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package coding.bat.solutions;

public class WarmUp_Solution_Two {

public static void main(String[] args) {

}

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

// Given a string and a non-negative int n, return a larger string that is n

// copies of the original string.

// stringTimes("Hi", 2) → "HiHi"

// stringTimes("Hi", 3) → "HiHiHi"

// stringTimes("Hi", 1) → "Hi"

public String stringTimes(String str, int n) {

String result = "";

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

result = result + str;

}

return result;

}

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

// Given a string and a non-negative int n, we'll say that the front of the

// string is the first 3 chars, or whatever is there if the string is less than

// length 3. Return n copies of the front;

// frontTimes("Chocolate", 2) → "ChoCho"

// frontTimes("Chocolate", 3) → "ChoChoCho"

// frontTimes("Abc", 3) → "AbcAbcAbc"

public String frontTimes(String str, int n) {

int frontLen = 3;

if (frontLen > str.length()) {

frontLen = str.length();

}

String front = str.substring(0, frontLen);

String result = "";

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

result = result + front;

}

return result;

}

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

// Count the number of "xx" in the given string. We'll say that overlapping is

// allowed, so "xxx" contains 2 "xx".

// countXX("abcxx") → 1

// countXX("xxx") → 2

// countXX("xxxx") → 3

int countXX(String str) {

int count = 0;

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

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

count++;

}

return count;

}

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

// Given a string, return true if the first instance of "x" in the string is

// immediately followed by another "x".

// doubleX("axxbb") → true

// doubleX("axaxax") → false

// doubleX("xxxxx") → true

boolean doubleX(String str) {

int i = str.indexOf("x");

if (i == -1)

return false;

if (i + 1 >= str.length())

return false;

return str.substring(i + 1, i + 2).equals("x");

}

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

// Given a string, return a new string made of every other char starting with

// the first, so "Hello" yields "Hlo".

// stringBits("Hello") → "Hlo"

// stringBits("Hi") → "H"

// stringBits("Heeololeo") → "Hello"

public String stringBits(String str) {

String result = "";

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

result = result + str.substring(i, i + 1);

}

return result;

}

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

// Given a non-empty string like "Code" return a string like "CCoCodCode".

// stringSplosion("Code") → "CCoCodCode"

// stringSplosion("abc") → "aababc"

// stringSplosion("ab") → "aab"

public String stringSplosion(String str) {

String result = "";

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

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

}

return result;

}

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

// Given a string, return the count of the number of times that a substring

// length 2 appears in the string and also as the last 2 chars of the string, so

// "hixxxhi" yields 1 (we won't count the end substring).

// last2("hixxhi") → 1

// last2("xaxxaxaxx") → 1

// last2("axxxaaxx") → 2

public int last2(String str) {

if (str.length() < 2)

return 0;

String end = str.substring(str.length() - 2);

int count = 0;

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

String sub = str.substring(i, i + 2);

if (sub.equals(end))

count++;

}

return count;

}

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

// Given an array of ints, return the number of 9's in the array.

// arrayCount9([1, 2, 9]) → 1

// arrayCount9([1, 9, 9]) → 2

// arrayCount9([1, 9, 9, 3, 9]) → 3

public int arrayCount9(int[] nums) {

int count = 0;

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

if (nums[i] == 9) {

count++;

}

return count;

}

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

// Given an array of ints, return true if one of the first 4 elements in the

// array is a 9. The array length may be less than 4.

// arrayFront9([1, 2, 9, 3, 4]) → true

// arrayFront9([1, 2, 3, 4, 9]) → false

// arrayFront9([1, 2, 3, 4, 5]) → false

public boolean arrayFront9(int[] nums) {

int end = nums.length;

if (end > 4)

end = 4;

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

if (nums[i] == 9)

return true;

}

return false;

}

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

// Given an array of ints, return true if the sequence of numbers 1, 2, 3

// appears in the array somewhere.

// array123([1, 1, 2, 3, 1]) → true

// array123([1, 1, 2, 4, 1]) → false

// array123([1, 1, 2, 1, 2, 3]) → true

public boolean array123(int[] nums) {

if (nums.length >= 3) {

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

if ((nums[i] == 1) && (nums[i + 1] == 2) && (nums[i + 2] == 3))

return true;

}

return false;

}

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

// Given 2 strings, a and b, return the number of the positions where they

// contain the same length 2 substring. So "xxcaazz" and "xxbaaz" yields 3,

// since the "xx", "aa", and "az" substrings appear in the same place in both

// strings.

// stringMatch("xxcaazz", "xxbaaz") → 3

// stringMatch("abc", "abc") → 2

// stringMatch("abc", "axc") → 0

public int stringMatch(String a, String b) {

int len = Math.min(a.length(), b.length());

int count = 0;

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

String aSub = a.substring(i, i + 2);

String bSub = b.substring(i, i + 2);

if (aSub.equals(bSub)) {

count++;

}

return count;

}

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

// Given a string, return a version where all the "x" have been removed. Except

// an "x" at the very start or end should not be removed.

// stringX("xxHxix") → "xHix"

// stringX("abxxxcd") → "abcd"

// stringX("xabxxxcdx") → "xabcdx"

public String stringX(String str) {

String result = "";

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

if (!(i > 0 && i < (str.length() - 1) && str.substring(i, i + 1).equals("x"))) {

result = result + str.substring(i, i + 1);

}

return result;

}

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

// Given a string, return a string made of the chars at indexes 0,1, 4,5, 8,9

// ... so "kittens" yields "kien".

// altPairs("kitten") → "kien"

// altPairs("Chocolate") → "Chole"

// altPairs("CodingHorror") → "Congrr"

public String altPairs(String str) {

String result = "";

for (int i = 0; i < str.length(); i += 4) {

int last = i + 2;

if (last > str.length())

last = str.length();

result = result + str.substring(i, last);

}

return result;

}

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

// 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.

// stringYak("yakpak") → "pak"

// stringYak("pakyak") → "pak"

// stringYak("yak123ya") → "123ya"

public String stringYak(String str) {

String result = "";

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

if (i + 2 < str.length() && str.charAt(i) == 'y' && str.charAt(i + 2) == 'k') {

i = i + 2;

} else {

result = result + str.charAt(i);

}

return result;

}

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

// Given an array of ints, return the number of times that two 6's are next to

// each other in the array. Also count instances where the second "6" is

// actually a 7.

// array667([6, 6, 2]) → 1

// array667([6, 6, 2, 6]) → 1

// array667([6, 7, 2, 6]) → 1

public int array667(int[] nums) {

int count = 0;

if (nums.length > 2) {

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

if ((nums[i] == 6) && ((nums[i + 1] == 6) || (nums[i + 1] == 7))) {

count++;

}

return count;

}

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

// Given an array of ints, we'll say that a triple is a value appearing 3 times

// in a row in the array. Return true if the array does not contain any triples.

// noTriples([1, 1, 2, 2, 1]) → true

// noTriples([1, 1, 2, 2, 2, 1]) → false

// noTriples([1, 1, 1, 2, 2, 2, 1]) → false

public boolean noTriples(int[] nums) {

if (nums.length > 2) {

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

if (nums[i] == nums[i + 1]) {

if (nums[i + 1] == nums[i + 2])

return false;

}

return true;

}

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

// Given an array of ints, return true if it contains a 2, 7, 1 pattern: a

// value, followed by the value plus 5, followed by the value minus 1.

// Additionally the 271 counts even if the "1" differs by 2 or less from the

// correct value.

// has271([1, 2, 7, 1]) → true

// has271([1, 2, 8, 1]) → false

// has271([2, 7, 1]) → true

public boolean has271(int[] nums) {

if (nums.length > 2) {

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

int first = nums[i];

if ((first + 5 == nums[i + 1]) && (Math.abs(nums[i + 2] - (first - 1)) <= 2))

return true;

}

return false;

}