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/**

package coding.bat.solutions;

/**

* @author Aman Shekhar

public class StringTwo {

/**

* @param args

public static void main(String[] args) {

// TODO Auto-generated method stub

}

// Given a string, return a string where for every char in the original, there

// are two chars.

// doubleChar("The") → "TThhee"

// doubleChar("AAbb") → "AAAAbbbb"

// doubleChar("Hi-There") → "HHii--TThheerree"

public String doubleChar(String str) {

int mLength = (str.length()) * 2;

StringBuilder mString = new StringBuilder(mLength);

for (int i = 0; i < str.length(); i++) {

for (int j = 0; j < 2; j++) {

mString.append(str.charAt(i));

}

return mString.toString();

}

// --------------------------------------------------------------------------------------------

// Return the number of times that the string "hi" appears anywhere in the given

// string.

// countHi("abc hi ho") → 1

// countHi("ABChi hi") → 2

// countHi("hihi") → 2

public int countHi(String str) {

String find = "hi";

int start = 0;

int count = 0;

while (true) {

int found = str.indexOf(find, start);

if (found == -1)

break;

count++;

start = found + find.length(); // move start up for next iteration

}

return count;

}

// --------------------------------------------------------------------------------------------

// Return true if the string "cat" and "dog" appear the same number of times in

// the given string.

// catDog("catdog") → true

// catDog("catcat") → false

// catDog("1cat1cadodog") → true

public boolean catDog(String str) {

int mCatsCounts = 0;

int mDogsCounts = 0;

for (int i = 0; i < str.length() - 2; i++) {

if (str.substring(i, i + 3).equals("cat"))

mCatsCounts += 1;

if (str.substring(i, i + 3).equals("dog"))

mDogsCounts += 1;

}

return mCatsCounts == mDogsCounts;

}

// Return the number of times that the string "code" appears anywhere in the

// given string, except we'll accept any letter for the 'd', so "cope" and

// "cooe" count.

// countCode("aaacodebbb") → 1

// countCode("codexxcode") → 2

// countCode("cozexxcope") → 2

// --------------------------------------------------------------------------------------------

public int countCode(String str) {

int count = 0;

for (int i = 0; i < str.length() - 3; i++) {

if (str.charAt(i) == 'c' && str.charAt(i + 1) == 'o' && str.charAt(i + 3) == 'e')

count += 1;

}

return count;

}

// Given two strings, return true if either of the strings appears at the very

// end of the other string, ignoring upper/lower case differences (in other

// words, the computation should not be "case sensitive"). Note:

// str.toLowerCase() returns the lowercase version of a string.

// endOther("Hiabc", "abc") → true

// endOther("AbC", "HiaBc") → true

// endOther("abc", "abXabc") → true

// --------------------------------------------------------------------------------------------

public boolean endOther(String a, String b) {

a = a.toLowerCase();

b = b.toLowerCase();

return a.endsWith(b) || b.endsWith(a);

}

// Return true if the given string contains an appearance of "xyz" where the xyz

// is not directly preceeded by a period (.). So "xxyz" counts but "x.xyz" does

// not.

// xyzThere("abcxyz") → true

// xyzThere("abc.xyz") → false

// xyzThere("xyz.abc") → true

// --------------------------------------------------------------------------------------------

public boolean xyzThere(String str) {

if (str.length() >= 3) {

if (str.substring(0, 3).equals("xyz"))

return true;

for (int i = 0; i < str.length() - 3; i++)

if (str.substring(i + 1, i + 4).equals("xyz") && str.charAt(i) != '.')

return true;

}

return false;

}

// Return true if the given string contains a "bob" string, but where the middle

// 'o' char can be any char.

// bobThere("abcbob") → true

// bobThere("b9b") → true

// bobThere("bac") → false

// --------------------------------------------------------------------------------------------

public boolean bobThere(String str) {

for (int i = 0; i < str.length() - 2; i++) {

if (str.charAt(i) == 'b' && str.charAt(i + 2) == 'b')

return true;

}

return false;

}

// We'll say that a String is xy-balanced if for all the 'x' chars in the

// string, there exists a 'y' char somewhere later in the string. So "xxy" is

// balanced, but "xyx" is not. One 'y' can balance multiple 'x's. Return true if

// the given string is xy-balanced.

// xyBalance("aaxbby") → true

// xyBalance("aaxbb") → false

// xyBalance("yaaxbb") → false

// --------------------------------------------------------------------------------------------

public boolean xyBalance(String str) {

int lastX = str.lastIndexOf("y");

int lastY = str.lastIndexOf("x");

return lastX == -1 && lastY == -1 || (lastX > lastY);

}

// Given two strings, a and b, create a bigger string made of the first char of

// a, the first char of b, the second char of a, the second char of b, and so

// on. Any leftover chars go at the end of the result.

// mixString("abc", "xyz") → "axbycz"

// mixString("Hi", "There") → "HTihere"

// mixString("xxxx", "There") → "xTxhxexre"

// --------------------------------------------------------------------------------------------

public String mixString(String a, String b) {

String mResult = "";

for (int i = 0; i < Math.min(a.length(), b.length()); i++)

mResult += "" + a.charAt(i) + b.charAt(i);

if (a.length() > b.length())

return mResult + a.substring(b.length());

return mResult + b.substring(a.length());

}

// Given a string and an int n, return a string made of n repetitions of the

// last n characters of the string. You may assume that n is between 0 and the

// length of the string, inclusive.

// repeatEnd("Hello", 3) → "llollollo"

// repeatEnd("Hello", 2) → "lolo"

// repeatEnd("Hello", 1) → "o"

// --------------------------------------------------------------------------------------------

public String repeatEnd(String str, int n) {

String mResult = "";

for (int i = 0; i < n; i++) {

mResult += "" + str.substring(str.length() - n);

}

return mResult;

}

// Given a string and an int n, return a string made of the first n characters

// of the string, followed by the first n-1 characters of the string, and so on.

// You may assume that n is between 0 and the length of the string, inclusive

// (i.e. n >= 0 and n <= str.length()).

// repeatFront("Chocolate", 4) → "ChocChoChC"

// repeatFront("Chocolate", 3) → "ChoChC"

// repeatFront("Ice Cream", 2) → "IcI"

// --------------------------------------------------------------------------------------------

public String repeatFront(String str, int n) {

String mResult = "";

for (int i = n; i > 0; i--) {

mResult += "" + str.substring(0, i);

}

return mResult;

}

// Given two strings, word and a separator sep, return a big string made of

// count occurrences of the word, separated by the separator string.

// repeatSeparator("Word", "X", 3) → "WordXWordXWord"

// repeatSeparator("This", "And", 2) → "ThisAndThis"

// repeatSeparator("This", "And", 1) → "This"

// --------------------------------------------------------------------------------------------

public String repeatSeparator(String word, String sep, int count) {

String res = "";

if (count == 0)

return "";

while (count > 1) {

res += word + sep;

count--;

}

return res + word;

}

// Given a string, consider the prefix string made of the first N chars of the

// string. Does that prefix string appear somewhere else in the string? Assume

// that the string is not empty and that N is in the range 1..str.length().

// prefixAgain("abXYabc", 1) → true

// prefixAgain("abXYabc", 2) → true

// prefixAgain("abXYabc", 3) → false

// --------------------------------------------------------------------------------------------

public boolean prefixAgain(String str, int n) {

String prefix = str.substring(0, n);

for (int i = n; i <= str.length() - prefix.length(); i++)

if (str.substring(i, i + prefix.length()).equals(prefix))

return true;

return false;

}

// Given a string, does "xyz" appear in the middle of the string? To define

// middle, we'll say that the number of chars to the left and right of the "xyz"

// must differ by at most one. This problem is harder than it looks.

// xyzMiddle("AAxyzBB") → true

// xyzMiddle("AxyzBB") → true

// xyzMiddle("AxyzBBB") → false

// --------------------------------------------------------------------------------------------

public boolean xyzMiddle(String str) {

int len = str.length();

if (len < 3)

return false;

if (len % 2 == 1)

return str.substring(len / 2 - 1, len / 2 + 2).equals("xyz");

return str.substring(len / 2 - 2, len / 2 + 1).equals("xyz")

|| str.substring(len / 2 - 1, len / 2 + 2).equals("xyz");

}

// A sandwich is two pieces of bread with something in between. Return the

// string that is between the first and last appearance of "bread" in the given

// string, or return the empty string "" if there are not two pieces of bread.

// getSandwich("breadjambread") → "jam"

// getSandwich("xxbreadjambreadyy") → "jam"

// getSandwich("xxbreadyy") → ""

// --------------------------------------------------------------------------------------------

public String getSandwich(String str) {

int mStart = str.indexOf("bread");

int mLast = str.lastIndexOf("bread");

if (mStart == mLast)

return "";

return str.substring(mStart + 5, mLast);

}

// Returns true if for every '*' (star) in the string, if there are chars both

// immediately before and after the star, they are the same.

// sameStarChar("xy*yzz") → true

// sameStarChar("xy*zzz") → false

// sameStarChar("*xa*az") → true

// --------------------------------------------------------------------------------------------

public boolean sameStarChar(String str) {

for (int i = 1; i < str.length() - 1; i++)

if (str.charAt(i) == '*' && str.charAt(i - 1) != str.charAt(i + 1))

return false;

return true;

}

// Given a string, compute a new string by moving the first char to come after

// the next two chars, so "abc" yields "bca". Repeat this process for each

// subsequent group of 3 chars, so "abcdef" yields "bcaefd". Ignore any group of

// fewer than 3 chars at the end.

// oneTwo("abc") → "bca"

// oneTwo("tca") → "cat"

// oneTwo("tcagdo") → "catdog"

// --------------------------------------------------------------------------------------------

public String oneTwo(String str) {

String string = "";

for (int i = 0; i < str.length() - 2; i += 3) {

string += str.substring(i + 1, i + 3) + str.charAt(i);

}

return string;

}

// Look for patterns like "zip" and "zap" in the string -- length-3, starting

// with 'z' and ending with 'p'. Return a string where for all such words, the

// middle letter is gone, so "zipXzap" yields "zpXzp".

// zipZap("zipXzap") → "zpXzp"

// zipZap("zopzop") → "zpzp"

// zipZap("zzzopzop") → "zzzpzp"

// --------------------------------------------------------------------------------------------

public String zipZap(String str) {

for (int i = 0; i < str.length() - 2; i++)

if (str.charAt(i) == 'z' && str.charAt(i + 2) == 'p')

str = str.substring(0, i + 1) + str.substring(i + 2);

return str;

}

// Return a version of the given string, where for every star (*) in the string

// the star and the chars immediately to its left and right are gone. So "ab*cd"

// yields "ad" and "ab**cd" also yields "ad".

// starOut("ab*cd") → "ad"

// starOut("ab**cd") → "ad"

// starOut("sm*eilly") → "silly"

// --------------------------------------------------------------------------------------------

public String starOut(String str) {

while (str.indexOf('*') != -1) {

if (str.equals("*"))

return "";

int mFirstStar = str.indexOf('*');

int mSecondStar = str.indexOf('*', mFirstStar + 1);

if (mSecondStar - mFirstStar == 1) {

str = str.substring(0, mFirstStar) + str.substring(mSecondStar);

continue;

}

if (mFirstStar == 0)

str = str.substring(mFirstStar + 2);

else if (mFirstStar == str.length() - 1)

str = str.substring(0, str.length() - 2);

else

str = str.substring(0, mFirstStar - 1) + str.substring(mFirstStar + 2);

}

return str;

}

// --------------------------------------------------------------------------------------------

// Given a string and a non-empty word string, return a version of the original

// String where all chars have been replaced by pluses ("+"), except for

// appearances of the word string which are preserved unchanged.

// plusOut("12xy34", "xy") → "++xy++"

// plusOut("12xy34", "1") → "1+++++"

// plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"

public String plusOut(String str, String word) {

int start = 0;

int end = str.indexOf(word);

while (end != -1) {

for (int i = start; i < end; i++)

str = str.substring(0, i) + "+" + str.substring(i + 1);

start = end + word.length();

end = str.indexOf(word, start);

}

for (int i = start; i < str.length(); i++)

str = str.substring(0, i) + "+" + str.substring(i + 1);

return str;

}

// --------------------------------------------------------------------------------------------

// Given a string and a non-empty word string, return a string made of each char

// just before and just after every appearance of the word in the string. Ignore

// cases where there is no char before or after the word, and a char may be

// included twice if it is between two words.

// wordEnds("abcXY123XYijk", "XY") → "c13i"

// wordEnds("XY123XY", "XY") → "13"

// wordEnds("XY1XY", "XY") → "11"

public String wordEnds(String str, String word) {

String res = "";

int pos = str.indexOf(word);

while (pos != -1) {

if (pos != 0)

res += str.charAt(pos - 1);

if (pos + word.length() == str.length())

break;

res += str.charAt(pos + word.length());

pos = str.indexOf(word, pos + 1);

}

return res;

}