return result; }
Here is a solution: public String longestSequenceOfChar(String str) { String result = ""; for (int i = 0; i < str.length(); i++) { int j = i; while(j < str.length() && str.charAt(j) == str.charAt(i)) { j++; } // If this one is longer than previous, then asign it to result. if(j - i > result.length()) { result = str.substring(i, j); } } return result; }
This can be solved easily using HashMap. Checkout this sample code: import java.util.Collections; import java.util.HashMap; import java.util.Map; import java.util.Map.Entry; public class MaximumOccuringCharUsingHashMap { public static void main(String[] args) { String test = "test samples"; MaximumOccuringCharUsingHashMap mc = new MaximumOccuringCharUsingHashMap(); System.out.println( mc.findMaximunOccurenceCharacter(test)); } char findMaximunOccurenceCharacter(String input){ Map<Character, Integer> countHash = new HashMap<Character, Integer>(); for(int i=0; i<input.length() ;i++ ){ char currentChar = input.charAt(i); if(countHash.get(currentChar)==null){ countHash.put(currentChar, 1); }else{ countHash. put(currentChar, countHash.get(currentChar)+1); } } int max = Collections.max(countHash.values()); char maxCharacter =0; for(Entry<Character, Integer> entry :countHash.entrySet()){ if(entry.getValue() == max){ maxCharacter = entry.getKey(); } } return maxCharacter; } } Above code will print s as output, which is occurring maximum number of times in the given String.
Try This... public class HelloWorld { public static void main(String[] args) { System.out.println(maxLen(null)); System.out.println(maxLen("")); System.out.println(maxLen("a")); System.out.println(maxLen("aa")); System.out.println(maxLen("abcddd")); System.out.println(maxLen("abcd")); System.out.println(maxLen("aabbba")); } public static String maxLen(String input) { // Avoid NPEs if (input == null) { return null; } int maxLen = 0; int tempLen = 0; char prevChar = 0; char c = 0; char repeatChar = 0; for (int i = 0; i < input.length(); i++) { c = input.charAt(i); if (c == prevChar) { tempLen++; if (tempLen > maxLen) repeatChar = c; } else { maxLen = (tempLen > maxLen) ? tempLen : maxLen; prevChar = c; tempLen = 1; } } maxLen = (tempLen > maxLen) ? tempLen : maxLen; if (maxLen == 0 || maxLen == 1) return "no sequence found"; else { String str = ""; for (int i = 1; i <= maxLen; i++) str += String.valueOf(repeatChar); return str; } } } This will pass all test cases.
Count words in a string method?
I was wondering how I would write a method to count the number of words in a java string only by using string methods like charAt, length, or substring. Loops and if statements are okay! I really appreciate any help I can get! Thanks!
This would work even with multiple spaces and leading and/or trailing spaces and blank lines: String trim = s.trim(); if (trim.isEmpty()) return 0; return trim.split("\\s+").length; // separate string around spaces More info about split here.
public static int countWords(String s){ int wordCount = 0; boolean word = false; int endOfLine = s.length() - 1; for (int i = 0; i < s.length(); i++) { // if the char is a letter, word = true. if (Character.isLetter(s.charAt(i)) && i != endOfLine) { word = true; // if char isn't a letter and there have been letters before, // counter goes up. } else if (!Character.isLetter(s.charAt(i)) && word) { wordCount++; word = false; // last word of String; if it doesn't end with a non letter, it // wouldn't count without this. } else if (Character.isLetter(s.charAt(i)) && i == endOfLine) { wordCount++; } } return wordCount; }
Hi I just figured out with StringTokenizer like this: String words = "word word2 word3 word4"; StringTokenizer st = new Tokenizer(words); st.countTokens();
Simply use , str.split("\\w+").length ;
public static int countWords(String str){ if(str == null || str.isEmpty()) return 0; int count = 0; for(int e = 0; e < str.length(); e++){ if(str.charAt(e) != ' '){ count++; while(str.charAt(e) != ' ' && e < str.length()-1){ e++; } } } return count; }
private static int countWordsInSentence(String input) { int wordCount = 0; if (input.trim().equals("")) { return wordCount; } else { wordCount = 1; } for (int i = 0; i < input.length(); i++) { char ch = input.charAt(i); String str = new String("" + ch); if (i+1 != input.length() && str.equals(" ") && !(""+ input.charAt(i+1)).equals(" ")) { wordCount++; } } return wordCount; }
Use myString.split("\\s+"); This will work.
There is a Simple Solution You can Try this code String s = "hju vg jhdgsf dh gg g g g "; String[] words = s.trim().split("\\s+"); System.out.println("count is = "+(words.length));
public static int countWords(String input) { int wordCount = 0; boolean isBlankSet = false; input = input.trim(); for (int j = 0; j < input.length(); j++) { if (input.charAt(j) == ' ') isBlankSet = true; else { if (isBlankSet) { wordCount++; isBlankSet = false; } } } return wordCount + 1; }
Algo in O(N) count : 0; if(str[0] == validChar ) : count++; else : for i = 1 ; i < sizeOf(str) ; i++ : if(str[i] == validChar AND str[i-1] != validChar) count++; end if; end for; end if; return count;
import com.google.common.base.Optional; import com.google.common.base.Splitter; import com.google.common.collect.HashMultiset; import com.google.common.collect.ImmutableSet; import com.google.common.collect.Multiset; String str="Simple Java Word Count count Count Program"; Iterable<String> words = Splitter.on(" ").trimResults().split(str); //google word counter Multiset<String> wordsMultiset = HashMultiset.create(); for (String string : words) { wordsMultiset.add(string.toLowerCase()); } Set<String> result = wordsMultiset.elementSet(); for (String string : result) { System.out.println(string+" X "+wordsMultiset.count(string)); } add at the pom.xml <dependency> <groupId>com.google.guava</groupId> <artifactId>guava</artifactId> <version>r09</version> </dependency>
Counting Words in a String: This might also help --> package data.structure.test; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class CountWords { public static void main(String[] args) throws IOException { // Couting number of words in a string BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); System.out.println("enter Your String"); String input = br.readLine(); char[] arr = input.toCharArray(); int i = 0; boolean notCounted = true; int counter = 0; while (i < arr.length) { if (arr[i] != ' ') { if (notCounted) { notCounted = false; counter++; } } else { notCounted = true; } i++; } System.out.println("words in the string are : " + counter); } }
public class TestStringCount { public static void main(String[] args) { int count=0; boolean word= false; String str = "how ma ny wo rds are th ere in th is sente nce"; char[] ch = str.toCharArray(); for(int i =0;i<ch.length;i++){ if(!(ch[i]==' ')){ for(int j=i;j<ch.length;j++,i++){ if(!(ch[j]==' ')){ word= true; if(j==ch.length-1){ count++; } continue; } else{ if(word){ count++; } word = false; } } } else{ continue; } } System.out.println("there are "+(count)+" words"); } }
import java.util.; import java.io.; public class Main { public static void main(String[] args) { File f=new File("src/MyFrame.java"); String value=null; int i=0; int j=0; int k=0; try { Scanner in =new Scanner(f); while(in.hasNextLine()) { String a=in.nextLine(); k++; char chars[]=a.toCharArray(); i +=chars.length; } in.close(); Scanner in2=new Scanner(f); while(in2.hasNext()) { String b=in2.next(); System.out.println(b); j++; } in2.close(); System.out.println("the number of chars is :"+i); System.out.println("the number of words is :"+j); System.out.println("the number of lines is :"+k); } catch (Exception e) { e.printStackTrace(); } } }
My idea of that program is that: package text; import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class CoutingWords { public static void main(String[] args) throws IOException { String str; int cWords = 1; char ch; BufferedReader buffor = new BufferedReader(new InputStreamReader(System.in)); System.out.println("Enter text: "); str = buffor.readLine(); for(int i =0; i<str.length(); i++){ ch = str.charAt(i); if(Character.isWhitespace(ch)){ cWords++; } } System.out.println("There are " + (int)cWords +" words."); } }
I'm new to stackoverflow but I hope my code helps: private int numOfWordsInLineCounter(String line){ int words = 0; for(int i = 1 ; i<line.length();i++){ Character ch = line.charAt(i-1); Character bch = line.charAt(i); if(Character.isLetterOrDigit(ch) == true && Character.isLetterOrDigit(bch)== false ) words++; if(i == line.length()-1 && Character.isLetterOrDigit(bch))words++; } return words; }
A string phrase normaly has words separated by space. Well you can split the phrase using the spaces as separating characters and count them as follows. import java.util.HashMap; import java.util.Map; public class WordCountMethod { public static void main (String [] args){ Map<String, Integer>m = new HashMap<String, Integer>(); String phrase = "hello my name is John I repeat John"; String [] array = phrase.split(" "); for(int i =0; i < array.length; i++){ String word_i = array[i]; Integer ci = m.get(word_i); if(ci == null){ m.put(word_i, 1); } else m.put(word_i, ci+1); } for(String s : m.keySet()){ System.out.println(s+" repeats "+m.get(s)); } } }
Taking the chosen answer as a starting point the following deals with a few English language issues including hyphenated words, apostrophes for possessives and shortenings, numbers and also any characters outside of UTF-16: public static int countWords(final String s) { int wordCount = 0; boolean word = false; final int endOfLine = s.length() - 1; for (int i = 0; i < s.length(); i++) { // if the char is a letter, word = true. if (isWordCharacter(s, i) && i != endOfLine) { word = true; // if char isn't a letter and there have been letters before, // counter goes up. } else if (!isWordCharacter(s, i) && word) { wordCount++; word = false; // last word of String; if it doesn't end with a non letter, it // wouldn't count without this. } else if (isWordCharacter(s, i) && i == endOfLine) { wordCount++; } } return wordCount; } private static boolean isWordCharacter(final String s, final int i) { final char ch = s.charAt(i); return Character.isLetterOrDigit(ch) || ch == '\'' || Character.getType(ch) == Character.DASH_PUNCTUATION || Character.isSurrogate(ch); }
I just put this together. The incrementer in the wordCount() method is a bit inelegant to me, but it works. import java.util.*; public class WordCounter { private String word; private int numWords; public int wordCount(String wrd) { StringTokenizer token = new StringTokenizer(wrd, " "); word = token.nextToken(); numWords = token.countTokens(); numWords++; return numWords; } public static void main(String[] args) { Scanner input = new Scanner(System.in); String userWord; WordCounter wc = new WordCounter(); System.out.println("Enter a sentence."); userWord = input.nextLine(); wc.wordCount(userWord); System.out.println("You sentence was " + wc.numWords + " words long."); } }
create variable count, state. initialize variables if space is present keep count as it is else increase count. for eg: if (string.charAt(i) == ' ' ) { state = 0; } else if (state == 0) { state = 1; count += 1;
lambda, in which splitting and storing of the counted words is dispensed withand only counting is done String text = "counting w/o apostrophe's problems or consecutive spaces"; int count = text.codePoints().boxed().collect( Collector.of( () -> new int[] {0, 0}, (a, c) -> { if( ".,; \t".indexOf( c ) >= 0 ) a[1] = 0; else if( a[1]++ == 0 ) a[0]++; }, (a, b) -> {a[0] += b[0]; return( a );}, a -> a[0] ) ); gets: 7 works as a status machine that counts the transitions from spacing characters .,; \t to words
if(str.isEmpty() || str.trim().length() == 0){ return 0; } return (str.trim().split("\\s+").length);
String a = "Some String"; int count = 0; for (int i = 0; i < a.length(); i++) { if (Character.isWhitespace(a.charAt(i))) { count++; } } System.out.println(count+1); It will count white spaces. However, If we add 1 in count , we can get exact words.