Author Archives: Mr. M
Array Fun 2 [Java]
Description: This method creates an array that is filled with all of the digits of
nums
.
Method Call | return value/output |
digitsToArray( 523 ) | {5,2,3} |
digitsToArray(1267 ) | { 1 , 2 , 6 , 7} |
Description: This method returns true if each element of
nums
is a factor of the sum ofnums
.
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) |
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”} |
Description: This method returns true if
num
is an elementnums
.
Method Call | return value/output |
isThere( {6,1,2,3}, 6) | true |
isThere( {6,1,2,3}, 5 ) | false |
Description: This returns the index of the first occurrence element ‘5’ or -1 if 5 does not exist.
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 |
Description: This method returns the index value of the first appearance of
num
or -1 ifnum
is not an element ofnums
.
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 |
Description: This method returns the index value of the last appearance of
num
or -1 ifnum
is not an element ofnums
.
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 |
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 |
isIncreasing( {-1, –1, -2, -3, -4 }) | false |
Description: This method returns the largest span of consecutive increasing numbers
Method Call | return value/output |
largestSpan( {4, 3, 1, 3, 6 , 1} ) | 3 |
largestSpan( {4 3 1 , 12 , 31 , 44 , 52 , 1} ) | 5 |
largestSpan( {1 , 7 8 , 12,4, 3, 0, 4 , 1) | 3 |
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 }) | {21 , 5 , 5, 4,13 } |
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} |
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:
12 if(overFlowAdditionOccurred)throw new ArithmeticException("Addition Overflow Error");
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 ) |
Array Fun 1 [Java]
Array Fun 1
Create a Class called ArrayFun that has the following methods Write the body for the methods described below.
public int[] swapFirstLast(int[] nums)
Description: This method returns a new version of the array with the first and last elements of
nums
swapped.
Method Call | return value/output |
swapFirstLast( {3, 2 ,5 } ) | {5,2,3} |
swapFirstLast( {7, 6 ,9 ,12} ) | { 12 , 6 , 9, 7} |
public String[] swapFirstMiddle(String[] strs)
Description: This method returns a new version of the array with the first and middle elements of
strs
swapped.
Method Call | return value/output |
swapFirstMiddle( {“a”,”b”,”c” } ) | { “b”, “a”, “c”} |
swapFirstMiddle( { “zoo”,”foo”, “xoo”} ) | { “foo”,”zoo”, “xoo”} |
Loops
public void printEveryOther(int[] nums)
Description: This method prints every other element of
nums
on new lines .
Method Call | return value/output |
printEveryOther( {1 , 2 , 3 , 4 } ) | 1 3 |
printEveryOther( {13 , 42 , 33 , 44 , 52} ) | 13 33 52 |
public int sum(int[] nums )
Description: This method returns the sum of
nums
..
Method Call | return value/output |
sum( { 2 , 8 }) | 10 ie 2 + 8 |
sum( { 2 , 8 , 4 } ) | 14 ie (2 + 8 + 4 ) |
public int sumOdds(int[] nums )
Description: This method returns the sum of all the odd element of
nums
.
Method Call | return value/output |
sumOdds( { 5 , 2, 4 , 6 , 7 } ) | 12 ( ie 5 + 7 ) |
sumOdds( { 13 , 3, 2, 6 , 7 } ) | 23 |
Description: This method returns the mean of
nums
. Pay close attention to the second sample call and make sure you get “5.25“
Method Call | return value/output |
mean( { 2 , 8 }) | 5.0 ie (2+8)/2 |
mean( { 2 , 8 , 4 , 7 } ) | 5.25 ie (2+8+4+7)/4 |
public double largest(double[] nums )
Description: This method returns the element of
nums
with the greatest value. Make sure that you test out the second call and get ‘-2‘.
Method Call | return value/output |
largest( { 2.6 , 8.2, 5.2}) | 8.2 |
largest( { -2,-11, -4} ) | -2 |
public int sumEveryN(int[] nums, int n)
Description: This method returns the sum of every
n
elements ofnum
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) |
*public int secondSmallest(int[] nums )
Description: This method returns the element of
nums
with the second smallest value.
Method Call | return value/output |
secondSmallest( { 2 , 18 , 22, 4 , 6 } ) | 4 |
secondSmallest( { 3 , 7 , 15 , 1 ,101} ) | 3 |
** boolean isPalindromic(int[] nums)
Description:This method returns true if the elements of
nums
are a palindrome..
Method Call | return value/output |
isPalindromic(( { 5 , 2, 7 , 2 , 5} ) | true |
isPalindromic( { 5 , 2, 7 , 3 , 5} )) | false |
isPalindromic(( { 1 , 2, 1} ) | true |
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:
12 if(overFlowAdditionOccurred)throw new ArithmeticException("Addition Overflow Error");
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. What you have to do is reproduce the additional algorithm and find a way to carry digits , programmatically.
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 ) |
Java String Loops 2
public int countF(String str)
Description: This method returns the number of times that the letter ‘f’ occurs in str.
Method Call | return value/output |
countF(“abcdef”) | 1 |
countF(“abcdfef”) | 2 |
countF(“fff”) | 3 |
countF(“xxx”) | 0 |
public boolean has2Fs(String str)
Description: This method returns true if str has exactly two occurrences of the letter ‘f’ in it.
Method Call | return value/output |
has2Fs(“foo”) | false |
has2Fs(“fofo”) | true |
has2Fs(“foffo”) | false |
public boolean has2Fs3Gs(String str)
Description: This method returns true if str has exactly two occurrences of the letter ‘f’ in it and 3 occurences ‘g’.
Method Call | return value/output |
has2Fs3Gs(“foog”) | false |
has2Fs3Gs(“foggfog”) | true |
has2Fs3Gs(“foffoggg”) | false |
public int momOrDad(String str)
Description: This method returns the number of times that the String “mom” or the string “dad” occurs in the parameter str .
Method Call | return value/output |
momOrDad(“foog”) | 0 |
momOrDad(“momf”) | 1 |
momOrDad(“momdadmo”) | 2 |
momOrDad(“momdadmom”) | 3 |
public String threeTimes(String str)
Description: This method returns the String str concatenated with itself three times
Method Call | return value/output |
threeTimes(“foog”) | “foogfoogfoog” |
threeTimes(“abc”) | “abcabcabc” |
threeTimes(“zt”) | “ztztzt” |
threeTimes(“”) | “” |
public String nTimes(String str, int n)
Description: This method returns the String str concatenated with itself n times
Method Call | return value/output |
nTimes(“fg” , 2 ) | “fgfg” |
nTimes(“abc” , 0 ) | “” |
nTimes(“zt” , 3 ) | “ztztzt” |
nTimes(“uiz” , 4 ) | “uizuizuizuiz” |
public String countMiddleChar(String str)
Precondition: str.length() ≥ 3.
Description: This method returns the number of times the middle letter of str appears in str.
Method Call | return value/output |
countMiddleChar(“acbcb” ) | 2 |
countMiddleChar(“acbcx” ) | 1 |
countMiddleChar(“bbbbb” ) | 5 |
countMiddleChar(“xytbtzy”) | 1 |
public int indexOf(String haystack, String needle)
Description: Write your own indexOf() method. Obviously, you cannot make use of the String’s build in indexOf() method. This method returns the index of the first occurrence of needle in haystack . (Full credit if you can get this to work with Strings whose length is greater than 1).
Method Call | return value/output |
indexOf(“acxb”, “x” ) | 2 |
indexOf(“zazt”, “z” ) | 0 |
indexOf(“tayv”, “g” ) | -1 |
indexOf(“hijklj”, “j” ) | 2 |
If you want the extra credit, then it must also be able to complete the following example calls
Method Call | return value/output |
indexOf(“acxb”, “cx” ) | 1 |
indexOf(“acxb”, “cxb” ) | 1 |
indexOf(“acxb”, “cxbt” ) | -1 |
public int countChars(String str, String chars)
Precondition:The length of chars is less than or equal to the length of str
Description: This method returns the number of times that charsoccurs in String str.
Method Call | return value/output |
countChars(“momdadmom” , “dad” ) | 1 |
countChars(“foobofoo” , “foo”) | 2 |
countChars(“foobofoofoo” , “foo”) | 3 |
countChars(“foobofoofoo” , “xy” ) | 0 |
Java String Loop Assingments 1
public void printTwos(String str)
Description: This method prints two letters from the String starting at with the first two letters and ending with the last two.
Method Call | return value/output |
printTwos(“abcdef”) | “ab” “bc” “de” “ef” |
Description: This method print every letter in reverse and on a new line.
Method Call | return value/output |
printReverse(“abcd”) | “d” “c” “b” “a” |
printReverse(“thefoo”) | “o” “o” “f” “e” “h” “t” |
Description: This method print every 2 letter in reverse and on a new line.
Method Call | return value/output |
printTwoReverse(“abcd”) | “dc” “cb” “ba” |
printTwoReverse(“thefoo”) | “oo” “of” “fe” “eh” “ht” |
public void printTwoReverseAgain(String str)
Description: This method print every 2 letter in reverse and on a new line and with no overlap of letters
Method Call | return value/output |
printTwoReverseAgain(“abcd”) | “dc” “ba” |
printTwoReverseAgain(“thefoo”) | “oo” “fe” “ht” |
Gridworld Bug Subclasses
SlowBug class
A SlowBug is a Bug that poops Rocks instead of Flowers when it moves. However, since Rocks take a little bit longer to get out than Flowers do, a SlowBug takes THREE timesteps (three calls to act) before it is able to move like a normal bug .
A SlowBug is a Bug has the following behaviors and attributes:
- a SlowBug has an instance variable called
counter
that is increased each time it acts - the default constructor for this class will set the bug’s color to green.
- It also has a constructor that takes a color parameter :
public SlowBug(java.awt.Color col)
- After a SlowBug has waited two turns without doing anything, it acts exactly like a normal bug..
At the end of each act, whether or not the bug has done anything, it incrementscounter
variable. - the move method works the exact same as a regular Bug’s move, except instead of pooping a Flower in the space it leaves, it poops a Rock the same color as itself.
If you want, you can download SlowBug.gif and drag it into Eclipse or Bluej to use that image instead of the default Bug image.
The final result of running SlowBug would be something like the picture below.
FastBug
- FastBug is exactly like a normal bug, except it acts twice each “timestep” instead of just once. For example, in one timestep it will:
- check if it can move, and if it can, it will move. If it can’t it will turn.
- After it moves or turns, it will check if it can move again, and if it can, it will move. If it can’t, it will turn.
A FastBug is a Bug and should contain the following behaviors:
- the default constructor for this class will set the bug’s color to red. It should also make it start out in a random direction (not always facing North like a normal bug does).
- the act method should behave as described above
- a FastBug is so fast, the flowers it leaves behind are already almost wilted when they come out. To do this, you will need to modify the move method so that the Flower it generates is not the same color as the FastBug, but five “shades” darker. To increase the darkness, use the Color class’s non-static
darker()
method five successive times on your color before passing it in to the Flower constructor.
If you want, you can download FastBug.gif and drag it into Eclipse to use that image instead of the default Bug image.
If you want an A, you must do FIsh:
Fish
Last year’s Marine Biology Case Study was similar to GridWorld, but it worked with Fish objects that have a 1/7 chance of breeding (or, more accurately, spontaneously generating children) each timestep and a 1/5 chance of dying each timestep. A Fish cannot move backwards, and our Fish will not turn, move, or generate children diagonally; it will only use the 4 cardinal directions North, South, East, and West. For this part of the lab, we are going to create a similar Fish in GridWorld.
A Fish is an Actor and should contain the following behaviors and attributes:
- a shared, static instance variable:
private static int nextAvailableID = 1;
A static variable is one that ALL instances of the class share. So the first fish that is made will initialize the static variable to 1; the others will keep using it. - a private int instance variable:
private int myID;
- a private static double instance variable for
probabilityOfBreeding
that is initialized to 1.0/7.0 and a private double instance variable forprobabilityOfDying
that is initialized to 1.0/5.0 - Also, your fish should not die during the first 5 steps of the program.
- the default constructor for this class should
- set the Fish’s color to a random color. Use
(Math.random()*256)
typecast as an int to get random numbers between 0 and 255 for the Color constructor.- import java.awt.* to be able to initialize a new Color(double r, double g, double b);where r,g,b are each ints between 0 and 255 inclusive and represent the red, green and blue values of the color
- The constructor should also make the Fish start out in a random direction (from North, South, East, or West…no other directions are allowed for a Fish). In the constructor, set
myID=nextAvailableID++;
This means the first fish will get the ID 1, the next will get 2, the next will get 3, etc. Since nextAvailableID is static, it is shared amongst all fish.
- set the Fish’s color to a random color. Use
- override the
public String toString()
method as follows:public String toString()
{
return myID + " facing " + getDirection();
} - a
protected boolean breed()
method should be written that:- First calls
Math.random()
and checks to see if the number is the random number generated is>= probabilityOfBreeding
. If it is, returnfalse
(your Fish did not breed this time). - Otherwise, if the random number was
< probabilityOfBreeding
, that means we will try to breed in all the empty locations in front, behind, to the left, and to the right of us.
To do this, first find out how many empty locations you have above you, to your right, to your left, and below you.- If you have zero empty spaces above you, to your right, to your left, and below you, return false (you didn’t have an open space to generate children to so you couldn’t breed).
- If you have one to four empty spaces, you will need create new Fish objects, then have them place themselves in the empty locations. When you are finished generating all the children you can, return
true
(you successfully bred).
- First calls
- a
protected void move()
method should be written that:- checks for all valid, empty spaces above you, to your right, and to your left (Fish don’t move backwards, so we don’t check below you), then picks one of those to move to and goes there, changing directions to face the spot you are moving to (for instance, if you are going to an empty spot to your left, you turn left after you go there). If there are no empty locations you can go to, just stay still.
Note: If you look through the GridWorld API in the Location class and Grid interface, you will find some methods very helpful in writing this method.
- checks for all valid, empty spaces above you, to your right, and to your left (Fish don’t move backwards, so we don’t check below you), then picks one of those to move to and goes there, changing directions to face the spot you are moving to (for instance, if you are going to an empty spot to your left, you turn left after you go there). If there are no empty locations you can go to, just stay still.
- the act method should be overridden with the following code:
public void act()
// Try to breed.
{
// Did not breed, so try to move.// Determine whether this fish will die in this timestep.
if ( ! breed() )
move();
if ( Math.random() < probabilityOfDying )
removeSelfFromGrid();
}
If you want, you can download Fish.gif and drag it into Eclipse to use that image instead of the default Actor image.
It can be kind of hard to tell what is happening, but if you move your mouse over a Fish in the grid you can see the toString()
printed out, which will tell you which unique fish id you are looking at. Since breeding, dying, and moving are random, there is no set way to guarantee what your grid will look like, but here is how mine looked after it ran a few times
while-if-elif-else Jeroo[Python]
Assignment if_elif_else_while_D
- start at (0,0)
- end at (0,23)
- use 1 loop
- pick up the 1 flower and use the following while loo
Use this code at the start of your program
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 |
tim = Jeroo() while not ( tim. isWater(AHEAD) and tim.hasFlower() ) : if tim.isFlower(AHEAD) : tim.hop() tim.pick() elif not tim.isClear(AHEAD) : tim.turn(RIGHT) tim.hop() tim.turn(LEFT) tim.hop(2) tim.turn(LEFT) tim.hop() tim.turn(RIGHT) else : tim.hop() |
The map
What the map and Jeroo looks like at the completion of the loop
1 |
Systematically Visit [python]
Parameters
- 1 while loop
- visit all cells that you can
- pick up all flowers
- your loop must terminate (ie your Jeroo cannot keep going around the field forever)
Map #1
Grade : 2/10
- start in (0,0) with 625 flowers!
- row by row, plant flowers until you to position (23,0) as shown in animation below.
Grade : 3/10
D:\My Drive\2021-22_gdrive\intro 2021-22\jeroo\maps-student\unit-5
Grade : 8/10
Note: Do not be fooled by how similar the next map is. You will probably have to rewrite some code to complete the next one.
Java String Assignments -intro-1
Java String Assignments
These exercises are introductory methods that require use of basic String methods including
- substring()
- length()
- “abc”.length() – >3
- concatenation
public boolean sameStrings(String string1 ,String string2)
Description: This method returns true if string1 and string2 are the same. Use the ‘.equals()’ method.
Method Call | return value/output |
sameStrings(“foo”,”f”) | false |
sameStrings(“foo”,”foo”) | true |
sameStrings(“abc”, “cba”) | false |
public boolean any2Same(String a,String b, String c)
Description: This method returns true if any 2 of the strings are the same. Remember: Use the ‘.equals()’ method.
Method Call | return value/output |
any2Same(“xz”,”f”, “xz”) | true |
any2Same(“xz”,”f”, “xt”) | false |
any2Same(“xz”,”xz”, “fff”) | true |
any2Same(“xtz”,”abc”, “abc”) | true |
any2Same(“xtz”,”a^c”, “a!c”) | false |
public String firstThirdLettters(String str)
Description: This method returns the first and third letters of str concatenated together.
Method Call | return value/output |
firstThirdLettters(“foo“) | “fo” |
firstThirdLettters(“abcdefg”) | “ac” |
firstThirdLettters(“ad!kjkj”) | “a!” |
public boolean sameFirst2Letters(String a, String b)
Description: This method returns the first 2 letters of a and of b are the same .
Method Call | return value/output |
sameFirst2Letters(“axt”, “axjjj”) | true |
sameFirst2Letters(“1%3″ , “3$1″) | false |
sameFirst2Letters(“a~dd” ,”~adt” ) | false |
public String concatTwice(String str)
Description: This method returns str concatenated with itself .
Method Call | return value/output |
concatTwice(“foo”) | “foofoo” |
concatTwice(“a”) | “aa” |
concatTwice(“abcdd”) | “abcddabcdd” |
public String concatWithComma(String str)
Description: This method returns str concatenated with itself and with a comma in between
Method Call | return value/output |
concatWithComma(“foo”) | “foo,foo” |
concatWithComma(“a”) | “a,a” |
concatWithComma(“abcdd”) | “abcdd,abcdd” |
public String sandwich(String bread, String meat)
Description: This method is easiest to understand by looking at the sample calls below
Method Call | return value/output |
sandwich(“a“,”b“) | “aba“ |
sandwich(“xy“,”ab“) | “xyabxy“ |
sandwich(“hi“,”bye“) | “hibyehi“ |
public int lengthTimesTwo(String str)
Description: This method returns the length of str times 2.
Method Call | return value/output |
lengthTimesTwo(“foo”) | 6 |
lengthTimesTwo(“a”) | 2 |
lengthTimesTwo(“abcdd”) | 10 |
String prePendFoo(String str)
Description: prepend “foo ” to the input and return the concatenation.
Method Call | return value/output |
prePendFoo(“abc”) | “foo abc” |
prePendFoo(“x”) | “foo x” |
prePendFoo(“abcdd”) | foo abcdd” |
public int sumOfLengths(String a, String a)
Description: This method returns the sum of the lengths of String a and String b .
Method Call | return value/output |
sumOfLengths(“ab”, “jk1”) | 5 ie ( 2 +3) |
sumOfLengths(“jj”, “”) | 2 (ie 2 + 0) |
sumOfLengths(“a~dd” ,”6″ ) | 5 ie ( 4 + 1) |
**public String concat5Times(String str)
Description: This method returns str concatenated with itself 5 times (Do this with a loop)
Method Call | return value/output |
concat5Times(“foo”) | “foofoofoofoofoo” |
concat5Times(“a”) | “aaaaa” |
concat5Times(“abcdd”) | “abcddabcddabcddabcdd” |