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Encode the string using stream api [closed]

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I am looking for solution using stream api for below requirement. basically we need to write a encryption logic to convert a string
aaabbbbaae --> a3b4a2e1
Solution without stream's api :
public static void main(String[] args) {
String sampleString = "aaabbbaaaaaee";
char charters[] = sampleString.toCharArray();
char currentChar = charters[0];
char prevChar = charters[0];
int count = 0;
String str = "";
for (char c : charters) {
if(c==currentChar) {
count++;
prevChar = c;
}
else {
str = str + prevChar + count;
currentChar = c;
count=1;
}
}
System.out.println(str+currentChar+count);
}

This would be pretty easy to do yourself
String encryptToLettersAndNumbers(String input) {
String product = "";
char last = input.charAt(0);
int streak = 0;
for(int i = 0; i < input.length(); i++) {
if(input.charAt(i) == last) {
streak++;
} else {
product = product + last + streak;
streak = 1;
last = input.charAt(i);
}
}
product = product + last + streak;
return product;
}
Note that this code is untested. Furthermore this will be a very unsafe method of encryption because h1e1l2o1 is pretty easy to guess.

This should get you started-
String sampleString = "aaabbbaaaaaee";
Arrays.asList(sampleString.chars().mapToObj(c->(char)c).toArray(Character[]::new)).stream().forEach(c->{
//You can put your logic here to process
System.out.print(c);
});
This will convert your String to a Character list and stream through it.

How can I write a java program to check if a string is a substring of another string without using indexOf(), contains(), etc? [closed]

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for (int i = 0; i < str1.length(); i++)
{
String c = str1.substring(i, i + 1);
for (int j = 0; j < str2.length(); j++)
{
String d = str1.substring(j, j + 1);
if (d != c)
{
System.out.println("no");
}
else
{
System.out.println("yes");
}
}
}
I know this is not even correct but I have no idea how to start.

Do you mean as you would do in the classical introductory programming exercise?
E.g. something like this?
String a = "Hello, world!";
String b = "world";
boolean contained;
contained = false;
for (int i = 0; i < a.length() - b.length() + 1 && !contained; i++) {
contained = true;
for (int j = 0; j < b.length() && contained; j++)
if (a.charAt(i + j) != b.charAt(j))
contained = false;
}
System.out.println(contained ? "b is a substring of a" : "b is not a substring of a");

How can I write a java program to check if a string is a substring of
another string without using indexOf(), contains(), etc?
There are many ways (also based on which functions/methods you use) e.g. the following solution uses String#length and String#equalsIgnoreCase in addition to String#substring (the method which you have already used in your code). The logic checks if the subject substring is equal to (ignoring case) the substring of the same length starting from an index in str.
public class Main {
public static void main(String[] args) {
// Test
System.out.println(isSubstring("World", "Hello World!"));
System.out.println(isSubstring("foo", "Hello World!"));
}
static boolean isSubstring(String substring, String str) {
if (substring == null || str == null) {
return false;
}
int len = substring.length();
for (int i = 0; i <= str.length() - len; i++) {
if (substring.equalsIgnoreCase(str.substring(i, i + len))) {
return true;
}
}
return false;
}
}
Output:
true
false

Why won the if statement execute? [closed]

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I wrote a simple program in Java which writes a word backwards. Trying to check if "hello" works. In if-statement I'm checking that string is equal to "olleh". Could anyone see why the if statement won't execute.
public class MyProgram {
public static void main(String[] args) {
String x = "hello";
System.out.println(back(x));
}
public static String back(String str) {
String y = " ";
String temp = " ";
for (int i = str.length() - 1; i >= 0; i--) {
char lets = str.charAt(i);
y += Character.toString(lets);
System.out.println(y);
if (y.equals("olleh")) {
System.out.println("nice");
}
}
return y;
}
}

Try this it will work
public class MyProgram
{
public static void main(String[] args)
{
String x = "hello";
System.out.println(back(x));
}
public static String back(String str )
{
String temp = "";
for (int i = str.length() - 1; i >= 0; i--) {
char lets = str.charAt(i);
temp = temp + lets;
}
if (temp.equals("olleh")) {
System.out.println("nice");
}
return temp;
}
}

If you will initialize y variable to empty string instead of space your if-statement will execute and print "nice". Also you do not need a temp string as you don't use it. You probably want to return you reverted string back (alternatively you can make your method void and remove the return statement).
public static String back(String str) {
String y = "";
for (int i = str.length() - 1; i >= 0; i--) {
char lets = str.charAt(i);
y += Character.toString(lets);
System.out.println(y);
if (y.equals("olleh")) {
System.out.println("nice");
}
}
return y;
}
By the way, it's better to use StringBuilder when you're concatenating strings in a loop.

Finding substring occurrence in string [closed]

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I am trying to find the number of substrings in a given string. Currently, My code doesn't take account of over lapping strings.
For example
substr = "cde"
str = "cdcde"
My code:
public static int ssCount(String str, String substr) {
int count = 0;
int strlen = str.length();
int substrlen = substr.length();
int numsubstr = 0;
int substrpointer = 0;
for (int i = 0; i < strlen; i++) {
if (str.charAt(i) == substr.charAt(substrpointer)) {
substrpointer++;
count++;
}
else {
count = 0;
substrpointer = 0;
}
if (count == substrlen) {
numsubstr++;
count = 0;
}
}
return numsubstr;
}
My attempt:
public static int ssCount(String str, String substr) {
int count = 0;
int strlen = str.length();
int substrlen = substr.length();
int numsubstr = 0;
int substrpointer = 0;
int firstchar = 0;
for (int i = 0; i < strlen; i++) {
if (str.charAt(i) == substr.charAt(substrpointer)) {
substrpointer++;
count++;
if (str.charAt(i) == substr.charAt(0)) {
firstchar = i;
}
}
else {
count = 0;
substrpointer = 0;
i = firstchar;
}
if (count == substrlen) {
numsubstr++;
count = 0;
}
}
return numsubstr;
}
I tried to add a 2nd pointer that will point to the next occurrence of the first character of the substring in order to continue the comparisons from that spot. However I am having trouble because I can run into some infinite loops.

This finds all overlapping sub-strings in a larger string. The non-regex way followed by the regex way. An interesting problem.
import java.util.regex.Pattern;
import java.util.regex.Matcher;
/**
<P>{#code java OverlappingSubstringsXmpl}</P>
**/
public class OverlappingSubstringsXmpl {
public static final void main(String[] igno_red) {
String sToFind = "cdc";
String sToSearch = "cdcdcdedcdc";
System.out.println("Non regex way:");
int iMinIdx = 0;
while(iMinIdx <= (sToSearch.length() - sToFind.length())) {
int iIdxFound = sToSearch.indexOf(sToFind, iMinIdx);
if(iIdxFound == -1) {
break;
}
System.out.println(sToFind + " found at index " + iIdxFound);
iMinIdx = iIdxFound + 1;
}
System.out.println("Regex way:");
Matcher m = Pattern.compile(sToFind, Pattern.LITERAL).matcher(sToSearch);
boolean bFound = m.find();
while (bFound) {
System.out.println(sToFind + " found at index " + m.start());
bFound = m.find(m.start() + 1);
}
}
}
Output:
[C:\java_code\]java OverlappingSubstringsXmpl
Non regex way:
cdc found at index 0
cdc found at index 2
cdc found at index 8
Regex way:
cdc found at index 0
cdc found at index 2
cdc found at index 8

Not sure what your question is, probably how to fix your code, but my recommendation is to look into standard approaches to solving this problem, such as the KMP algorithm. It efficiently takes into account overlaps too.

Can anybody help me to correct the following code?

Please help me to identify my mistakes in this code. I am new to Java. Excuse me if I have done any mistake. This is one of codingbat java questions. I am getting Timed Out error message for some inputs like "xxxyakyyyakzzz". For some inputs like "yakpak" and "pakyak" this code is working fine.
Question:
Suppose the string "yak" is unlucky. Given a string, return a version where all the "yak" are removed, but the "a" can be any char. The "yak" strings will not overlap.
public String stringYak(String str) {
String result = "";
int yakIndex = str.indexOf("yak");
if (yakIndex == -1)
return str; //there is no yak
//there is at least one yak
//if there are yaks store their indexes in the arraylist
ArrayList<Integer> yakArray = new ArrayList<Integer>();
int length = str.length();
yakIndex = 0;
while (yakIndex < length - 3) {
yakIndex = str.indexOf("yak", yakIndex);
yakArray.add(yakIndex);
yakIndex += 3;
}//all the yak indexes are stored in the arraylist
//iterate through the arraylist. skip the yaks and get non-yak substrings
for(int i = 0; i < length; i++) {
if (yakArray.contains(i))
i = i + 2;
else
result = result + str.charAt(i);
}
return result;
}

Shouldn't you be looking for any three character sequence starting with a 'y' and ending with a 'k'? Like so?
public static String stringYak(String str) {
char[] chars = (str != null) ? str.toCharArray()
: new char[] {};
StringBuilder sb = new StringBuilder();
for (int i = 0; i < chars.length; i++) {
if (chars[i] == 'y' && chars[i + 2] == 'k') { // if we have 'y' and two away is 'k'
// then it's unlucky...
i += 2;
continue; //skip the statement sb.append
} //do not append any pattern like y1k or yak etc
sb.append(chars[i]);
}
return sb.toString();
}
public static void main(String[] args) {
System.out.println(stringYak("1yik2yak3yuk4")); // Remove the "unlucky" strings
// The result will be 1234.
}

It looks like your programming assignment. You need to use regular expressions.
Look at http://www.vogella.com/articles/JavaRegularExpressions/article.html#regex for more information.
Remember, that you can not use contains. Your code maybe something like
result = str.removeall("y\wk")

you can try this
public static String stringYak(String str) {
for (int i = 0; i < str.length(); i++) {
if(str.charAt(i)=='y'){
str=str.replace("yak", "");
}
}
return str;
}