# 2 Player Button Presser V2 -Loops

After making our 2 player Button Presser Game, let’s make some improvements.

Objective: modify the code so that the game repeats until there are no ties!

Hint : use a while True loop and break when someone wins

# Java Assignment Create a Class called ArrayFun2  that has the following methods Write the body for the methods described below.

Description: This method creates an array that is filled with all of the digits of `nums`.

 Method Call return value/output digitsToArray( 523 ) {3, 2, 5} digitsToArray(1267 ) { 7 , 6 , 2,  1 }

public boolean allFactorsOfSum(int[] nums)

Description: This method returns true if each element of`nums` is a factor of the sum of `nums`.

 Method Call return value/output allFactorsOfSum( {6,1,2,3} ) true (because all of the elements are factors of the sum which is12) allFactorsOfSum( 1,4,7) false (because 7 is not a factor of the sum of this array which is 12)

public String[] doubleArr(String[] strs)

Description: This method returns a new version of `strs` with each element appearing twice.

 Method Call return value/output doubleArr( {“a”,”b”,”c”} ) {“a”,”a”,”b”,”b”, “c” , “c”} doubleArr( {“math”,”ware”,”house”,”.com” }) {“math”,”math”,”ware”,”ware”,”house”,”house”,”.com”, “.com”}

public boolean isThere(int[] nums, int num)

Description: This method returns true if `num `is an element `nums`.

 Method Call return value/output isThere( {6,1,2,3}, 6) true isThere( {6,1,2,3}, 5 ) false

public int indexOf(int[] nums, int num)

Description: This method returns the index value of the first appearance of `num` or -1 if `num` is not an element of `nums` .

 Method Call return value/output indexOf( {6,4 ,7,3, 4 }, 4) 1 indexOf( {6,4 7 ,3,2,7}, 7) 2 indexOf( {6,4 ,2,3}, 22) -1

public int lastIndexOf(int[] nums, int num)

Description: This method returns the index value of the last appearance of `num` or -1 if `num` is not an element of `nums` .

 Method Call return value/output lastIndexOf( {6, 4 ,7 ,3, 11, ,4}, 4) 5 lastIndexOf( {7, 6,4 , 7 ,3}, 7) 3 lastIndexOf( {6,4 ,2,3}, 22) -1

public boolean isIncreasing(double[] nums)

Description: This method returns true if each element in nums is greater than the element to its left.

 Method Call return value/output isIncreasing( {1,2,3,4 }) true isIncreasing( {1,0 ,3,4 }) false isIncreasing( {1,1, 2,3,4 }) false

public int largestSpan(int[] nums)

Description: This method returns the largest span of consecutive increasing numbers

 Method Call return value/output largestSpan( {4, 3, 1, 2, 3 , 1} ) 3 largestSpan( {4  , 3,  1 , 12 , 31 , 44 , 52 , 1} ) 5 largestSpan( {9, 7, 8 , 12,4, 3, 0, 4 , 1) 3 largestSpan( {1, 0, 4, 5 , 3,  2, 8 , 9, 10, 12, 4, 3,  1 }) 5

** int[] invert(int[] nums)

Description: This method ‘inverts’ an array by spliting the array in halves and ‘inverting’ each half. See sample calls to understand.

 Method Call return value/output invert( {5 , 21, 5, 13,4 }) {4, 13, 5, 21, 5  } invert( { 1, 2, 3, 4, 5,6  }) { 6, 5, 4, 3, 2, 1}

** int[] shiftByN(int[] nums, int delta)

Description: This method returns the array with all digits shifted by `delta` . Any digits that are circulated off the end of the array should be returned to the other side. Note: delta could be positive or negative. Please keep in mind that Math.abs(delta) > nums.length could be possible 😉

 Method Call return value/output shiftByN( {5 , 21, 13,4 } 1) { 4 , 5 , 21, 13 } shiftByN( {5 , 21, 13, 4 , 11} 2) { 4 , 11 , 5 , 21, 13} shiftByN( {5 , 21, 13, 4 } -1) { 21, 13, 4 , 5} shiftByN( {5 , 21, 13, 4 ,7 } -2) { 13, 4 ,7 , 5 , 21}
** int[] add(int[] num1, int[] num2, int base) throws ArithmeticException@precondition `1 < base < 11`
@precondition `num1.length = 5 and num2.length = 5`
@postcondition: the returned array has 5 elements representing the sum or an ArithmeticException is thrown

Description: This method attempts to replicate addition. Consider both num1 and num2 represent the five digits of a number. Each element in the array stores one of the digits. For instance, the number 143 would be represented as {0,0,1,4,3 }. Numbers can be in any base between 2 and 10 inclusive , and the number 101102 in binary would appear in an arrays as : {1, 0, 1 , 1, 0 } . Let’s assume the numbers are positive.You should return an array representing the digits of the sum of num1 and num2. If the number of digits in the sum exceeds the maximum number of digits (5) , you should throw an arithmetic exception as shown in the code below:

Note: You may not use any kind of helper or utility methods that are built into pre-defined Java classes for converting numbers between bases. Only use techniques taught in this class.

 Method Call return value/output add( {0,0,0 ,4,2},{0,0,0,5,1}, 10) {0, 0, 0, 9, 3} ie (42 + 51 = 93) add( {0, 0, 0 ,7,2},{0,0,0,5,1}, 10) {0, 0 , 1 , 2, 3} ie (72 + 51 = 123) add( {0, 0 , 0 , 1 , 1}, { 0 , 0 , 0 , 1 }, 2) {0, 0 , 1 , 0 , 0} ie ( 112 + 12= 1002 )