Develop Reference

Palindrome in java

Here what I tried
sample input is "aabaa"
eg: in if condition val[0] = a[4]
if it is equal i stored it in counter variable if it is half of the length it original string it is palindrome
if it is not it is not a palindrome
I tried with my basic knowledge in java if there is any errors let me know
boolean solution(String inputString) {
int val = inputString.length();
int count = 0;
for (int i = 0; i<inputString.length(); i++) {
if(inputString.charAt(i) == inputString.charAt(val-i)) {
count = count++;
if (count>0) {
return true;
}
}
}
return true;
}

How about
public boolean isPalindrome(String text) {
String clean = text.replaceAll("\\s+", "").toLowerCase();
int length = clean.length();
int forward = 0;
int backward = length - 1;
while (backward > forward) {
char forwardChar = clean.charAt(forward++);
char backwardChar = clean.charAt(backward--);
if (forwardChar != backwardChar)
return false;
}
return true;
}
From here

In your version you compare first element with last, second with second last etc.
last element in this case is inputString.length()-1(so need to use 'inputString.charAt(val-i-1)' . If you iterate till end, then the count should be equal to length of the string.
for(int i = 0; i<inputString.length(); i++){
if(inputString.charAt(i) == inputString.charAt(val-i-1)){
count ++;
}
}
return (count==val); //true when count=val
Or alternatlively iterate till the mid point of the array, then count value is val/2.
for(int i = 0; i<inputString.length()/2; i++){
if(inputString.charAt(i) == inputString.charAt(val-i-1)){
count ++;
}
}
return (count==val/2); //true when count=val/2

There's no constraints in the question so let me throw in a more cheesy solution.
boolean isPalindrome(String in)
final String inl = in.toLowerCase();
return new StringBuilder(inl).reverse().toString().equals(inl);
}

A palindrome is a word, sentence, verse, or even a number that reads the same forward and backward. In this java solution, we’ll see how to figure out whether the number or the string is palindrome in nature or not.
Method - 1
class Main {
public static void main(String[] args) {
String str = "Nitin", revStr = "";
int strLen = str.length();
for (int i = (strLen - 1); i >=0; --i) {
revStr = revStr + str.charAt(i);
}
if (str.toLowerCase().equals(revStr.toLowerCase())) {
System.out.println(str + " is a Palindrome String.");
}
else {
System.out.println(str + " is not a Palindrome String.");
}
Method - 2
class Main {
public static void main(String[] args) {
int n = 3553, revNum = 0, rem;
// store the number to the original number
int orgNum = n;
/* get the reverse of original number
store it in variable */
while (n != 0) {
remainder = n % 10;
revNum = revNum * 10 + rem;
n /= 10;
}
// check if reversed number and original number are equal
if (orgNum == revNum) {
System.out.println(orgNum + " is Palindrome.");
}
else {
System.out.println(orgNum + " is not Palindrome.");
}

Splitting this string to get the max count to a corresponding character

I am currently implementing Run Length Encoding for text compression and my algorithm does return Strings of the following form:
Let's say we have a string as input
"AAAAABBBBCCCCCCCC"
then my algorithm returns
"1A2A3A4A5A1B2B3B4B1C2C3C4C5C6C7C8C"
Now I want to apply Java String split to solve this, because I want to get the highest number corresponding to character. For our example it would be
"5A4B8C"
My function can be seen below
public String getStrfinal(){
String result = "";
int counter = 1;
StringBuilder sb = new StringBuilder();
sb.append("");
for (int i=0;i<str.length()-1;i++) {
char c = str.charAt(i);
if (str.charAt(i)==str.charAt(i+1)) {
counter++;
sb.append(counter);
sb.append(c);
}
else {
counter = 1;
continue;
}
}
result = sb.toString();
return result;
}

public static String getStrfinal(){
StringBuilder sb = new StringBuilder();
char last = 0;
int count = 0;
for(int i = 0; i < str.length(); i++) {
if(i > 0 && last != str.charAt(i)) {
sb.append(count + "" + last);
last = 0;
count = 1;
}
else {
count++;
}
last = str.charAt(i);
}
sb.append(count + "" + last);
return sb.toString();
}

Here is one possible solution. It starts with the raw string and simply iterates thru the string.
public static void main(String[] args) {
String input = "AAAABBBCCCCCCCDDDEAAFBBCD";
int index = 0;
StringBuilder sb = new StringBuilder();
while (index < input.length()) {
int count = 0;
char c = input.charAt(index);
for (; index < input.length(); index++) {
if (c != input.charAt(index)) {
count++;
}
else {
break;
}
}
sb.append(Integer.toString(count));
sb.append(c);
count = 0;
}
System.out.println(sb.toString());
}
But one problem with this method and others is what happens if there are digits in the text? For example. What if the string is AAABB999222AAA which would compress to 3A2B39323A. That could also mean AAABB followed by 39 3's and 23 A's

Instead of string Buffer you can use a map it will be much easier and clean to do so.
public static void main(String[] args) {
String input = "AAAAABBBBCCCCCCCCAAABBBDDCCCC";
int counter=1;
for(int i=1; i<input.length(); i++) {
if(input.charAt(i-1)==input.charAt(i)) {
counter=counter+1;
}else if(input.charAt(i-1)!=input.charAt(i)){
System.out.print(counter+Character.toString(input.charAt(i-1)));
counter=1;
}if(i==input.length()-1){
System.out.print(counter+Character.toString(input.charAt(i)));
}
}
}
This will gives
5A4B8C3A3B2D4C
UPDATES
I Agree with #WJS if the string contains number the out put becomes messy
hence if the System.out in above code will be exchange with below i.e.
System.out.print(Character.toString(input.charAt(i-1))+"="+counter+" ");
then for input like
AAAAABBBBCCCCCCCCAAABBBDD556677CCCCz
we get out put as below
A=5 B=4 C=8 A=3 B=3 D=2 5=2 6=2 7=2 C=4 z=1

This is one of the possible solutions to your question. We can use a LinkedHashMap data structure which is similar to HashMap but it also maintains the order. So, we can traverse the string and store the occurrence of each character as Key-value pair into the map and retrieve easily with its maximum occurrence.
public String getStrFinal(String str){
if(str==null || str.length()==0) return str;
LinkedHashMap<Character,Integer> map = new LinkedHashMap<>();
StringBuilder sb=new StringBuilder(); // to store the final string
for(char ch:str.toCharArray()){
map.put(ch,map.getOrDefault(ch,0)+1); // put the count for each character
}
for(Map.Entry<Character,Integer> entry:map.entrySet()){ // iterate the map again and append each character's occurence into stringbuilder
sb.append(entry.getValue());
sb.append(entry.getKey());
}
System.out.println("String = " + sb.toString()); // here you go, we got the final string
return sb.toString();
}

Counting words in a string using methods

I have a project to make a program that takes a string as input and then prints the number of words in the string as output. We are supposed to use 3 methods in this, one to read input, one to print the output, and one to count the words.
I know I am missing something basic but I have spent hours on this and cannot figure out why the program wont run as it should. I need to keep the program pretty simple so I dont want to edit it too much, just find the issue and fix it so it will run correctly.
Example: Enter a string of text: The quick brown fox jumped over the lazy dog. 9 words
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.print("Enter text: ");
String words = wordInput(in);
int count = wordCount(words);
System.out.println(words);
System.out.println(count);
printLine(count);
}
private static String wordInput(Scanner in)
{
String words = in.nextLine();
return words;
}
private static int wordCount(String words)
{
int length = words.length();
int ctr = 0;
int spot = 0;
int stop = 1;
char space = ' ';
char end = '.';
char com = ',';
char yes = '!';
char question = '?';
while (length > 0 && stop > 0)
{
if (words.charAt(spot) == space)
{
ctr++;
spot++;
}
else if (words.charAt(spot) == com)
{
spot++;
}
else if (words.charAt(spot) == yes || words.charAt(spot) == end || words.charAt(spot) == question)
{
stop = -1;
}
else if (spot > length)
{
stop = -1;
}
else
spot++;
}
return ctr + 1;
}
private static void printLine(int ctr)
{
System.out.println(ctr + " words");
}

Here is a rewritten wordCount that does as you request, minimal changes to the code. However, I am not sure it produces the answers you expect.
private static int wordCount(String words)
{
int length = words.length();
int ctr = 0;
int spot = 0;
char space = ' ';
char end = '.';
char com = ',';
char yes = '!';
char question = '?';
while (spot < length)
{
if (words.charAt(spot) == space)
{
ctr++;
spot++;
}
else if (words.charAt(spot) == com)
{
spot++;
}
else if (words.charAt(spot) == yes || words.charAt(spot) == end || words.charAt(spot) == question)
{
break;
}
else
spot++;
}
return ctr + 1;
}
However, fundamentally you are trying to count words in a string, and that is a known art. A few options and further reading:
http://www.quickprogrammingtips.com/java/find-number-of-words-in-a-string-in-java.html
Count words in a string method?
Which leads to a simpler result of:
private static int wordCount(String words) {
return words.split("\\s+").length;
}

Going on the assumption that whoever enters the string is not going to make a grammar error, try split the string at all spaces and set it equal to a string[]. Here is an example:
String temp = JOptionPane.showInputDialog(null, "Enter some text here");
String[] words = temp.split(" ");
JOptionPane.showMessage(null, String.format("Words used %d", words.length));

String Compression loop logic

My for loop for my string compression is a bit off. I have been working on this assignment the past 5 days and I can't figure out for the life of me what is wrong. Can someone help me out?
For example, I passed over the string "TTTTrrrEe" and instead of getting T4r3Ee, I'm getting T4r3EeTT. I don't know why it jumps back to the beginning of the string like that, but I am getting closer.We can only use charAt,equals,length, and substring from the string class.
Can someone help guide me in the right direction by helping to correct my logic? I still want to try and code this myself, seeing as how it is an assignment.
public static String compress(String s){
int count = 0;
String temp = s.substring(0,1);
for(int i = 0; i < s.length(); i++){
if(i !=s.length()-1){
if(temp.equals(s.substring(i,i+1))){
count++;
}else{
if(count < 1){
System.out.print(s.substring(i,i+2));
System.out.print(temp.substring(0,1) );
}else{
System.out.print("" + temp.substring(0,1) + count);
i--;
temp = s.substring(count,count+1);
System.out.println(" temp is now " + temp);
count = 0;
//i--;
}
}
}
}
System.out.println(temp);
return temp;
}

Since this is a learning exercise, I wouldn't try fixing your code, just point out a few things to work on to get it right:
The if (i !=s.length()-1) condition inside the loop becomes unnecessary if you change your for loop condition to i < s.length()-1
Comparing individual characters is easier (and faster) than comparing substrings. You get a character at position i by calling char ch1 = s.charAt(i), and compare two characters using == operator, rather than calling equals() on them.
When count is zero (your count < 1 condition is equivalent to count == 0) you print both the current character and the character after it, in addition to the first character of temp followed by the count. This does not look correct.
Rather than growing temp as you go through the loop, you set it on each iteration. This does not look correct.
A better way of growing temp as you go through the loop is using StringBuilder and append(), instead of using a plain String, and performing concatenations.

Try using some logic like this;
int count = 0;
for(int i =0; i < string.length()-1; i++){
if(string.charAt(i) == string.charAt(i + 1)){
count++;
// DO SOME OPERATION
}
}

temp = s.substring(count,count+1); does not relate to a position (i), but a size.
In fact I would try to rewrite it afresh, with externally sensible names:
char repeatedChar = `\u0000`; // Not present.
int repetitions = 0;
Because of the no-namer count you got into trouble.

Working code:
public class HelloWorld {
public static void compress(String s){
StringBuilder buff = new StringBuilder();
char tmp = '\0';
int index = 1;
for(int i = 0; i < s.length(); i++){
char curr = s.charAt(i);
if(buff.length() == 0){
tmp = curr;
buff.append(tmp);
continue;
}
if(curr == tmp){
index++;
}
else{
if(index > 1){
buff.append(index);
index = 1;
tmp = curr;
}
buff.append(curr);
}
}
System.out.println(buff.toString());
}
public static void main(String args[]){
compress("TTTTrrrEe");
}
}
Output: T4r3Ee
For compress("TTsssssssssssTTrrrEe");
Output: T2s11T2r3Ee

String temp = s.substring(0,1);
temp.equals(s.substring(i,i+1))
In case of these 2 sentences you should have used a char instead of String, as such:
char temp = s.charAt(0)
temp == s.charAt(i)
I would start with 3 variables:
char lastCharacter = inputString.charAt(0);
int count = 1;
String result = "";
then proceed to process the input string in a loop:
if (length <= 1) return inputString;
for i = 1 ; i < length;i++
if (inputString.charAt(i) == lastCharacter && i != length-1)
count++
else
if count == 1
result += lastCharacter
else
result = result + lastCharacter + count;
count = 1;
end if
lastCharacter = inputString.charAt(i);
end if
end for
return result;

TRY THIS
public class Compress {
/**
* #param args
* #author Rakesh KR
*/
public static String encode(String source) {
StringBuffer dest = new StringBuffer();
for (int i = 0; i < source.length(); i++) {
int runLength = 1;
while (i+1 < source.length() && source.charAt(i) == source.charAt(i+1)) {
runLength++;
i++;
}
dest.append(source.charAt(i));
dest.append(runLength);
}
return dest.toString();
}
public static void main(String[] args) {
// TODO Auto-generated method stub
String example = "aaaaaaBBBBccc";
System.out.println("Encode::"+encode(example));
}
}

interviewstreet.com - String similarity

I'm trying to solve the string similarity question on interviewstreet.com. My code is working for 7/10 cases (and it is exceeding the time limit for the other 3).
Here's my code -
public class Solution {
public static void main(String[] args) {
Scanner user_input = new Scanner(System.in);
String v1 = user_input.next();
int number_cases = Integer.parseInt(v1);
String[] cases = new String[number_cases];
for(int i=0;i<number_cases;i++)
cases[i] = user_input.next();
for(int k=0;k<number_cases;k++){
int similarity = solve(cases[k]);
System.out.println(similarity);
}
}
static int solve(String sample){
int len=sample.length();
int sim=0;
for(int i=0;i<len;i++){
for(int j=i;j<len;j++){
if(sample.charAt(j-i)==sample.charAt(j))
sim++;
else
break;
}
}
return sim;
}
}
Here's the question -
For two strings A and B, we define the similarity of the strings to be the length of the longest prefix common to both strings. For example, the similarity of strings "abc" and "abd" is 2, while the similarity of strings "aaa" and "aaab" is 3.
Calculate the sum of similarities of a string S with each of it's suffixes.
Input:
The first line contains the number of test cases T. Each of the next T lines contains a string each.
Output:
Output T lines containing the answer for the corresponding test case.
Constraints:
1 <= T <= 10
The length of each string is at most 100000 and contains only lower case characters.
Sample Input:
2
ababaa
aa
Sample Output:
11
3
Explanation:
For the first case, the suffixes of the string are "ababaa", "babaa", "abaa", "baa", "aa" and "a". The similarities of each of these strings with the string "ababaa" are 6,0,3,0,1,1 respectively. Thus the answer is 6 + 0 + 3 + 0 + 1 + 1 = 11.
For the second case, the answer is 2 + 1 = 3.
How can I improve the running speed of the code. It becomes harder since the website does not provide a list of test cases it uses.

I used char[] instead of strings. It reduced the running time from 5.3 seconds to 4.7 seconds and for the test cases and it worked. Here's the code -
static int solve(String sample){
int len=sample.length();
char[] letters = sample.toCharArray();
int sim=0;
for(int i=0;i<len;i++){
for(int j=i;j<len;j++){
if(letters[j-i]==letters[j])
sim++;
else
break;
}
}
return sim;
}

used a different algorithm. run a loop for n times where n is equals to length the main string. for each loop generate all the suffix of the string starting for ith string and match it with the second string. when you find unmatched character break the loop add j's value to counter integer c.
import java.io.BufferedReader;
import java.io.InputStreamReader;
class Solution {
public static void main(String args[]) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int T = Integer.parseInt(in.readLine());
for (int i = 0; i < T; i++) {
String line = in.readLine();
System.out.println(count(line));
}
}
private static int count(String input) {
int c = 0, j;
char[] array = input.toCharArray();
int n = array.length;
for (int i = 0; i < n; i++) {
for (j = 0; j < n - i && i + j < n; j++)
if (array[i + j] != array[j])
break;
c+=j;
}
return c;
}
}

I spent some time to resolve this question, and here is an example of my code (it works for me, and pass thru all the test-cases):
static long stringSimilarity(String a) {
int len=a.length();
char[] letters = a.toCharArray();
char localChar = letters[0];
long sim=0;
int sameCharsRow = 0;
boolean isFirstTime = true;
for(int i=0;i<len;i++){
if (localChar == letters[i]) {
for(int j = i + sameCharsRow;j<len;j++){
if (isFirstTime && letters[j] == localChar) {
sameCharsRow++;
} else {
isFirstTime = false;
}
if(letters[j-i]==letters[j])
sim++;
else
break;
}
if (sameCharsRow > 0) {
sameCharsRow--;
sim += sameCharsRow;
}
isFirstTime = true;
}
}
return sim;
}
The point is that we need to speed up strings with the same content, and then we will have better performance with test cases 10 and 11.

Initialize sim with the length of the sample string and start the outer loop with 1 because we now in advance that the comparison of the sample string with itself will add its own length value to the result.

import java.util.Scanner;
public class StringSimilarity
{
public static void main(String args[])
{
Scanner user_input = new Scanner(System.in);
int count = Integer.parseInt(user_input.next());
char[] nextLine = user_input.next().toCharArray();
try
{
while(nextLine!= null )
{
int length = nextLine.length;
int suffixCount =length;
for(int i=1;i<length;i++)
{
int j =0;
int k=i;
for(;k<length && nextLine[k++] == nextLine[j++]; suffixCount++);
}
System.out.println(suffixCount);
if(--count < 0)
{
System.exit(0);
}
nextLine = user_input.next().toCharArray();
}
}
catch (Exception e)
{
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}