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These are the simple solutions of the kotlin example problems ON LINE. If you want to add your answer, you can make a PR.
Sum - online
Your task is to implement the sum() function so that it computes the sum of
all elements in the given array a.
Solution 1
fun sum(a: IntArray): Int {
return a.filter { it != null }.sum()
}Solution 2
fun sum(a: IntArray): Int {
var _sum = 0
for(element in a)
_sum += element
return _sumSolution 3
fun sum(a: IntArray): Int {
val iterator = a.iterator()
var _sum: Int = 0
while (iterator.hasNext()){
_sum += iterator.next()
}
return _sum
}Index of Maximum - online
Your task is to implement the indexOfMax() function so that it returns
the index of the largest element in the array, or null if the array is empty
Solution
fun _indexOfMax(a: IntArray): Int? {
var maxIndex = 0
for(elem in a.indices){
val newElem = a[elem]
if (newElem >= a[maxIndex]){
maxIndex = elem;
}
}
return maxIndex
}
return if(a.size != 0) _indexOfMax(a) else nullRuns - online
Any array may be viewed as a number of "runs" of equal numbers.
For example, the following array has two runs:
1, 1, 1, 2, 2
Three 1's in a row form the first run, and two 2's form the second.
This array has two runs of length one:
3, 4
And this one has five runs:
1, 0, 1, 1, 1, 2, 0, 0, 0, 0, 0, 0, 0
Your task is to implement the runs() function so that it returns the number
of runs in the given array.
Solution
fun runs(a: IntArray): Int {
var res = 0;
if (a.size == res)
return res
var base = a[res]
a.forEach({
if(it != base){
res ++
base = it
}
})
return (++res);
}Palindrome - online
Your task is to implement a palindrome test.
A string is called a palindrome when it reads the same way left-to-right
and right-to-left.
Solution 1
fun isPalindrome(s: String): Boolean {
return s == s.reversed()
}Solution 2
fun isPalindrome(s: String): Boolean {
var reversed = ""
for(i in s.indices.reversed())
reversed = "$reversed${s.charAt(i)}"
return s == reversed
}Solution 3
fun isPalindrome(s: String): Boolean {
val iterator = s.iterator()
var _reversed = ""
while(iterator.hasNext()){
val _char = iterator.next()
_reversed = "$_char$_reversed"
}
return s == _reversed
}Pairless - online
Think of a perfect world where everybody has a soulmate.
Now, the real world is imperfect: there is exactly one number in the array
that does not have a pair. A pair is an element with the same value.
For example in this array:
1, 2, 1, 2
every number has a pair, but in this one:
1, 1, 1
one of the ones is lonely.
Your task is to implement the findPairless() function so that it finds the
lonely number and returns it.
A hint: there's a solution that looks at each element only once and uses no
data structures like collections or trees.
Solution
INPUT OUTPUT
A B A XOR B
0 0 0 OK
0 1 1
1 0 1
1 1 0 OK
fun findPairless(a: IntArray): Int {
return a.reduce { a, b -> a xor b }
}- @Author: Victor Bolinches Marin
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