Given an unsorted linked list of N nodes. The task is to remove duplicate elements from this unsorted Linked List. When a value appears in multiple nodes, the node which appeared first should be kept, all others duplicates are to be removed.
Example 1:
Input:
N = 4
value[] = {5,2,2,4}
Output: 5 2 4
Explanation: Given linked list elements are
5->2->2->4, in which 2 is repeated only.
So, we will delete the extra repeated
elements 2 from the linked list and the
resultant linked list will contain 5->2->4
It's a question on GFG and I passed 405 test cases but on 406th test case it shows 'Time limit exceeded Error'. Is it like ArrayList is taking too long to traverse every element again and again?
Then I tried using HashSet with the same logic and it worked perfectly and passes all test cases.
class Solution
{
//Function to remove duplicates from unsorted linked list.
public Node removeDuplicates(Node head)
{
// Your code here
Node prev = head;
Node temp = head.next;
ArrayList<Integer> val = new ArrayList<>();
val.add(head.data);
while(temp != null){
if(val.contains(temp.data)){
prev.next = temp.next;
temp = temp.next;
}
else{
val.add(temp.data);
temp = temp.next;
prev = prev.next;
}
}
return head;
}
}
While your logic is indeed correct, it is not the most optimal solution.
Consider this example:
Input: 5 -> 2 -> 2 -> 2 -> 2 -> 3
Your Result steps would be:
5 -> 2 -> 2 -> 2 -> 2 -> 3
5 -> 2 -> 2 -> 2 -> 3
5 -> 2 -> 2 -> 3
5 -> 2 -> 3
Instead of doing the pointer writes so many times, you could keep going forward till you encounter the first not-seen-so-far node and make your pointers point to the new node directly.
Moreover, instead of using an ArrayList, you could use a Set. The .contains() method in ArrayList has a time complexity of O(n) where n is the number of elements in the ArrayList. As you do this operation M times for each loop, your order is going to be O(mn). Set has a time complexity of O(1) (Constant lookup time), So if you the do the operation M times, it will be M times O(1) which is whole lot faster.
Perhaps this code might help:
class Solution
{
//Function to remove duplicates from unsorted linked list.
public Node removeDuplicates(Node head)
{
if(head == NULL) { // BASE CASE
return head;
}
Node prev = head;
Node temp = head.next;
boolean shouldDelete = false;
Set<Integer> set = new HashSet<>();
set.add(head.data);
while(temp != null){
if(set.contains(temp.data)){
shouldDelete = true; // Set this to true to make sure any duplicates at the end is also taken care of.
}
else{
shouldDelete = false; // Set this to false as we have encountered a not-seen-so-far node.
set.add(temp.data);
prev.next = temp;
prev = temp;
}
temp = temp.next;
}
if(shouldDelete) { // There is duplicate elements in the end that has not been trimmed because we never encountered a not-seen-so-far element. So set it to NULL
prev.next = NULL;
}
return head;
}
}
Update: I completely overlooked the complexity added by arr.sort() method. So in Kotlin for array of Int, It compiles to use java.util.DualPivotQuicksort see this which in turn has complexity of O(n^2). see this. Other than that, this is also a valid approach.
I know It can be solved by keeping multiple arrays or using collections (which is what I ended up submitting), I want to know what I missed in the following approach
fun migratoryBirds(arr: Array<Int>): Int {
var maxCount = 0
var maxType = 0
var count = 0
var type = 0
arr.sort()
println(arr.joinToString(" "))
for (value in arr){
if (type != value){
if (count > maxCount){
maxCount = count
maxType = type
}
// new count values
type = value
count = 1
} else {
count++
}
}
return maxType
}
This code passes every scenario except for Test case 2 which has 73966 items for array. On my local machine, that array of 73k+ elements was causing timeout but I did test for array up-to 20k+ randomly generated value 1..5 and every time it succeeded. But I couldn't manage to pass Test case 2 with this approach. So even though I ended up submitting an answer with collection stream approach, I would really like to know what could I be missing in above logic.
I am running array loop only once Complexity should be O(n), So that could not be reason for failing. I am pre-sorting array in ascending order, and I am checking for > not >=, therefore, If two types end up having same count, It will still return the lower of the two types. And this approach is working correctly even for array of 20k+ elements ( I am getting timeout for anything above 25k elements).
The reason it is failing is this line
arr.sort()
Sorting an array takes O(n logn) time. However using something like a hash map this can be solved in O(n) time.
Here is a quick python solution I made to give you the general idea
# Complete the migratoryBirds function below.
def migratoryBirds(arr):
ans = -1
count = -1
dic = {}
for x in arr:
if x in dic:
dic[x] += 1
else:
dic[x] = 1
if dic[x] > count or dic[x] == count and x < ans:
ans = x
count = dic[x]
return ans
Is it bad practice to use mutable objects as Hashmap keys? What happens when you try to retrieve a value from a Hashmap using a key that has been modified enough to change its hashcode?
For example, given
class Key
{
int a; //mutable field
int b; //mutable field
public int hashcode()
return foo(a, b);
// setters setA and setB omitted for brevity
}
with code
HashMap<Key, Value> map = new HashMap<Key, Value>();
Key key1 = new Key(0, 0);
map.put(key1, value1); // value1 is an instance of Value
key1.setA(5);
key1.setB(10);
What happens if we now call map.get(key1)? Is this safe or advisable? Or is the behavior dependent on the language?
It has been noted by many well respected developers such as Brian Goetz and Josh Bloch that :
If an object’s hashCode() value can change based on its state, then we
must be careful when using such objects as keys in hash-based
collections to ensure that we don’t allow their state to change when
they are being used as hash keys. All hash-based collections assume
that an object’s hash value does not change while it is in use as a
key in the collection. If a key’s hash code were to change while it
was in a collection, some unpredictable and confusing consequences
could follow. This is usually not a problem in practice — it is not
common practice to use a mutable object like a List as a key in a
HashMap.
This is not safe or advisable. The value mapped to by key1 can never be retrieved. When doing a retrieval, most hash maps will do something like
Object get(Object key) {
int hash = key.hashCode();
//simplified, ignores hash collisions,
Entry entry = getEntry(hash);
if(entry != null && entry.getKey().equals(key)) {
return entry.getValue();
}
return null;
}
In this example, key1.hashcode() now points to the wrong bucket of the hash table, and you will not be able to retrieve value1 with key1.
If you had done something like,
Key key1 = new Key(0, 0);
map.put(key1, value1);
key1.setA(5);
Key key2 = new Key(0, 0);
map.get(key2);
This will also not retrieve value1, as key1 and key2 are no longer equal, so this check
if(entry != null && entry.getKey().equals(key))
will fail.
Hash maps use hash code and equality comparisons to identify a certain key-value pair with a given key. If the has map keeps the key as a reference to the mutable object, it would work in the cases where the same instance is used to retrieve the value. Consider however, the following case:
T keyOne = ...;
T keyTwo = ...;
// At this point keyOne and keyTwo are different instances and
// keyOne.equals(keyTwo) is true.
HashMap myMap = new HashMap();
myMap.push(keyOne, "Hello");
String s1 = (String) myMap.get(keyOne); // s1 is "Hello"
String s2 = (String) myMap.get(keyTwo); // s2 is "Hello"
// because keyOne equals keyTwo
mutate(keyOne);
s1 = myMap.get(keyOne); // returns "Hello"
s2 = myMap.get(keyTwo); // not found
The above is true if the key is stored as a reference. In Java usually this is the case. In .NET for instance, if the key is a value type (always passed by value), the result will be different:
T keyOne = ...;
T keyTwo = ...;
// At this point keyOne and keyTwo are different instances
// and keyOne.equals(keyTwo) is true.
Dictionary myMap = new Dictionary();
myMap.Add(keyOne, "Hello");
String s1 = (String) myMap[keyOne]; // s1 is "Hello"
String s2 = (String) myMap[keyTwo]; // s2 is "Hello"
// because keyOne equals keyTwo
mutate(keyOne);
s1 = myMap[keyOne]; // not found
s2 = myMap[keyTwo]; // returns "Hello"
Other technologies might have other different behaviors. However, almost all of them would come to a situation where the result of using mutable keys is not deterministic, which is very very bad situation in an application - a hard to debug and even harder to understand.
If key’s hash code changes after the key-value pair (Entry) is stored in HashMap, the map will not be able to retrieve the Entry.
Key’s hashcode can change if the key object is mutable. Mutable keys in HahsMap can result in data loss.
This will not work. You are changing the key value, so you are basically throwing it away. Its like creating a real life key and lock, and then changing the key and trying to put it back in the lock.
As others explained, it is dangerous.
A way to avoid that is to have a const field giving explicitly the hash in your mutable objects (so you would hash on their "identity", not their "state"). You might even initialize that hash field more or less randomly.
Another trick would be to use the address, e.g. (intptr_t) reinterpret_cast<void*>(this) as a basis for hash.
In all cases, you have to give up hashing the changing state of the object.
There are two very different issues that can arise with a mutable key depending on your expectation of behavior.
First Problem: (probably most trivial--but hell it gave me problems that I didn't think about!)
You are attempting to place key-value pairs into a map by updating and modifying the same key object. You might do something like Map<Integer, String> and simply say:
int key = 0;
loop {
map.put(key++, newString);
}
I'm reusing the "object" key to create a map. This works fine in Java because of autoboxing where each new value of key gets autoboxed to a new Integer object. What would not work is if I created my own (mutable) Integer object:
MyInteger {
int value;
plusOne(){
value++;
}
}
Then tried the same approach:
MyInteger key = new MyInteger(0);
loop{
map.put(key.plusOne(), newString)
}
My expectation is that, for instance, I map 0 -> "a" and 1 -> "b". In the first example, if I change int key = 0, the map will (correctly) give me "a". For simplicity let's assume MyInteger just always returns the same hashCode() (if you can somehow manage to create unique hashCode values for all possible states of an object, this will not be an issue, and you deserve an award). In this case, I call 0 -> "a", so now the map holds my key and maps it to "a", I then modify key = 1 and try to put 1 -> "b". We have a problem! The hashCode() is the same, and the only key in the HashMap is my MyInteger key object which has just been modified to be equal to 1, so It overwrites that key's value so that now, instead of a map with 0 -> "a" and 1 -> "b", I have 1 -> "b" only! Even worse, if I change back to key = 0, the hashCode points to 1 -> "b", but since the HashMap's only key is my key object, it satisfied the equality check and returns "b", not "a" as expected.
If, like me, you fall prey to this type of issue, it's incredibly difficult to diagnose. Why? Because if you have a decent hashCode() function it will generate (mostly) unique values. The hash value will largely take care of the inequality problem when structuring the map but if you have enough values, eventually you'll get a collision on the hash value and then you get unexpected and largely inexplicable results. The resultant behavior is that it works for small runs but fails for larger ones.
Advice:
To find this type of issue, modify the hashCode() method, even trivially (i.e. = 0--obviously when doing this, keep in mind that the hash values should be the same for two equal objects*), and see if you get the same results--because you should and if you don't, there's likely a semantic error with your implementation that's using a hash table.
*There should be no danger (if there is--you have a semantic problem) in always returning 0 from a hashCode() (although it would defeat the purpose of a Hash Table). But that's sort of the point: the hashCode is a "quick and easy" equality measure that's not exact. So two very different objects could have the same hashCode() yet not be equal. On the other hand, two equal objects must always have the same hashCode() value.
p.s. In Java, from my understanding, if you do such a terrible thing (as have many hashCode() collisions), it will start using a red-black-tree as opposed to ArrayList. So when you expect O(1) lookup, you'll get O(log(n))--which is better than the ArrayList which would give O(n).
Second Problem:
This is the one that most others seem to be focusing on, so I'll try to be brief. In this use case, I try to map a key-value pair and then I do some work on the key and then want to come back and get my value.
Expectation: key -> value is mapped, I then modify key and try to get(key). I expect that will give me value.
It seems kind of obvious to me that this wouldn't work but I'm not above having tried to use things like Collections as a key before (and quite quickly realizing it doesn't work). It doesn't work because it's quite likely that the hash value of key has changed so you won't even be looking in the correct bucket.
This is why it's very inadvisable to use collections as keys. I would assume, if you were doing this, you're trying to establish a many-to-one relationship. So I have a class (as in teaching) and I want two groups to do two different projects. What I want is that given a group, what is their project? Simple, I divide the class in two, and I have group1 -> project1 and group2 -> project2. But wait! A new student arrives so I place them in group1. The problem is that group1 has now been modified and likely its hash value has changed, therefore trying to do get(group1) is likely to fail because it will look in a wrong or non-existent bucket of the HashMap.
The obvious solution to the above is to chain things--instead of using the groups as keys, give them labels (that don't change) that point to the group and therefore the project: g1 -> group1 and g1 -> project1, etc.
p.s.
Please make sure to define a hashCode() and equals(...) method for any object you expect to use as a key (eclipse and, I'm assuming, most IDE's can do this for you).
Code Example:
Here is a class which exhibits the two different "problem" behaviors. In this case, I attempt to map 0 -> "a", 1 -> "b", and 2 -> "c" (in each case). In the first problem, I do that by modifying the same object, in the second problem, I use unique objects, and in the second problem "fixed" I clone those unique objects. After that I take one of the "unique" keys (k0) and modify it to attempt to access the map. I expect this will give me a, b, c and null when the key is 3.
However, what happens is the following:
map.get(0) map1: 0 -> null, map2: 0 -> a, map3: 0 -> a
map.get(1) map1: 1 -> null, map2: 1 -> b, map3: 1 -> b
map.get(2) map1: 2 -> c, map2: 2 -> a, map3: 2 -> c
map.get(3) map1: 3 -> null, map2: 3 -> null, map3: 3 -> null
The first map ("first problem") fails because it only holds a single key, which was last updated and placed to equal 2, hence why it correctly returns "c" when k0 = 2 but returns null for the other two (the single key doesn't equal 0 or 1). The second map fails twice: the most obvious is that it returns "b" when I asked for k0 (because it's been modified--that's the "second problem" which seems kind of obvious when you do something like this). It fails a second time when it returns "a" after modifying k0 = 2 (which I would expect to be "c"). This is more due to the "first problem": there's a hash code collision and the tiebreaker is an equality check--but the map holds k0, which it (apparently for me--could theoretically be different for someone else) checked first and thus returned the first value, "a" even though had it kept checking, "c" would have also been a match. Finally, the 3rd map works perfectly because I'm enforcing that the map holds unique keys no matter what else I do (by cloning the object during insertion).
I want to make clear that I agree, cloning is not a solution! I simply added that as an example of why a map needs unique keys and how enforcing unique keys "fixes" the issue.
public class HashMapProblems {
private int value = 0;
public HashMapProblems() {
this(0);
}
public HashMapProblems(final int value) {
super();
this.value = value;
}
public void setValue(final int i) {
this.value = i;
}
#Override
public int hashCode() {
return value % 2;
}
#Override
public boolean equals(final Object o) {
return o instanceof HashMapProblems
&& value == ((HashMapProblems) o).value;
}
#Override
public Object clone() {
return new HashMapProblems(value);
}
public void reset() {
this.value = 0;
}
public static void main(String[] args) {
final HashMapProblems k0 = new HashMapProblems(0);
final HashMapProblems k1 = new HashMapProblems(1);
final HashMapProblems k2 = new HashMapProblems(2);
final HashMapProblems k = new HashMapProblems();
final HashMap<HashMapProblems, String> map1 = firstProblem(k);
final HashMap<HashMapProblems, String> map2 = secondProblem(k0, k1, k2);
final HashMap<HashMapProblems, String> map3 = secondProblemFixed(k0, k1, k2);
for (int i = 0; i < 4; ++i) {
k0.setValue(i);
System.out.printf(
"map.get(%d) map1: %d -> %s, map2: %d -> %s, map3: %d -> %s",
i, i, map1.get(k0), i, map2.get(k0), i, map3.get(k0));
System.out.println();
}
}
private static HashMap<HashMapProblems, String> firstProblem(
final HashMapProblems start) {
start.reset();
final HashMap<HashMapProblems, String> map = new HashMap<>();
map.put(start, "a");
start.setValue(1);
map.put(start, "b");
start.setValue(2);
map.put(start, "c");
return map;
}
private static HashMap<HashMapProblems, String> secondProblem(
final HashMapProblems... keys) {
final HashMap<HashMapProblems, String> map = new HashMap<>();
IntStream.range(0, keys.length).forEach(
index -> map.put(keys[index], "" + (char) ('a' + index)));
return map;
}
private static HashMap<HashMapProblems, String> secondProblemFixed(
final HashMapProblems... keys) {
final HashMap<HashMapProblems, String> map = new HashMap<>();
IntStream.range(0, keys.length)
.forEach(index -> map.put((HashMapProblems) keys[index].clone(),
"" + (char) ('a' + index)));
return map;
}
}
Some Notes:
In the above it should be noted that map1 only holds two values because of the way I set up the hashCode() function to split odds and evens. k = 0 and k = 2 therefore have the same hashCode of 0. So when I modify k = 2 and attempt to k -> "c" the mapping k -> "a" gets overwritten--k -> "b" is still there because it exists in a different bucket.
Also there are a lot of different ways to examine the maps in the above code and I would encourage people that are curious to do things like print out the values of the map and then the key to value mappings (you may be surprised by the results you get). Do things like play with changing the different "unique" keys (i.e. k0, k1, and k2), try changing the single key k. You could also see how even the secondProblemFixed isn't actually fixed because you could also gain access to the keys (for instance via Map::keySet) and modify them.
I won't repeat what others have said. Yes, it's inadvisable. But in my opinion, it's not overly obvious where the documentation states this.
You can find it on the JavaDoc for the Map interface:
Note: great care must be exercised if mutable objects are used as map
keys. The behavior of a map is not specified if the value of an object
is changed in a manner that affects equals comparisons while the
object is a key in the map
Behaviour of a Map is not specified if value of an object is changed in a manner that affects equals comparision while object(Mutable) is a key. Even for Set also using mutable object as key is not a good idea.
Lets see a example here :
public class MapKeyShouldntBeMutable {
/**
* #param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Map<Employee,Integer> map=new HashMap<Employee,Integer>();
Employee e=new Employee();
Employee e1=new Employee();
Employee e2=new Employee();
Employee e3=new Employee();
Employee e4=new Employee();
e.setName("one");
e1.setName("one");
e2.setName("three");
e3.setName("four");
e4.setName("five");
map.put(e, 24);
map.put(e1, 25);
map.put(e2, 26);
map.put(e3, 27);
map.put(e4, 28);
e2.setName("one");
System.out.println(" is e equals e1 "+e.equals(e1));
System.out.println(map);
for(Employee s:map.keySet())
{
System.out.println("key : "+s.getName()+":value : "+map.get(s));
}
}
}
class Employee{
String name;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#Override
public boolean equals(Object o){
Employee e=(Employee)o;
if(this.name.equalsIgnoreCase(e.getName()))
{
return true;
}
return false;
}
public int hashCode() {
int sum=0;
if(this.name!=null)
{
for(int i=0;i<this.name.toCharArray().length;i++)
{
sum=sum+(int)this.name.toCharArray()[i];
}
/*System.out.println("name :"+this.name+" code : "+sum);*/
}
return sum;
}
}
Here we are trying to add mutable object "Employee" to a map. It will work good if all keys added are distinct.Here I have overridden equals and hashcode for employee class.
See first I have added "e" and then "e1". For both of them equals() will be true and hashcode will be same. So map sees as if the same key is getting added so it should replace the old value with e1's value. Then we have added e2,e3,e4 we are fine as of now.
But when we are changing the value of an already added key i.e "e2" as one ,it becomes a key similar to one added earlier. Now the map will behave wired. Ideally e2 should replace the existing same key i.e e1.But now map takes this as well. And you will get this in o/p :
is e equals e1 true
{Employee#1aa=28, Employee#1bc=27, Employee#142=25, Employee#142=26}
key : five:value : 28
key : four:value : 27
key : one:value : 25
key : one:value : 25
See here both keys having one showing same value also. So its unexpected.Now run the same programme again by changing e2.setName("diffnt"); which is e2.setName("one"); here ...Now the o/p will be this :
is e equals e1 true
{Employee#1aa=28, Employee#1bc=27, Employee#142=25, Employee#27b=26}
key : five:value : 28
key : four:value : 27
key : one:value : 25
key : diffnt:value : null
So by adding changing the mutable key in a map is not encouraged.
To make the answer compact:
The root cause is that HashMap calculates an internal hash of the user's key object hashcode only once and stores it inside for own needs.
All other operations for data navigation inside the map are doing by this pre-calculated internal hash.
So if you change the hashcode of the key object (mutate) it will be still stored nicely inside the map with the changed key object's hashcode (you could even observe it via HashMap.keySet() and see the altered hashcode).
But HashMap internal hash will not be recalculated of course and it will be the old stored one and the map won't be able to locate your data by the provided mutated key object new hashcode. (e.g. by HashMap.get() or HashMap.containsKey()).
Your key-value pairs will be still inside the map but to get it back you will need that old hash code value that was given when you put your data into the map.
Notice that you also will be unable to get data back by the mutated key object taken right from the HashMap.keySet().
I am pretty confused with both functions fold() and reduce() in Kotlin, can anyone give me a concrete example that distinguishes both of them?
fold takes an initial value, and the first invocation of the lambda you pass to it will receive that initial value and the first element of the collection as parameters.
For example, take the following code that calculates the sum of a list of integers:
listOf(1, 2, 3).fold(0) { sum, element -> sum + element }
The first call to the lambda will be with parameters 0 and 1.
Having the ability to pass in an initial value is useful if you have to provide some sort of default value or parameter for your operation. For example, if you were looking for the maximum value inside a list, but for some reason want to return at least 10, you could do the following:
listOf(1, 6, 4).fold(10) { max, element ->
if (element > max) element else max
}
reduce doesn't take an initial value, but instead starts with the first element of the collection as the accumulator (called sum in the following example).
For example, let's do a sum of integers again:
listOf(1, 2, 3).reduce { sum, element -> sum + element }
The first call to the lambda here will be with parameters 1 and 2.
You can use reduce when your operation does not depend on any values other than those in the collection you're applying it to.
The major functional difference I would call out (which is mentioned in the comments on the other answer, but may be hard to understand) is that reduce will throw an exception if performed on an empty collection.
listOf<Int>().reduce { x, y -> x + y }
// java.lang.UnsupportedOperationException: Empty collection can't be reduced.
This is because .reduce doesn't know what value to return in the event of "no data".
Contrast this with .fold, which requires you to provide a "starting value", which will be the default value in the event of an empty collection:
val result = listOf<Int>().fold(0) { x, y -> x + y }
assertEquals(0, result)
So, even if you don't want to aggregate your collection down to a single element of a different (non-related) type (which only .fold will let you do), if your starting collection may be empty then you must either check your collection size first and then .reduce, or just use .fold
val collection: List<Int> = // collection of unknown size
val result1 = if (collection.isEmpty()) 0
else collection.reduce { x, y -> x + y }
val result2 = collection.fold(0) { x, y -> x + y }
assertEquals(result1, result2)
Another difference that none of the other answers mentioned is the following:
The result of a reduce operation will always be of the same type (or a super type) as the data that is being reduced.
We can see that from the definition of the reduce method:
public inline fun <S, T : S> Iterable<T>.reduce(operation: (acc: S, T) -> S): S {
val iterator = this.iterator()
if (!iterator.hasNext()) throw UnsupportedOperationException("Empty collection can't be reduced.")
var accumulator: S = iterator.next()
while (iterator.hasNext()) {
accumulator = operation(accumulator, iterator.next())
}
return accumulator
}
On the other hand, the result of a fold operation can be anything, because there are no restrictions when it comes to setting up the initial value.
So, for example, let us say that we have a string that contains letters and digits. We want to calculate the sum of all the digits.
We can easily do that with fold:
val string = "1a2b3"
val result: Int = string.fold(0, { currentSum: Int, char: Char ->
if (char.isDigit())
currentSum + Character.getNumericValue(char)
else currentSum
})
//result is equal to 6
reduce - The reduce() method transforms a given collection into a single result.
val numbers: List<Int> = listOf(1, 2, 3)
val sum: Int = numbers.reduce { acc, next -> acc + next }
//sum is 6 now.
fold - What would happen in the previous case of an empty list? Actually, there’s no right value to return, so reduce() throws a RuntimeException
In this case, fold is a handy tool. You can put an initial value by it -
val sum: Int = numbers.fold(0, { acc, next -> acc + next })
Here, we’ve provided initial value. In contrast, to reduce(), if the collection is empty, the initial value will be returned which will prevent you from the RuntimeException.
Simple Answer
Result of both reduce and fold is "a list of items will be transformed into a single item".
In case of fold,we provide 1 extra parameter apart from list but in case of reduce,only items in list will be considered.
Fold
listOf("AC","Fridge").fold("stabilizer") { freeGift, itemBought -> freeGift + itemBought }
//output: stabilizerACFridge
In above case,think as AC,fridge bought from store & they give stabilizer as gift(this will be the parameter passed in the fold).so,you get all 3 items together.Please note that freeGift will be available only once i.e for the first iteration.
Reduce
In case of reduce,we get items in list as parameters and can perform required transformations on it.
listOf("AC","Fridge").reduce { itemBought1, itemBought2 -> itemBought1 + itemBought2 }
//output: ACFridge
The difference between the two functions is that fold() takes an initial value and uses it as the accumulated value on the first step, whereas the first step of reduce() uses the first and the second elements as operation arguments on the first step.