# Array Fun 2 

Complete all the methods below. When you are done, copy and paste the testmethods.v4  into your class and run it to see if you have any obvious errors.

int sumEveryN(int[] nums, int n)

Description: This method returns the sum of every `n` elements of `num`s..

 Method Call return value/output sumEveryN( {1 , 2 , 3 , 4 }, 2 ) 4( ie 1 +3) sumEveryN( {13 , 42, 15, 33 , 44 , 16 , 52} ,3) 98 ( ie 13 + 33+ 52)

String[] doubleArr(String[] strs)

Description: This method returns a new version of `strs` in which each element now appears twice. This can be done with a for-each loop, which I believe is easier and more intuitive.

 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”}

int  indexOf5(int[] nums )

Description:This returns the index of the first occurrence element 5  or -1 if 5 does not appear anywhere in the array.

 Method Call return value/output indexOf5( { 2 , 3 , 5 , 4 } ) 2 indexOf5( { 2 , 3 , 5 , 4, 5  } ) 2 indexOf5( { 2 , 3 ,7  , 4, 3,   } ) -1

More Sample calls and return vals 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

int[] randos(int start, int end, int howMany)

Description:  This method returns an array of random numbers between [start,end] . Note make sure that each element in the new array attempts to make a new random int.

More example calls and returns :

double meanBetween(int[] nums, int min, int max)

Description:  This method returns the mean of nums ; however, this method only counts values within the range (min,max) as shown in the examples below:

int secondSmallest(int[] nums )

Description: This method returns the element of `nums` with the second smallest value.

Note: You may not modify the input array. For instance, you may not put nums in order,  which would be bad because you were not asked to modify the array.

@precondition: nums.length >= 2

Note: You will lose credit if you use a constant to represent the smallest or second smallest number. See pseudocode

 Method Call return value/output secondSmallest( { 2 , 18 , 22, 4 , 6 } ) 4 secondSmallest( { 3 , 7 , 15 , 1 ,101} ) 3

The question below is extra credit .
** 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.  You may not do any math besides addition. You may not use any external libraries for any math or base conversions.   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 )

Old versions:

Array Fun 1

Array F un 2 (v 1)

Array Fun 3 (resizing included)

# 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 )