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/**

package coding.bat.solutions;

/**

* @author Aman Shekhar

public class StringThree {

/**

* @param args

public static void main(String[] args) {

// TODO Auto-generated method stub

}

// Given a string, count the number of words ending in 'y' or 'z' -- so the 'y'

// in "heavy" and the 'z' in "fez" count, but not the 'y' in "yellow" (not case

// sensitive). We'll say that a y or z is at the end of a word if there is not

// an alphabetic letter immediately following it. (Note:

// Character.isLetter(char) tests if a char is an alphabetic letter.)

// countYZ("fez day") → 2

// countYZ("day fez") → 2

// countYZ("day fyyyz") → 2

public int countYZ(String str) {

int count = 0;

str = str.toLowerCase() + " ";

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

if ((str.charAt(i) == 'y' || str.charAt(i) == 'z') && !Character.isLetter(str.charAt(i + 1))) {

count++;

}

return count;

}

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

// Given two strings, base and remove, return a version of the base string where

// all instances of the remove string have been removed (not case sensitive).

// You may assume that the remove string is length 1 or more. Remove only

// non-overlapping instances, so with "xxx" removing "xx" leaves "x".

// withoutString("Hello there", "llo") → "He there"

// withoutString("Hello there", "e") → "Hllo thr"

// withoutString("Hello there", "x") → "Hello there"

public String withoutString(String base, String remove) {

String result = "";

int index = base.toLowerCase().indexOf(remove.toLowerCase());

while (index != -1) {

result += base.substring(0, index);

base = base.substring(index + remove.length());

index = base.toLowerCase().indexOf(remove.toLowerCase());

}

result += base;

return result;

}

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

// Given a string, return true if the number of appearances of "is" anywhere in

// the string is equal to the number of appearances of "not" anywhere in the

// string (case sensitive).

// equalIsNot("This is not") → false

// equalIsNot("This is notnot") → true

// equalIsNot("noisxxnotyynotxisi") → true

public boolean equalIsNot(String str) {

int countIs = 0;

int countNot = 0;

str = str + "X";

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

if (str.substring(i, i + 2).equals("is"))

countIs++;

if (str.substring(i, i + 3).equals("not"))

countNot++;

}

return (countIs == countNot);

}

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

// We'll say that a lowercase 'g' in a string is "happy" if there is another 'g'

// immediately to its left or right. Return true if all the g's in the given

// string are happy.

// gHappy("xxggxx") → true

// gHappy("xxgxx") → false

// gHappy("xxggyygxx") → false

public boolean gHappy(String str) {

str = "X" + str + "X";

for (int i = 0; i < str.length() - 1; i++)

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

return false;

return true;

}

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

// We'll say that a "triple" in a string is a char appearing three times in a

// row. Return the number of triples in the given string. The triples may

// overlap.

// countTriple("abcXXXabc") → 1

// countTriple("xxxabyyyycd") → 3

// countTriple("a") → 0

public int countTriple(String str) {

int count = 0;

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

if ((str.charAt(i) == str.charAt(i + 1)) && str.charAt(i + 1) == str.charAt(i + 2))

count++;

}

return count;

}

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

// Given a string, return the sum of the digits 0-9 that appear in the string,

// ignoring all other characters. Return 0 if there are no digits in the string.

// (Note: Character.isDigit(char) tests if a char is one of the chars '0', '1',

// .. '9'. Integer.parseInt(string) converts a string to an int.)

// sumDigits("aa1bc2d3") → 6

// sumDigits("aa11b33") → 8

// sumDigits("Chocolate") → 0

public int sumDigits(String str) {

int sum = 0;

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

if (Character.isDigit(str.charAt(i)))

sum += Integer.parseInt(str.substring(i, i + 1));

}

return sum;

}

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

// Given a string, return the longest substring that appears at both the

// beginning and end of the string without overlapping. For example,

// sameEnds("abXab") is "ab".

// sameEnds("abXYab") → "ab"

// sameEnds("xx") → "x"

// sameEnds("xxx") → "x"

public String sameEnds(String string) {

String result = "";

int len = string.length();

for (int i = 0; i <= len / 2; i++)

for (int j = len / 2; j < len; j++)

if (string.substring(0, i).equals(string.substring(j)))

result = string.substring(0, i);

return result;

}

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

// Given a string, look for a mirror image (backwards) string at both the

// beginning and end of the given string. In other words, zero or more

// characters at the very begining of the given string, and at the very end of

// the string in reverse order (possibly overlapping). For example, the string

// "abXYZba" has the mirror end "ab".

// mirrorEnds("abXYZba") → "ab"

// mirrorEnds("abca") → "a"

// mirrorEnds("aba") → "aba"

public String mirrorEnds(String string) {

String result = "";

int len = string.length();

for (int i = 0, j = len - 1; i < len; i++, j--)

if (string.charAt(i) == string.charAt(j))

result += string.charAt(i);

else

break;

return result;

}

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

// Given a string, return the length of the largest "block" in the string. A

// block is a run of adjacent chars that are the same.

// maxBlock("hoopla") → 2

// maxBlock("abbCCCddBBBxx") → 3

// maxBlock("") → 0

public int maxBlock(String str) {

int mMax = 0;

int mLength = str.length();

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

int mCount = 0;

for (int j = i; j < mLength; j++) {

if (str.charAt(i) == str.charAt(j)) {

mCount++;

} else {

break;

}

if (mCount > mMax)

mMax = mCount;

}

return mMax;

}

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

// Given a string, return the sum of the numbers appearing in the string,

// ignoring all other characters. A number is a series of 1 or more digit chars

// in a row. (Note: Character.isDigit(char) tests if a char is one of the chars

// '0', '1', .. '9'. Integer.parseInt(string) converts a string to an int.)

// sumNumbers("abc123xyz") → 123

// sumNumbers("aa11b33") → 44

// sumNumbers("7 11") → 18

public int sumNumbers(String str) {

int sum = 0;

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

if (Character.isDigit(str.charAt(i))) {

int count = 0;

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

if (Character.isDigit(str.charAt(j)))

count++;

else

break;

}

sum += Integer.parseInt(str.substring(i, i + count));

i += count;

}

return sum;

}

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

// Given a string, return a string where every appearance of the lowercase word

// "is" has been replaced with "is not". The word "is" should not be immediately

// preceeded or followed by a letter -- so for example the "is" in "this" does

// not count. (Note: Character.isLetter(char) tests if a char is a letter.)

// notReplace("is test") → "is not test"

// notReplace("is-is") → "is not-is not"

// notReplace("This is right") → "This is not right"

public String notReplace(String str) {

String result = "";

str = " " + str + " "; // avoid issues with corner cases

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

if (str.charAt(i) == 'i') {

if (str.charAt(i + 1) == 's' && !Character.isLetter(str.charAt(i + 2))

&& !Character.isLetter(str.charAt(i - 1))) {

result += "is not";

i += 1;

} else

result += "i";

} else

result += str.charAt(i);

}

return result.substring(1);

}