question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
design-authentication-manager | Hash Map | Simple | hash-map-simple-by-prashant404-ym2l | T/S:
constructor: O(1)/O(1)
generate: O(1)/O(1)
renew: O(1)/O(1)
countUnexpiredTokens: O(n)/O(1), where n = number of unexpired tokens
Please upvote if this he | prashant404 | NORMAL | 2025-03-16T19:30:14.437606+00:00 | 2025-03-16T19:30:14.437606+00:00 | 9 | false | >T/S:
$$constructor:$$ O(1)/O(1)
$$generate:$$ O(1)/O(1)
$$renew:$$ O(1)/O(1)
$$countUnexpiredTokens:$$ O(n)/O(1), where n = number of unexpired tokens
```java []
class AuthenticationManager {
private final int timeToLive;
private final Map<String, Integer> tokenToExpireTime = new HashMap<>();
/**
* constructs the AuthenticationManager
*
* @param timeToLive set the timeToLive.
*/
public AuthenticationManager(int timeToLive) {
this.timeToLive = timeToLive;
}
/**
* @param tokenId generate a new token with the given tokenId
* @param currentTime generate a new token at the given currentTime in seconds.
*/
public void generate(String tokenId, int currentTime) {
tokenToExpireTime.put(tokenId, currentTime + timeToLive);
}
/**
* If there are no unexpired tokens with the given tokenId, the request is ignored, and nothing happens.
*
* @param tokenId renews the unexpired token with the given tokenId
* @param currentTime renews the unexpired token at the given currentTime in seconds
*/
public void renew(String tokenId, int currentTime) {
var expireTime = tokenToExpireTime.remove(tokenId);
if (expireTime == null || expired(currentTime, expireTime))
return;
tokenToExpireTime.put(tokenId, currentTime + timeToLive);
}
/**
* @param currentTime currentTime
* @return returns the number of unexpired tokens at the given currentTime.
*/
public int countUnexpiredTokens(int currentTime) {
tokenToExpireTime.entrySet()
.removeIf(e -> expired(currentTime, e.getValue()));
return tokenToExpireTime.size();
}
private boolean expired(int currentTime, Integer expireTime) {
return expireTime <= currentTime;
}
}
```
***Please upvote if this helps*** | 0 | 0 | ['Java'] | 0 |
design-authentication-manager | HashMap | hashmap-by-cx1z0-8ofq | null | Cx1z0 | NORMAL | 2025-03-08T02:39:26.672327+00:00 | 2025-03-08T02:39:26.672327+00:00 | 2 | false | ```javascript []
class AuthenticationManager {
constructor(timeToLive) {
this.timeToLive = timeToLive;
this.tokens = new Map();
}
generate(tokenId, currentTime) {
this.tokens.set(tokenId, currentTime + this.timeToLive);
}
renew(tokenId, currentTime) {
if (this.tokens.has(tokenId) && this.tokens.get(tokenId) > currentTime) {
this.tokens.set(tokenId, currentTime + this.timeToLive);
}
}
countUnexpiredTokens(currentTime) {
for (const [token, expiry] of this.tokens) {
if (expiry <= currentTime) {
this.tokens.delete(token);
}
}
return this.tokens.size;
}
}
``` | 0 | 0 | ['Hash Table', 'Design', 'JavaScript'] | 0 |
design-authentication-manager | Solution with hashmap and doubly linkedlist | solution-with-hashmap-and-doubly-linkedl-e0xp | Sharing my solution with hashmap and doubly linkedlist | atsiz | NORMAL | 2025-03-07T17:43:33.469058+00:00 | 2025-03-07T17:43:33.469058+00:00 | 6 | false | Sharing my solution with hashmap and doubly linkedlist
```cpp []
class Node{
public:
int time;
string token;
Node*next, *prev;
Node(string token, int n){
this->token = token;
time = n;
prev = next = NULL;
}
};
class AuthenticationManager {
public:
int ttl;
int size;
unordered_map<string, Node*>m;
Node*head, *tail;
AuthenticationManager(int timeToLive) {
ttl = timeToLive;
head = new Node("", 0);
tail = head;
size = 0;
}
void moveNodeToTop(Node* n){
n->prev = head;
n->next = head->next;
if(head->next) head->next->prev = n;
else tail = n;
head->next = n;
}
void generate(string tokenId, int ct) {
m[tokenId] = new Node(tokenId, ct+ttl);
size++;
moveNodeToTop(m[tokenId]);
}
void renew(string tokenId, int currentTime) {
deleteExpiredTokens(currentTime); // delete expired tokens
if(m.find(tokenId) == m.end()) return;
m[tokenId]->time = currentTime + ttl;
// move this node to first
Node* n = m[tokenId];
// exit if its already a first node
if(head->next == n) return;
// adjust tail or middle nodes
if(tail == n) {
tail = tail->prev;
tail->next = NULL;
} else {
n->prev->next = n->next;
if(n->next) n->next->prev = n->prev;
}
moveNodeToTop(n);
}
int countUnexpiredTokens(int currentTime) {
deleteExpiredTokens(currentTime);
return size;
}
void deleteExpiredTokens(int time){
while(tail != head){
if(tail->time > time) break;
else {
Node* d = tail;
tail = tail->prev;
m.erase(d->token);
delete d;
size--;
}
}
if(tail == head){
head->next = NULL;
}
}
};
/**
* Your AuthenticationManager object will be instantiated and called as such:
* AuthenticationManager* obj = new AuthenticationManager(timeToLive);
* obj->generate(tokenId,currentTime);
* obj->renew(tokenId,currentTime);
* int param_3 = obj->countUnexpiredTokens(currentTime);
*/
``` | 0 | 0 | ['C++'] | 0 |
design-authentication-manager | Optimized (1) Double LinkedList (2) HashTable with Explanation | optimized-hashtable-with-explanation-by-nq0qq | IntuitionApproachDouble LinkedList.Complexity
Time complexity:
Space complexity:
CodeIntuitionApproachHashTable.Complexity
Time complexity:
Space complexity: | stalebii | NORMAL | 2025-03-01T18:49:23.230265+00:00 | 2025-03-02T22:19:16.330766+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
Double LinkedList.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python []
class ListNode:
def __init__(self, token='', currentTime=-1, prev=None, next=None):
self.token = token
self.insertTime = currentTime
self.prev = prev
self.next = next
class AuthenticationManager:
def __init__(self, timeToLive):
self.ttl = timeToLive # set the conant time to live into ttl
self.cache = dict() # stores {tokenId: node}
self.head = ListNode() # two nodes for head and tail
self.tail = ListNode()
self.head.next = self.tail # update their pointers
self.tail.prev = self.head
self.size = 0 # store the length of linked list
def generate(self, tokenId, currentTime):
# 1. if already exists delete the node
if tokenId in self.cache:
self.delete(tokenId)
# 2. create a new node
node = ListNode(tokenId, currentTime)
# 3. add the new node
self.cache[tokenId] = node
# 4. add the node to the end of double linked list
self.addToEnd(node)
def renew(self, tokenId, currentTime):
# base
if tokenId not in self.cache:
return
# get the node
node = self.cache[tokenId]
# expired; remove node
if node.insertTime + self.ttl <= currentTime:
self.delete(node.token)
# otherwise; re-generate the tokenId at currentTime
else:
self.generate(tokenId, currentTime)
def countUnexpiredTokens(self, currentTime):
# start from the head
curr = self.head.next
# clean up linked-list if needed till reaching to tail
while curr and curr.insertTime != -1 and curr.insertTime + self.ttl <= currentTime:
# copy the next node after curr
next_curr = curr.next
# remove curr node
self.delete(curr.token)
# got to next
curr = next_curr
return self.size
def find(self, tokenId):
# pull out the node from the cache
node = self.cache[tokenId]
# x (node.prev), node , y (node.next)
# skip node between x & y
node.prev.next = node.next # x -> y
node.next.prev = node.prev # y -> x
# disable its prev and next pointers
node.next = None
node.prev = None
# return the node
return node
def addToEnd(self, node):
# tail.prev -> tail -> None transforms into
# tail.prev -> node -> tail -> None
# work on node pointers
node.prev = self.tail.prev
node.next = self.tail
# work on tail.prev pointer
self.tail.prev.next = node
# work on tail pointer
self.tail.prev = node
# update the size
self.size += 1
def delete(self, tokenId):
# find the node
deleteNode = self.find(tokenId)
# since it does not have any pointer delete it
del deleteNode
# remove it from the cache
del self.cache[tokenId]
# update the size
self.size -= 1
```
# Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
HashTable.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python []
class AuthenticationManager(object):
def __init__(self, timeToLive):
self.tokens = dict() # stores {tokenId: expiry time}
self.ttl = timeToLive # set the conant time to live into ttl
def generate(self, tokenId, currentTime):
# calculate the expiry time
expiry = currentTime + self.ttl
# update the map
self.tokens[tokenId] = expiry
def renew(self, tokenId, currentTime):
# base1; never generated
if tokenId not in self.tokens:
return
# base2; expired a while ago; remove from the map
if self.tokens[tokenId] <= currentTime:
del self.tokens[tokenId]
return
# otherwise; update expiry for tokenId
expiry = currentTime + self.ttl
self.tokens[tokenId] = expiry
def countUnexpiredTokens(self, currentTime):
res = 0
for tokenId, expiry in self.tokens.items():
if currentTime < expiry:
res += 1
return res
# Your AuthenticationManager object will be instantiated and called as such:
# obj = AuthenticationManager(timeToLive)
# obj.generate(tokenId,currentTime)
# obj.renew(tokenId,currentTime)
# param_3 = obj.countUnexpiredTokens(currentTime)
``` | 0 | 0 | ['Hash Table', 'Linked List', 'Design', 'Doubly-Linked List'] | 0 |
design-authentication-manager | A simple java solution using linked list | a-simple-java-solution-using-linked-list-avpg | null | yucan29 | NORMAL | 2025-02-17T12:00:30.836070+00:00 | 2025-02-17T12:00:30.836070+00:00 | 13 | false |
```java []
class AuthenticationManager {
class Node
{
String tokenId;
int createOrRenewtime;
Node next;
Node(String tokenId,int time)
{
this.tokenId = tokenId;
this.createOrRenewtime = time;
}
}
int timeToLive;
Node dummyHead, last;
public AuthenticationManager(int timeToLive) {
this.timeToLive = timeToLive;
dummyHead = new Node("dummy",0);
last = dummyHead;
}
public void generate(String tokenId, int currentTime) {
last.next= new Node(tokenId, currentTime);
last = last.next;
}
public void renew(String tokenId, int currentTime) {
Node curr = dummyHead.next;
while(curr!=null)
{
if(curr.tokenId.equals(tokenId))
{
if(curr.createOrRenewtime + timeToLive > currentTime)
curr.createOrRenewtime = currentTime;
}
curr=curr.next;
}
}
public int countUnexpiredTokens(int currentTime) {
Node curr = dummyHead.next;
int cnt=0;
while(curr!=null)
{
if(curr.createOrRenewtime + timeToLive > currentTime) cnt++;
curr=curr.next;
}
return cnt;
}
}
``` | 0 | 0 | ['Linked List', 'Java'] | 0 |
design-authentication-manager | Design Authentication Manager | design-authentication-manager-by-ubaidha-6uup | IntuitionThe problem requires managing authentication tokens with a specific time-to-live (TTL). Each token has an expiration time, and we need to handle operat | UbaidHabib0 | NORMAL | 2025-02-05T17:32:05.552800+00:00 | 2025-02-05T17:32:05.552800+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires managing authentication tokens with a specific time-to-live (TTL). Each token has an expiration time, and we need to handle operations like generating new tokens, renewing existing ones, and counting unexpired tokens at any given time. The key challenge is to efficiently manage the tokens and their expiration times while ensuring that operations like renewal and counting are performed correctly.
# Approach
<!-- Describe your approach to solving the problem. -->
1. Data Structure:
Use a dictionary (hash map) to store tokens and their expiration times. The key is the tokenId, and the value is the expiration time.
2. Methods:
generate(tokenId, currentTime):
. Add a new token to the dictionary with an expiration time of currentTime + timeToLive.
. renew(tokenId, currentTime):
.. Check if the token exists and is unexpired (i.e., its expiration time is greater than currentTime).
.. If valid, update its expiration time to currentTime + timeToLive.
. countUnexpiredTokens(currentTime):
.. Iterate through the dictionary and count tokens with expiration times greater than currentTime.
3. Edge Cases:
. Handle cases where tokens are renewed or counted exactly at their expiration time.
. Ensure that expired tokens are not counted or renewed.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
. generate: O(1) (dictionary insertion).
. renew: O(1) (dictionary lookup and update).
. countUnexpiredTokens: O(n), where n is the number of tokens (iterates through all tokens).
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
O(n), where n is the number of tokens (stored in the dictionary).
# Code
```python3 []
class AuthenticationManager:
def __init__(self, timeToLive: int):
self.timeToLive = timeToLive # Time-to-live for tokens
self.tokens = {} # Dictionary to store tokenId: expirationTime
def generate(self, tokenId: str, currentTime: int) -> None:
# Generate a new token with expiration time = currentTime + timeToLive
self.tokens[tokenId] = currentTime + self.timeToLive
def renew(self, tokenId: str, currentTime: int) -> None:
# Renew the token only if it exists and is unexpired
if tokenId in self.tokens and self.tokens[tokenId] > currentTime:
self.tokens[tokenId] = currentTime + self.timeToLive
def countUnexpiredTokens(self, currentTime: int) -> int:
# Count tokens with expiration time > currentTime
count = 0
for expirationTime in self.tokens.values():
if expirationTime > currentTime:
count += 1
return count
``` | 0 | 0 | ['Python3'] | 0 |
design-authentication-manager | Ruby Not Super Fast but seems to be one of few solutions | ruby-not-super-fast-but-seems-to-be-one-j2y0k | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | zacharylupstein | NORMAL | 2025-02-05T16:11:54.172154+00:00 | 2025-02-05T16:11:54.172154+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```ruby []
class AuthenticationManager
=begin
:type time_to_live: Integer
=end
def initialize(time_to_live)
@timeToLive = time_to_live
@tokens = Hash.new
end
=begin
:type token_id: String
:type current_time: Integer
:rtype: Void
=end
def generate(token_id, current_time)
endTime = current_time + @timeToLive
@tokens[token_id] = endTime
end
=begin
:type token_id: String
:type current_time: Integer
:rtype: Void
=end
def renew(token_id, current_time)
if @tokens.has_key?(token_id)
if @tokens[token_id] > current_time
newEnd = current_time + @timeToLive
@tokens[token_id] = newEnd
end
end
end
=begin
:type current_time: Integer
:rtype: Integer
=end
def count_unexpired_tokens(current_time)
result = 0
@tokens.each_key do |tokenId|
if @tokens[tokenId] > current_time
result = result + 1
end
end
return result
end
end
# Your AuthenticationManager object will be instantiated and called as such:
# obj = AuthenticationManager.new(time_to_live)
# obj.generate(token_id, current_time)
# obj.renew(token_id, current_time)
# param_3 = obj.count_unexpired_tokens(current_time)
``` | 0 | 0 | ['Ruby'] | 0 |
design-authentication-manager | Python (Simple Hashmap) | python-simple-hashmap-by-rnotappl-woml | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | rnotappl | NORMAL | 2025-02-04T08:47:09.316533+00:00 | 2025-02-04T08:47:09.316533+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class AuthenticationManager:
def __init__(self, timeToLive):
self.timeToLive = timeToLive
self.dict1 = {}
def generate(self, tokenId, currentTime):
self.dict1[tokenId] = currentTime + self.timeToLive
def renew(self, tokenId, currentTime):
if tokenId in self.dict1:
if currentTime < self.dict1[tokenId]:
self.dict1[tokenId] = currentTime + self.timeToLive
def countUnexpiredTokens(self, currentTime):
count = 0
for key,val in self.dict1.items():
if val > currentTime:
count += 1
return count
``` | 0 | 0 | ['Python3'] | 0 |
design-authentication-manager | Golang solution | golang-solution-by-sudarshan_a_m-v14f | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Sudarshan_A_M | NORMAL | 2025-01-30T13:21:47.274550+00:00 | 2025-01-30T13:21:47.274550+00:00 | 14 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```golang []
type AuthenticationManager struct {
TimeToLive int
expire map[string]int
}
func Constructor(timeToLive int) AuthenticationManager {
x := AuthenticationManager{
TimeToLive: timeToLive,
expire: make(map[string]int),
}
return x
}
func (this *AuthenticationManager) Generate(tokenId string, currentTime int) {
if this.expire[tokenId] == 0 {
this.expire[tokenId] = this.TimeToLive + currentTime
}
return
}
func (this *AuthenticationManager) Renew(tokenId string, currentTime int) {
if this.expire[tokenId] > currentTime {
this.expire[tokenId] = currentTime + this.TimeToLive
}
}
func (this *AuthenticationManager) CountUnexpiredTokens(currentTime int) int {
var count int
for _, v := range this.expire {
if v > currentTime {
count++
}
}
return count
}
/**
* Your AuthenticationManager object will be instantiated and called as such:
* obj := Constructor(timeToLive);
* obj.Generate(tokenId,currentTime);
* obj.Renew(tokenId,currentTime);
* param_3 := obj.CountUnexpiredTokens(currentTime);
*/
``` | 0 | 0 | ['Hash Table', 'Go'] | 0 |
design-authentication-manager | C# based simple approach | c-based-simple-approach-by-gakash5839-imvq | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | gakash5839 | NORMAL | 2025-01-21T06:16:42.682922+00:00 | 2025-01-21T06:16:42.682922+00:00 | 19 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```csharp []
public class AuthenticationManager {
public Dictionary<string, int> tokenttl;
public PriorityQueue<string, int> tokenQueue;
public int timeToLive;
public AuthenticationManager(int timeToLive) {
this.timeToLive = timeToLive;
tokenttl = new Dictionary<string,int>();
tokenQueue = new PriorityQueue<string, int>();
}
public void Generate(string tokenId, int currentTime) {
var temp = currentTime + timeToLive;
tokenttl[tokenId] = temp;
tokenQueue.Enqueue(tokenId,temp);
}
public void Renew(string tokenId, int currentTime) {
if(tokenttl.ContainsKey(tokenId))
{
if(tokenttl[tokenId] > currentTime)
{
var temp = currentTime + timeToLive;
tokenttl[tokenId] = temp;
tokenQueue.Enqueue(tokenId,temp);
}
}
}
public int CountUnexpiredTokens(int currentTime) {
int count =0;
string item;
int priority =0;
if(tokenQueue.Count == 0)
return 0;
tokenQueue.TryPeek(out item, out priority);
while(tokenQueue.Count > 0 && priority <= currentTime)
{
var temp = tokenQueue.Dequeue();
if(tokenttl[item] <= priority)
tokenttl.Remove(item);
tokenQueue.TryPeek(out item, out priority);
}
return tokenttl.Count;
}
}
/**
* Your AuthenticationManager object will be instantiated and called as such:
* AuthenticationManager obj = new AuthenticationManager(timeToLive);
* obj.Generate(tokenId,currentTime);
* obj.Renew(tokenId,currentTime);
* int param_3 = obj.CountUnexpiredTokens(currentTime);
*/
``` | 0 | 0 | ['C#'] | 0 |
design-authentication-manager | 1797. Design Authentication Manager | 1797-design-authentication-manager-by-g8-tonn | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-15T03:41:25.435915+00:00 | 2025-01-15T03:41:25.435915+00:00 | 5 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```python3 []
class AuthenticationManager:
def __init__(self, timeToLive: int):
self.timeToLive = timeToLive
self.tokens = {}
def generate(self, tokenId: str, currentTime: int) -> None:
self.tokens[tokenId] = currentTime + self.timeToLive
def renew(self, tokenId: str, currentTime: int) -> None:
if tokenId in self.tokens and self.tokens[tokenId] > currentTime:
self.tokens[tokenId] = currentTime + self.timeToLive
def countUnexpiredTokens(self, currentTime: int) -> int:
return sum(1 for expiry in self.tokens.values() if expiry > currentTime)
``` | 0 | 0 | ['Python3'] | 0 |
design-authentication-manager | Multi-set and HashMap | multi-set-and-hashmap-by-rudra201-coyc | Complexity
Time complexity:
generate O(log(n))
renew O(log(n))
CountUnexpiredTokens O(log(n))
Space complexity:O(n)
Code | Rudra201 | NORMAL | 2024-12-29T09:10:21.431881+00:00 | 2024-12-29T09:10:58.318464+00:00 | 4 | false |
# Complexity
- Time complexity:
1. generate $$O(log(n))$$
2. renew $$O(log(n))$$
3. CountUnexpiredTokens $$O(log(n))$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:$$O(n)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class AuthenticationManager {
unordered_map<string, int> tokenStore;
multiset<int> tokenLifetimes;
int lifetime = 0;
public:
AuthenticationManager(int timeToLive) { lifetime = timeToLive; }
void generate(string tokenId, int currentTime) {
tokenStore[tokenId] = currentTime + lifetime;
tokenLifetimes.insert(currentTime + lifetime);
}
void renew(string tokenId, int currentTime) {
if (tokenStore.find(tokenId) == tokenStore.end())
return;
int oldTime = tokenStore[tokenId];
if (oldTime <= currentTime)
return;
tokenStore[tokenId] = currentTime + lifetime;
auto tokenTime = tokenLifetimes.find(oldTime);
tokenLifetimes.erase(tokenTime);
tokenLifetimes.insert(currentTime + lifetime);
}
int countUnexpiredTokens(int currentTime) {
return distance(tokenLifetimes.upper_bound(currentTime),
tokenLifetimes.end());
}
};
/**
* Your AuthenticationManager object will be instantiated and called as such:
* AuthenticationManager* obj = new AuthenticationManager(timeToLive);
* obj->generate(tokenId,currentTime);
* obj->renew(tokenId,currentTime);
* int param_3 = obj->countUnexpiredTokens(currentTime);
*/
``` | 0 | 0 | ['Hash Table', 'Ordered Set', 'C++'] | 0 |
design-authentication-manager | Simple, intuitive and easy to understand Java solution! | simple-intuitive-and-easy-to-understand-cgc9m | Code | coderpanda_ | NORMAL | 2024-12-26T04:08:32.695155+00:00 | 2024-12-26T04:08:32.695155+00:00 | 12 | false |
# Code
```java []
class AuthenticationManager {
class Pair implements Comparable<Pair>{
int expTime;
String tokenId;
Pair(int eT, String t){
expTime=eT;
tokenId=t;
}
public int compareTo(Pair other){
return other.expTime-this.expTime;
}
}
private int ttl;
private HashMap<String, Integer> tokens;
private PriorityQueue<Pair> tokenPeriod;
private HashMap<String, Pair> tokenPair;
public AuthenticationManager(int timeToLive) {
ttl = timeToLive;
tokens = new HashMap<>();
tokenPeriod = new PriorityQueue<>();
tokenPair = new HashMap<>();
}
public void generate(String tokenId, int currentTime) {
tokens.put(tokenId, ttl+currentTime);
Pair p = new Pair(ttl+currentTime, tokenId);
tokenPeriod.add(p);
tokenPair.put(tokenId, p);
}
public void renew(String tokenId, int currentTime) {
if(tokens.containsKey(tokenId)){
if(currentTime < tokens.get(tokenId)){
tokens.put(tokenId, currentTime+ttl);
Pair oldPair = tokenPair.get(tokenId);
tokenPeriod.remove(oldPair);
Pair newPair = new Pair(currentTime+ttl, tokenId);
tokenPeriod.add(newPair);
tokenPair.put(tokenId, newPair);
}
}
}
public int countUnexpiredTokens(int currentTime) {
System.out.println("start --- currTime = "+currentTime);
int unexpiredTokenCount=0;
Iterator itr = tokenPeriod.iterator();
while(itr.hasNext()){
Pair p = (Pair)itr.next();
if(p.expTime > currentTime)
unexpiredTokenCount++;
}
return unexpiredTokenCount;
}
}
/**
* Your AuthenticationManager object will be instantiated and called as such:
* AuthenticationManager obj = new AuthenticationManager(timeToLive);
* obj.generate(tokenId,currentTime);
* obj.renew(tokenId,currentTime);
* int param_3 = obj.countUnexpiredTokens(currentTime);
*/
``` | 0 | 0 | ['Java'] | 0 |
design-authentication-manager | Java create an arrayList of Custom class | java-create-an-arraylist-of-custom-class-9ozy | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | saptarshichatterjee1 | NORMAL | 2024-12-18T02:08:27.189604+00:00 | 2024-12-18T02:08:27.189604+00:00 | 3 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass AuthenticationManager { //20.30\n\n int ttl;\n ArrayList<Token> list = new ArrayList();\n \n class Token{\n String token;\n int exp;\n public Token(String token, int exp){\n this.token = token;\n this.exp = exp;\n }\n }\n\n public AuthenticationManager(int timeToLive) {\n this.ttl = timeToLive;\n }\n \n public void generate(String tokenId, int currentTime) {\n Token t = new Token(tokenId, currentTime + this.ttl);\n list.add(t);\n }\n \n public void renew(String tokenId, int currentTime) {\n Token thz = null;\n int ind = 0;\n for(; ind < list.size(); ind++){\n if(list.get(ind).token.equals(tokenId)){\n thz = list.get(ind);\n break;\n }\n }\n if(thz != null){\n list.remove(ind);\n if(thz.exp <= currentTime) return;\n thz.exp = currentTime + this.ttl;\n list.add(thz);\n }\n }\n \n public int countUnexpiredTokens(int currentTime) {\n ListIterator<Token> lit = list.listIterator();\n while(lit.hasNext()){\n Token thz = lit.next();\n if(thz.exp <= currentTime){\n lit.remove();\n }\n }\n return list.size();\n }\n}\n\n/**\n * Your AuthenticationManager object will be instantiated and called as such:\n * AuthenticationManager obj = new AuthenticationManager(timeToLive);\n * obj.generate(tokenId,currentTime);\n * obj.renew(tokenId,currentTime);\n * int param_3 = obj.countUnexpiredTokens(currentTime);\n */\n``` | 0 | 0 | ['Java'] | 0 |
design-authentication-manager | Python Doubly Linked List + Hashmap Solution beats 99% | python-doubly-linked-list-hashmap-soluti-rkxi | We use a doubly linked list in order to efficiently clear tokens that have expired. We use a hashmap in order to efficiently handle the "renew" operation, so th | noahantisseril | NORMAL | 2024-12-15T02:22:55.407406+00:00 | 2024-12-15T02:22:55.407406+00:00 | 7 | false | We use a doubly linked list in order to efficiently clear tokens that have expired. We use a hashmap in order to efficiently handle the "renew" operation, so that we don\'t have to check our entire list in order to find the token we want to renew.\n\n# Code\n```python3 []\nclass Node:\n def __init__(self, token, time):\n self.token = token\n self.time = time\n self.prev = None\n self.next = None\n\nclass AuthenticationManager:\n\n def __init__(self, timeToLive: int):\n self.manager = {}\n self.time_to_live = timeToLive\n self.head = Node("head", 0)\n self.tail = Node("tail", 0)\n self.head.next = self.tail\n self.tail.prev = self.head\n \n def add(self, node):\n prev_node = self.tail.prev\n prev_node.next = node\n node.next = self.tail\n node.prev = prev_node\n self.tail.prev = node\n self.manager[node.token] = node\n \n def remove(self, node):\n prev_node = node.prev\n next_node = node.next\n prev_node.next = next_node\n next_node.prev = prev_node\n del self.manager[node.token]\n \n def clear_expired(self, time):\n curr = self.head.next\n while curr != self.tail and curr.time <= time:\n next_node = curr.next\n self.remove(curr)\n curr = next_node\n\n def generate(self, tokenId: str, currentTime: int) -> None:\n self.clear_expired(currentTime)\n node = Node(tokenId, currentTime + self.time_to_live)\n self.add(node)\n\n def renew(self, tokenId: str, currentTime: int) -> None:\n self.clear_expired(currentTime)\n if tokenId not in self.manager:\n return\n self.remove(self.manager[tokenId])\n node = Node(tokenId, currentTime + self.time_to_live)\n self.add(node)\n\n def countUnexpiredTokens(self, currentTime: int) -> int:\n self.clear_expired(currentTime)\n return len(self.manager)\n``` | 0 | 0 | ['Python3'] | 0 |
design-authentication-manager | Java Easy Solution for dummies (One HashMap) | java-easy-solution-for-dummies-one-hashm-llrj | \n# Code\njava []\nclass AuthenticationManager {\n class Token{\n String id;\n int startTime;\n int expiryTime;\n public Token(St | coder11 | NORMAL | 2024-12-06T21:02:05.004327+00:00 | 2024-12-06T21:02:05.004350+00:00 | 4 | false | \n# Code\n```java []\nclass AuthenticationManager {\n class Token{\n String id;\n int startTime;\n int expiryTime;\n public Token(String id,int st, int ex)\n {\n this.id=id;\n this.startTime=st;\n this.expiryTime=ex;\n }\n }\n\n int ttl;\n Map<String,Token> tokens;\n\n public AuthenticationManager(int timeToLive) {\n this.ttl=timeToLive;\n this.tokens = new HashMap();\n }\n \n public void generate(String tokenId, int currentTime) {\n Token t = new Token(tokenId, currentTime, currentTime+ttl);\n tokens.put(tokenId,t);\n }\n \n public void renew(String tokenId, int currentTime) {\n if(tokens.containsKey(tokenId))\n {\n if(!isExpired(tokens,tokenId, currentTime))\n {\n Token currToken = tokens.get(tokenId);\n currToken.expiryTime = currentTime + ttl;\n tokens.put(tokenId,currToken);\n }\n }\n }\n public int countUnexpiredTokens(int currentTime) {\n int res=0;\n Iterator<Map.Entry<String,Token>> iterator = tokens.entrySet().iterator();\n while(iterator.hasNext())\n {\n Map.Entry<String, Token> entry = iterator.next();\n Token curr = entry.getValue();\n if(currentTime > curr.startTime && currentTime < curr.expiryTime)\n {\n res++;\n }\n else\n {\n iterator.remove();\n }\n }\n return res;\n }\n\n private boolean isExpired(Map<String,Token> tokens, String tokenId, int currTime)\n {\n Token token = tokens.get(tokenId);\n if(currTime>=token.expiryTime)\n {\n return true;\n }\n return false;\n }\n}\n\n/**\n * Your AuthenticationManager object will be instantiated and called as such:\n * AuthenticationManager obj = new AuthenticationManager(timeToLive);\n * obj.generate(tokenId,currentTime);\n * obj.renew(tokenId,currentTime);\n * int param_3 = obj.countUnexpiredTokens(currentTime);\n */\n``` | 0 | 0 | ['Java'] | 0 |
design-authentication-manager | Solution using One HashMap | solution-using-one-hashmap-by-coder11-l5xp | \n\n# Code\njava []\nclass AuthenticationManager {\n class Token{\n String id;\n int startTime;\n int expiryTime;\n public Token( | coder11 | NORMAL | 2024-12-06T20:27:24.272280+00:00 | 2024-12-06T20:27:24.272316+00:00 | 2 | false | \n\n# Code\n```java []\nclass AuthenticationManager {\n class Token{\n String id;\n int startTime;\n int expiryTime;\n public Token(String id,int st, int ex)\n {\n this.id=id;\n this.startTime=st;\n this.expiryTime=ex;\n }\n }\n int ttl;\n Map<String,Token> tokens;\n public AuthenticationManager(int timeToLive) {\n this.ttl=timeToLive;\n this.tokens = new HashMap();\n }\n \n public void generate(String tokenId, int currentTime) {\n Token t = new Token(tokenId, currentTime, currentTime+ttl);\n tokens.put(tokenId,t);\n }\n \n public void renew(String tokenId, int currentTime) {\n if(tokens.containsKey(tokenId))\n {\n if(!isExpired(tokens,tokenId, currentTime))\n {\n Token currToken = tokens.get(tokenId);\n currToken.expiryTime = currentTime + ttl;\n tokens.put(tokenId,currToken);\n }\n }\n }\n \n public int countUnexpiredTokens(int currentTime) {\n int res=0;\n for(Map.Entry<String,Token> map:tokens.entrySet())\n {\n Token curr = tokens.get(map.getKey());\n if(currentTime > curr.startTime && currentTime< curr.expiryTime)\n {\n res++;\n }\n }\n return res;\n }\n\n private boolean isExpired(Map<String,Token> tokens, String tokenId, int currTime)\n {\n Token token = tokens.get(tokenId);\n if(currTime>=token.expiryTime)\n {\n return true;\n }\n return false;\n }\n}\n\n/**\n * Your AuthenticationManager object will be instantiated and called as such:\n * AuthenticationManager obj = new AuthenticationManager(timeToLive);\n * obj.generate(tokenId,currentTime);\n * obj.renew(tokenId,currentTime);\n * int param_3 = obj.countUnexpiredTokens(currentTime);\n */\n``` | 0 | 0 | ['Java'] | 0 |
design-authentication-manager | Easy using Maps. | easy-using-maps-by-starc_208-svng | \n\n# Complexity\n- Time complexity:\n Renew - O(1), Generate - O(1), countUnexpiredTokens - O(n). Where n is the number of tokens\n\n- Space complexity:\n O( | starc_208 | NORMAL | 2024-11-30T17:50:04.747001+00:00 | 2024-11-30T17:50:04.747027+00:00 | 3 | false | \n\n# Complexity\n- Time complexity:\n Renew - O(1), Generate - O(1), countUnexpiredTokens - O(n). Where n is the number of tokens\n\n- Space complexity:\n O(n)\n\n# Code\n```javascript []\n/**\n * @param {number} timeToLive\n */\nvar AuthenticationManager = function(timeToLive) {\n this.map = new Map();\n this.ttl = timeToLive;\n};\n\n/** \n * @param {string} tokenId \n * @param {number} currentTime\n * @return {void}\n */\nAuthenticationManager.prototype.generate = function(tokenId, currentTime) {\n this.map.set(tokenId,currentTime);\n};\n\n/** \n * @param {string} tokenId \n * @param {number} currentTime\n * @return {void}\n */\nAuthenticationManager.prototype.renew = function(tokenId, currentTime) {\n let time = this.map.get(tokenId);\n if(time && time+this.ttl>currentTime){\n this.map.set(tokenId, currentTime);\n }\n};\n\n/** \n * @param {number} currentTime\n * @return {number}\n */\nAuthenticationManager.prototype.countUnexpiredTokens = function(currentTime) {\n let cnt = 0;\n for(const [key,value] of this.map){\n if(value + this.ttl > currentTime){\n cnt++;\n }\n }\n return cnt;\n};\n\n/** \n * Your AuthenticationManager object will be instantiated and called as such:\n * var obj = new AuthenticationManager(timeToLive)\n * obj.generate(tokenId,currentTime)\n * obj.renew(tokenId,currentTime)\n * var param_3 = obj.countUnexpiredTokens(currentTime)\n */\n``` | 0 | 0 | ['JavaScript'] | 0 |
design-authentication-manager | C# simple solution using Dictionary only | c-simple-solution-using-dictionary-only-emfn2 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Pradeep_2000 | NORMAL | 2024-11-21T10:26:26.025842+00:00 | 2024-11-21T10:28:49.227981+00:00 | 12 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: 36 ms, Beats 73.68%\n\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: Memory 125.26 MB, Beats 55.56%\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```csharp []\npublic class AuthenticationManager {\n\n public Dictionary<string,int> tokens;\n public int expiry=0;\n\n public AuthenticationManager(int timeToLive) {\n tokens = new Dictionary<string,int>();\n expiry = timeToLive;\n }\n \n public void Generate(string tokenId, int currentTime) {\n tokens.Add(tokenId, expiry+currentTime);\n }\n \n public void Renew(string tokenId, int currentTime) {\n if(tokens.ContainsKey(tokenId)){\n if(tokens[tokenId] > currentTime){\n tokens[tokenId] = expiry + currentTime;\n }\n }\n }\n \n public int CountUnexpiredTokens(int currentTime) {\n int unexptokens=0;\n foreach(var elm in tokens){\n if(elm.Value > currentTime){\n unexptokens++;\n }\n }\n return unexptokens;\n }\n}\n\n/**\n * Your AuthenticationManager object will be instantiated and called as such:\n * AuthenticationManager obj = new AuthenticationManager(timeToLive);\n * obj.Generate(tokenId,currentTime);\n * obj.Renew(tokenId,currentTime);\n * int param_3 = obj.CountUnexpiredTokens(currentTime);\n */\n``` | 0 | 0 | ['C#'] | 0 |
design-authentication-manager | C# Hashing + Sliding window | c-hashing-sliding-window-by-junkmann-ho8c | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n- _token keeps a log of generated tokens (new or renewed).\n- After Adjus | junkmann | NORMAL | 2024-11-13T17:41:13.629763+00:00 | 2024-11-13T17:41:13.629787+00:00 | 7 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- `_token` keeps a log of generated tokens (new or renewed).\n- After `AdjustWindow` is called at time `t`, `_windowIndex` points to the oldest unexpired token at time `t`.\n- After `AdjustWindow` is called at time `t`, `_window` contains all unexpired tokens at time `t`. The reason we need this is twofold. First, it allows to tell in O(1) if a token can be renewed. Second, it allows to answer `CountUnexpiredTokens`. `CountUnexpiredTokens` cannot simply be answered with `_tokens.Count - _windowIndex`, because tokens can be renewed and the same token may appear multiple times in this window.\n\n# Complexity\n- Time complexity:\n`Generate`: $$O(1)$$\n`Renew`: $$O(1)$$\n`CountUnexpiredTokens`: $$O(1)$$ amortized on `Generate` and `Renew`\n\n- Space complexity:\n$$O(n)$$ where `n` is the sum of the number of newly generated tokens and succesfully renewed tokens.\n\n# Code\n```csharp []\npublic class AuthenticationManager {\n private int _ttl;\n private List<(string Token, int Time)> _tokens;\n private Dictionary<string, int> _window;\n private int _windowIndex;\n\n public AuthenticationManager(int timeToLive) {\n _ttl = timeToLive;\n _tokens = new();\n _window = new();\n _windowIndex = 0;\n }\n \n public void Generate(string tokenId, int currentTime) {\n _tokens.Add((tokenId, currentTime));\n _window[tokenId] = _window.GetValueOrDefault(tokenId) + 1;\n }\n \n public void Renew(string tokenId, int currentTime) {\n AdjustWindow(currentTime);\n\n if (_window.ContainsKey(tokenId)) {\n Generate(tokenId, currentTime);\n }\n }\n \n public int CountUnexpiredTokens(int currentTime) {\n AdjustWindow(currentTime);\n return _window.Count;\n }\n\n private void AdjustWindow(int currentTime) {\n for (; _windowIndex < _tokens.Count && _tokens[_windowIndex].Time <= currentTime - _ttl; ++_windowIndex) {\n var cnt = _window[_tokens[_windowIndex].Token];\n\n if (cnt == 1) {\n _window.Remove(_tokens[_windowIndex].Token);\n } else {\n _window[_tokens[_windowIndex].Token] = cnt - 1;\n }\n }\n }\n}\n\n/**\n * Your AuthenticationManager object will be instantiated and called as such:\n * AuthenticationManager obj = new AuthenticationManager(timeToLive);\n * obj.Generate(tokenId,currentTime);\n * obj.Renew(tokenId,currentTime);\n * int param_3 = obj.CountUnexpiredTokens(currentTime);\n */\n``` | 0 | 0 | ['C#'] | 0 |
design-authentication-manager | Java solution with map | java-solution-with-map-by-milkbiki-jo1w | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | milkbiki | NORMAL | 2024-11-08T03:24:06.352360+00:00 | 2024-11-08T03:24:06.352388+00:00 | 0 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass AuthenticationManager {\n\n Map<String, Integer> m;\n int ttl;\n public AuthenticationManager(int timeToLive) {\n this.m = new HashMap();\n this.ttl = timeToLive;\n }\n\n private void expire(int currentTime) {\n List<String> expiredTokens = new ArrayList();\n for (Map.Entry e : m.entrySet()) {\n String token = (String) e.getKey();\n int expiryTime = (int) e.getValue();\n if (expiryTime <= currentTime) {\n expiredTokens.add(token);\n }\n }\n\n for (String token : expiredTokens) {\n m.remove(token);\n }\n }\n \n public void generate(String tokenId, int currentTime) {\n expire(currentTime);\n m.put(tokenId, currentTime + ttl);\n }\n \n public void renew(String tokenId, int currentTime) {\n expire(currentTime);\n if (!m.containsKey(tokenId)) {\n //System.out.println("Nothing to do");\n return;\n }\n m.put(tokenId, currentTime + ttl);\n }\n \n public int countUnexpiredTokens(int currentTime) {\n expire(currentTime);\n return m.size();\n }\n}\n\n/**\n * Your AuthenticationManager object will be instantiated and called as such:\n * AuthenticationManager obj = new AuthenticationManager(timeToLive);\n * obj.generate(tokenId,currentTime);\n * obj.renew(tokenId,currentTime);\n * int param_3 = obj.countUnexpiredTokens(currentTime);\n */\n\n /*\ncreate a map of token and expiry time\n\nexpire(time t):\ngo through the map and expire all those tokens that have a expiry => t \ncreate a list of expired tokens, then remove them from the map\n\nrenew(token, time):\ninvoke expire\nif token does not exist or is expired then do nothing\nif token exists then set expiry to time + ttl\n\ncountUnexpired(time t) \ninvoke expire\nreturn map.size\n\n\n\n */\n``` | 0 | 0 | ['Java'] | 0 |
design-authentication-manager | Hold values with an expiry time in dictionary | hold-values-with-an-expiry-time-in-dicti-jybv | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | belka | NORMAL | 2024-11-04T19:54:50.982635+00:00 | 2024-11-04T19:54:50.982670+00:00 | 1 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass AuthenticationManager:\n\n def __init__(self, timeToLive: int):\n self.t=timeToLive\n self.d={}\n\n def generate(self, tokenId: str, currentTime: int) -> None:\n self.d[tokenId]=currentTime+self.t\n\n def renew(self, tokenId: str, currentTime: int) -> None:\n if tokenId in self.d and self.d[tokenId]>currentTime:\n self.d[tokenId]=currentTime+self.t\n\n def countUnexpiredTokens(self, currentTime: int) -> int:\n cnt=0\n for t in self.d.values():\n if t>currentTime: cnt+=1\n return cnt\n\n\n# Your AuthenticationManager object will be instantiated and called as such:\n# obj = AuthenticationManager(timeToLive)\n# obj.generate(tokenId,currentTime)\n# obj.renew(tokenId,currentTime)\n# param_3 = obj.countUnexpiredTokens(currentTime)\n``` | 0 | 0 | ['Python3'] | 0 |
design-authentication-manager | Simple AuthenticationManager in Swift. Prune tokens only on renewal / counting | simple-authenticationmanager-in-swift-pr-dbc0 | \n\n# Code\nswift []\n\nclass AuthenticationManager {\n let timeToLive: Int\n var tokens = [String: Int]()\n\n init(_ timeToLive: Int) {\n self. | arrb | NORMAL | 2024-10-30T22:04:29.016074+00:00 | 2024-10-30T22:04:29.016095+00:00 | 4 | false | \n\n# Code\n```swift []\n\nclass AuthenticationManager {\n let timeToLive: Int\n var tokens = [String: Int]()\n\n init(_ timeToLive: Int) {\n self.timeToLive = timeToLive\n }\n \n func generate(_ tokenId: String, _ currentTime: Int) {\n tokens[tokenId] = currentTime + self.timeToLive\n }\n \n func renew(_ tokenId: String, _ currentTime: Int) {\n pruneExpiredTokens(currentTime)\n guard tokens[tokenId] != nil else { return }\n tokens[tokenId] = currentTime + self.timeToLive\n }\n \n func countUnexpiredTokens(_ currentTime: Int) -> Int {\n pruneExpiredTokens(currentTime)\n return tokens.count\n }\n\n private func pruneExpiredTokens(_ currentTime: Int) {\n tokens = tokens.filter { $0.value > currentTime } \n }\n}\n\n/**\n * Your AuthenticationManager object will be instantiated and called as such:\n * let obj = AuthenticationManager(timeToLive)\n * obj.generate(tokenId, currentTime)\n * obj.renew(tokenId, currentTime)\n * let ret_3: Int = obj.countUnexpiredTokens(currentTime)\n */\n``` | 0 | 0 | ['Swift'] | 0 |
maximum-xor-for-each-query | Easiest Solution | Beats 100% ✅| C++ | Java | Python3 | Javascript | easiest-solution-beats-100-c-java-python-m68i | \n\n\n# Intuition\n1) Understanding the Goal:\n\n- We are given a sorted array nums and an integer maximumBit.\n- We need to find an integer k less than 2^maxim | jay_singla | NORMAL | 2024-11-08T00:38:21.989047+00:00 | 2024-11-08T00:38:21.989079+00:00 | 23,831 | false | \n\n\n# Intuition\n1) Understanding the Goal:\n\n- We are given a sorted array nums and an integer maximumBit.\n- We need to find an integer k less than 2^maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized for each query.\n- After each query, we remove the last element of nums and repeat the process on the remaining array.\n2) Maximizing XOR:\n\n- To maximize XOR, we need to find a k that when XORed with the cumulative XOR of the array (xor_all) gives the largest possible result.\n- The largest value that can be formed with maximumBit bits is 2^maximumBit - 1. This number has all bits set to 1 up to maximumBit (e.g., for maximumBit = 3, the number is 111 in binary or 7 in decimal).\n- Thus, to maximize the XOR with xor_all, we can choose k to be xor_all XOR maxXor, where maxXor = 2^maximumBit - 1.\n3) Reversing from the End:\n\n- We\u2019re asked to start from the entire array and progressively "remove" the last element for each query.\n- For each query, we adjust xor_all by XORing out the last element in the current nums array, giving us an updated cumulative XOR for the remaining array.\n\n# Approach\n1) Initialize Cumulative XOR and Maximum XOR Mask:\n\n- First, compute the cumulative XOR of all elements in nums (xor_all). This gives us the XOR of the full array.\n- Calculate maxXor = 2^maximumBit - 1, which will be used to find k in each query.\n2) Compute Each Answer Iteratively:\n\n- For each query, find k by computing xor_all XOR maxXor, which maximizes the XOR value for the current cumulative XOR.\n- Append this result to ans.\n- Update xor_all by XORing out the last element of the current nums, effectively removing it from consideration for the next query.\n3) Return the Result in Reverse Order:\n\n- Since we processed the array from end to beginning, ans will contain the answers in reverse order, so we return it as-is.\n\n# Complexity\n- Time complexity:\nO(n) -> to traverse the array\n\n- Space complexity:\nO(1) but O(n) if we consider space used for answer.\n\n# Code\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size(),xorr = nums[0],maxxorr = pow(2,maximumBit)-1;\n for(int i=1;i<n;i++)xorr ^= nums[i];\n vector<int>ans(n);\n for(int i=0;i<n;i++){\n ans[i] = xorr^maxxorr;\n xorr ^= nums[n-1-i];\n }\n return ans;\n }\n};\n```\n```Java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length;\n int xorr = nums[0];\n int maxXor = (1 << maximumBit) - 1;\n\n for (int i = 1; i < n; i++) {\n xorr ^= nums[i];\n }\n\n int[] ans = new int[n];\n for (int i = 0; i < n; i++) {\n ans[i] = xorr ^ maxXor;\n xorr ^= nums[n - 1 - i];\n }\n\n return ans;\n }\n}\n```\n```Python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n n = len(nums)\n xorr = nums[0]\n max_xor = (1 << maximumBit) - 1\n \n for i in range(1, n):\n xorr ^= nums[i]\n \n ans = []\n for i in range(n):\n ans.append(xorr ^ max_xor)\n xorr ^= nums[n - 1 - i]\n \n return ans\n\n \n```\n```Javascript []\n/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function(nums, maximumBit) {\n const n = nums.length;\n let xorr = nums[0];\n const maxXor = (1 << maximumBit) - 1;\n\n for (let i = 1; i < n; i++) {\n xorr ^= nums[i];\n }\n\n const ans = [];\n for (let i = 0; i < n; i++) {\n ans.push(xorr ^ maxXor);\n xorr ^= nums[n - 1 - i];\n }\n\n return ans;\n};\n```\n\n | 128 | 1 | ['Array', 'Bit Manipulation', 'C++', 'Java', 'Python3', 'JavaScript'] | 12 |
maximum-xor-for-each-query | ✅ Easy O(N) Solution w/ Explanation | Max XOR = 2^maximumBit - 1 | easy-on-solution-w-explanation-max-xor-2-upyg | \u2714\uFE0F Solution\n\nWe need to realize that maximum XOR for a query can always be made equal to 2^maximumBit - 1. For this it\'s also necessary to notice t | archit91 | NORMAL | 2021-04-17T16:01:10.341233+00:00 | 2021-04-18T15:15:33.717864+00:00 | 7,445 | false | \u2714\uFE0F ***Solution***\n\nWe need to realize that maximum XOR for a query can always be made equal to `2^maximumBit - 1`. For this it\'s also necessary to notice this constraint -\n\n* `0 <= nums[i] < 2^maximumBit`\n\n***How is maximum always `2^maximumBit - 1` ?***\n\nIt is given that all elements of the `nums` are in the range `[0, 2^maximumBit - 1]`. And, after xor-ing the whole `nums` array, we will always have a final value (lets call it `XOR`) in between `[0, 2^maximumBit - 1]`.\n\nNow, we can xor this value with any number from the range `[0, 2^maximumBit - 1]`. To maximize it, we can always choose a value from that range which when xor-ed with `XOR`(above obtained value), will maximize it to `2^maximumBit - 1`.\n\n```\nSome Examples:\nHere, XOR will denote the xor of all elements of original nums array.\n\nLet maximumBit = 3.\nOur range will be from [0, 7]. Now, if - \n\n1. XOR = 1 (001) : we can choose to xor it with 6 so that maximum xor for query will become \'2^maximumBit - 1 = 7\'\n\t ^ 6 (110)\n\t ----------\n\t\t 111\n\t\t\t\n2. XOR = 2 (010) : we can choose to xor it with 5 so that maximum xor for query will become \'2^maximumBit - 1 = 7\'\n\t ^ 5 (101)\n\t ----------\n\t\t 111\n\n3. XOR = 3 (011) : we can choose to xor it with 4 \n\t ^ 4 (100)\n\t ----------\n\t\t 111\n\t\t\t\n4. XOR = 4 (100) : we can choose to xor it with 3\n\t ^ 3 (011)\n\t ----------\n\t\t 111\t\n\t\t\t\nAnd so on for any other cases, we can choose a value to maximize xor to \'2^maximumBit-1\'\n```\n\nAlso, after we first calculate the XOR for original `nums` array, the only thing that changes in subsequent queries is that the last element of `nums` is removed. \n\nSo, we can just use the previous **`XOR`** and xor it with element which is to be removed. The element to be removed will be xor-ed twice and cancel out each other and we will get the final answer of that query.\n**C++**\n```\nvector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n\tint n = size(nums), XOR = 0;\n\tvector<int> ans(n);\n\tfor(auto& num : nums) XOR ^= num;\n\tans[0] = XOR ^= (1 << maximumBit) - 1; // 2^maximumBit - 1\n\tfor(int i = 1; i < n; i++)\n\t\tans[i] = XOR ^= nums[n - i]; \n\treturn ans;\n}\n```\n\n---\n\n**Python**\n\nThe same idea in python with slightly different implementation approach -\n\n```\ndef getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n\tmaxXOR, arrayXOR, ans = (1 << maximumBit) - 1, reduce(ixor, nums), []\n\tfor i in range(len(nums) - 1, -1, -1):\n\t\tans.append(arrayXOR ^ maxXOR)\n\t\tarrayXOR ^= nums[i]\n\treturn ans\n```\n\n***Time Complexity :*** **`O(N)`**\n\n---\n--- | 123 | 12 | ['C'] | 19 |
maximum-xor-for-each-query | C++/Java One-Pass | cjava-one-pass-by-votrubac-4hgv | The maximum value we can possibly get is (1 << maximumBit) - 1. That is, all maximumBit bits are set.\n\nFor the first i elements, k would be equal to ((1 << m | votrubac | NORMAL | 2021-04-17T16:07:51.977601+00:00 | 2021-04-17T16:29:26.952107+00:00 | 4,583 | false | The maximum value we can possibly get is `(1 << maximumBit) - 1`. That is, all `maximumBit` bits are set.\n\nFor the first `i` elements, `k` would be equal to `((1 << maximumBit) - 1) ^ n[0] ^ ... ^ n[i - 1]`.\n\nWe can start from the first element, and the result is `((1 << maximumBit) - 1) ^ n[0]`. Then we can just do XOR with the next element and record the result.\n\n> Note that we should record our result in the reverse order.\n\n**C++**\n```cpp\nvector<int> getMaximumXor(vector<int>& n, int maximumBit) {\n vector<int> res(n.size());\n int val = (1 << maximumBit) - 1;\n for (int i = 0; i < n.size(); ++i)\n res[n.size() - i - 1] = val ^= n[i];\n return res;\n}\n```\n**Java**\n```java\npublic int[] getMaximumXor(int[] n, int maximumBit) {\n int[] res = new int[n.length];\n int val = (1 << maximumBit) - 1;\n for (int i = 0; i < n.length; ++i)\n res[n.length - i - 1] = val ^= n[i]; \n return res;\n}\n``` | 67 | 9 | [] | 9 |
maximum-xor-for-each-query | [JAVA] O(N) Very Detailed Explanation For Starters | java-on-very-detailed-explanation-for-st-mj79 | Basic XOR Operations\n\nx ^ x = 0 0 ^ x = x ----------- (1)\n111111 ^ 101010 = 010101 (flip all bits) \n\nlet xorVal = a ^ b ^ c if we w | Zudas | NORMAL | 2021-04-17T16:04:03.743565+00:00 | 2021-04-18T14:35:13.107805+00:00 | 2,104 | false | <b>Basic XOR Operations</b>\n```\nx ^ x = 0 0 ^ x = x ----------- (1)\n111111 ^ 101010 = 010101 (flip all bits) \n\nlet xorVal = a ^ b ^ c if we want to remove any number just again xor it with xorVal\nExample We want to remove c\nxorVal ^ c = a ^ b ^ c ^ c (xoring both sides with c)\nxorVal ^ c = a ^ b (using (1) c ^ c = 0 and a ^ b ^ 0 = a ^ b)\nwe have removed c\n```\nLet `xor` be the <b>xor</b> of all elements. We want to maximize `xor ^ k`.\nLet\'s say if we have `maximumBit` = `3 `and the number is `5` => `101` \nnow to get a <b>maximum</b> after xoring `5` with another number. It needs to be have <b>opposite</b> bits at the corresponding place to get all set bits. `101` ----opposite---> `010`\nAnd we can find it by doing \n`111 ^ 101 = 010 => 7 ^ 5 = 2 (k)`\nAnd `5 ^ 2 = 7` (max result)\nBy applying the same we can find `k` by doing `xor ^ ((1 << maximumBit) - 1) `\n```((1 << maximumBit) - 1) = Math.pow(2, maximumBit) - 1```\nFor example `maximumBit = 3 => Math.pow(2, 3) - 1 = 7 => 111(all set bits)`\n\nIn simple words we are finding `k` such that \n`xor ^ k = max `(let max be a number with <b>maximum</b> number of allowed set bits) \n`xor ^ xor ^ k = max ^ xor` (xoring both sides with `xor`)\n`k = max ^ xor`\n\n```\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int xor = 0, n = nums.length;\n int max = (1 << maximumBit) - 1;\n int res[] = new int[n];\n \n for(int e : nums) xor ^= e;//xor all elements\n \n for(int i = 0; i < n; i++){\n res[i] = xor ^ max;// k = xor ^ max\n xor ^= nums[n - i - 1];//remove last element from xor\n }\n return res;\n }\n}\n\nExample\n[0, 1, 1, 3] and maximumBit = 2 \nxor of all elements = 3 and max = 4 -1 = 3(111)\n\nwhen i = 0 => res[i] = 3 ^ 3 = 0 and xor = 3 ^ 3 = 0\nwhen i = 1 => res[i] = 0 ^ 3 = 3 and xor = 0 ^ 1 = 1\nwhen i = 2 => res[i] = 1 ^ 3 = 2 and xor = 1 ^ 1 = 0\nwhen i = 3 => res[i] = 0 ^ 3 = 3 and xor = 0 ^ 0 = 0\n```\n<b>TC:</b> `O(N)` `N `is the length of the array\n<b>SC:</b> `O(N)` | 52 | 1 | ['Bit Manipulation', 'Java'] | 5 |
maximum-xor-for-each-query | [Java/Python 3] 1 pass bit manipulation O(n) prefix xor code w/ explanation and analysis. | javapython-3-1-pass-bit-manipulation-on-6qkar | Tips: For any integer x, 0 ^ x = x and x ^ x = 0\n\n----\n\n1. The maximum value less than 2 ^ maximumBit is 2 ^ maximumBit - 1, which is (1 << maximumBit) - 1; | rock | NORMAL | 2021-04-17T16:11:01.497002+00:00 | 2024-11-08T09:36:19.651985+00:00 | 1,259 | false | **Tips**: For any integer `x`, `0 ^ x = x` and `x ^ x = 0`\n\n----\n\n1. The maximum value less than `2 ^ maximumBit` is `2 ^ maximumBit - 1`, which is `(1 << maximumBit) - 1`;\n2. Use `mx` to represent the afore-mentioned maximum value; For a given value `xor`, if \n```\nxor ^ k = mx\n```\nthen \n```\nxor ^ xor ^ k = xor ^ mx\n``` \nthat is,\n```\n0 ^ k = xor ^ mx\n```\nwe can conclude\n```\nk = xor ^ mx\n```\n.\n```java\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length, i = n - 1, mx = (1 << maximumBit) - 1, xor = 0;\n int[] ans = new int[n];\n for (int num : nums) {\n xor ^= num;\n ans[i--] = xor ^ mx;\n }\n return ans;\n }\n```\n```python\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n ans, mx, xor = [], (1 << maximumBit) - 1, 0\n for num in nums:\n xor ^= num\n ans.append(xor ^ mx)\n return ans[:: -1]\n```\n**Analysis:**\nTime & space: `O(n)`, where `n = nums.length`. | 26 | 2 | [] | 8 |
maximum-xor-for-each-query | ✅Beats 100% | Very Short & Simple Solution | beats-100-very-short-simple-solution-by-74750 | \npython3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n mask = (1 << maximumBit) - 1\n n = le | Piotr_Maminski | NORMAL | 2024-11-08T00:21:11.635231+00:00 | 2024-11-08T17:32:53.304514+00:00 | 5,368 | false | \n```python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n mask = (1 << maximumBit) - 1\n n = len(nums)\n res = [0] * n\n curr = 0\n \n for i in range(n):\n curr ^= nums[i]\n res[n-i-1] = ~curr & mask\n \n return res\n\n```\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int mask = (1LL << maximumBit) - 1, n = nums.size();\n vector<int> res(n);\n for(int i = 0, curr = 0; i < n; i++) {\n curr ^= nums[i];\n res[n-i-1] = ~curr & mask;\n }\n return res;\n }\n};\n```\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int mask = (1 << maximumBit) - 1;\n int n = nums.length;\n int[] res = new int[n];\n int curr = 0;\n \n for(int i = 0; i < n; i++) {\n curr ^= nums[i];\n res[n-i-1] = ~curr & mask;\n }\n \n return res;\n }\n}\n\n```\n```csharp []\npublic class Solution {\n public int[] GetMaximumXor(int[] nums, int maximumBit) {\n int mask = (1 << maximumBit) - 1;\n int n = nums.Length;\n int[] res = new int[n];\n int curr = 0;\n \n for(int i = 0; i < n; i++) {\n curr ^= nums[i];\n res[n-i-1] = ~curr & mask;\n }\n \n return res;\n }\n}\n\n```\n```golang []\nfunc getMaximumXor(nums []int, maximumBit int) []int {\n mask := (1 << maximumBit) - 1\n n := len(nums)\n res := make([]int, n)\n curr := 0\n \n for i := 0; i < n; i++ {\n curr ^= nums[i]\n res[n-i-1] = ^curr & mask\n }\n \n return res\n}\n\n```\n```swift []\nclass Solution {\n func getMaximumXor(_ nums: [Int], _ maximumBit: Int) -> [Int] {\n let mask = (1 << maximumBit) - 1\n let n = nums.count\n var res = Array(repeating: 0, count: n)\n var curr = 0\n \n for i in 0..<n {\n curr ^= nums[i]\n res[n-i-1] = ~curr & mask\n }\n \n return res\n }\n}\n\n```\n```javascript [JS]\n// JavaScript\n\nvar getMaximumXor = function(nums, maximumBit) {\n const mask = (1 << maximumBit) - 1;\n const n = nums.length;\n const res = new Array(n);\n let curr = 0;\n \n for(let i = 0; i < n; i++) {\n curr ^= nums[i];\n res[n-i-1] = (~curr & mask);\n }\n \n return res;\n};\n```\n```typescript [TS]\n// TypeScript\n\nfunction getMaximumXor(nums: number[], maximumBit: number): number[] {\n const mask: number = (1 << maximumBit) - 1;\n const n: number = nums.length;\n const res: number[] = new Array(n);\n let curr: number = 0;\n \n for(let i = 0; i < n; i++) {\n curr ^= nums[i];\n res[n-i-1] = (~curr & mask);\n }\n \n return res;\n}\n```\n```rust []\nimpl Solution {\n pub fn get_maximum_xor(nums: Vec<i32>, maximum_bit: i32) -> Vec<i32> {\n let mask = (1 << maximum_bit) - 1;\n let n = nums.len();\n let mut res = vec![0; n];\n let mut curr = 0;\n \n for i in 0..n {\n curr ^= nums[i];\n res[n-i-1] = !curr & mask;\n }\n \n res\n }\n}\n\n```\n```ruby []\ndef get_maximum_xor(nums, maximum_bit)\n mask = (1 << maximum_bit) - 1\n n = nums.length\n res = Array.new(n)\n curr = 0\n \n n.times do |i|\n curr ^= nums[i]\n res[n-i-1] = ~curr & mask\n end\n \n res\nend\n```\n### Complexity\n- Time complexity: O(n)\n\n\n- Space complexity: O(n)\n\n\n\n# Explanation\n---\n\n## Intuition\nFind the maximum XOR value for each prefix of an array by using a bit manipulation approach. \n\n\n## Aproach\n1. First, it creates a mask of 1s based on maximumBit. For example, if maximumBit is 3, mask becomes 7 (111 in binary)\n\n2. It creates a result array (res) of the same length as the input array\n\n3. Then it goes through each number in the array and:\n - Keeps track of running XOR in \'curr\'\n - For each position, finds a number that would give maximum XOR with current running XOR\n - Stores results in reverse order (that\'s why we use n-i-1)\n4. The magic happens with ~curr & mask:\n\n - ~curr flips all bits\n - mask keeps only the relevant bits (up to maximumBit)\n - This combination gives us the number that produces maximum XOR\n\nFor example:\n\n- If curr = 101 and maximumBit = 3\n- ~curr = 010 (after masking with 111)\n- This 010 will give maximum XOR with 101\n\nWorks bcs uses the fact that to get maximum XOR, you want opposite bits wherever possible within your bit limit.\n\nSimple\n\n---\n\n# Most common Array Interview Problems\nList based on my research (interviews, reddit, github) Probably not 100% accurate but very close to my recruitments. Leetcode Premium is probably more accurate [I don\'t have]\n\n\n**Easy:** [[1. Two Sum]](https://leetcode.com/problems/two-sum/solutions/5999466/beats-100-explained-step-by-step-list-most-common-array-inverview) **(common warm-up)** [[121. Best Time to Buy and Sell Stock]](https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description/?envType=problem-list-v2&envId=atbyk0mj) [[88. Merge Sorted Array]](https://leetcode.com/problems/merge-sorted-array/description/) [[27. Remove Element]](https://leetcode.com/problems/remove-element/description/) [[929. Unique Email Addresses]](https://leetcode.com/problems/unique-email-addresses/description/) [[482. License Key Formatting]](https://leetcode.com/problems/license-key-formatting/description/)\n\n\n**Medium/Hard:** [[15. 3Sum]](https://leetcode.com/problems/3sum/description/) [[42. Trapping Rain Water]](https://leetcode.com/problems/trapping-rain-water/?envType=problem-list-v2&envId=atbyk0mj) [[56. Merge Intervals]](https://leetcode.com/problems/merge-intervals/description/) [[57. Insert Interval]](https://leetcode.com/problems/insert-interval/description/) [[46. Permutations]](https://leetcode.com/problems/permutations/description/) [[53. Maximum Subarray]](https://leetcode.com/problems/maximum-subarray/description/) [[73. Set Matrix Zeroes]](https://leetcode.com/problems/set-matrix-zeroes/description/) [[284. Peeking Iterator]](https://leetcode.com/problems/peeking-iterator/description/)\n\n\n#### [Interview Questions and Answers Repository](https://github.com/RooTinfinite/Interview-questions-and-answers)\n\n\n\n\n\n | 25 | 6 | ['Swift', 'C++', 'Java', 'Go', 'TypeScript', 'Python3', 'Rust', 'Ruby', 'JavaScript', 'C#'] | 5 |
maximum-xor-for-each-query | Prefix sum 1-pass vs partial_sum->1-line||beats 100% | prefix-sum-1-pass-vs-partial_sum-1-lineb-lef7 | Intuition\n Describe your first thoughts on how to solve this problem. \n1-pass solution using prefix sum\n\nNote that y^x^x=y^(0)=y \nx^k=y=> x^k^x=k=y^x\n2nd | anwendeng | NORMAL | 2024-11-08T00:17:23.056690+00:00 | 2024-11-08T06:23:12.850064+00:00 | 1,602 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1-pass solution using prefix sum\n\nNote that `y^x^x=y^(0)=y` \n`x^k=y=> x^k^x=k=y^x`\n2nd C++ is using partial_sum for the exercise.\nPython 1-liner is made. \nThis question is an example for Boolean Algebra; since only xor is considered, just use the facts from the abelian group.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n[Please turn on English subtitles if necessary]\n[https://youtu.be/WgH192Tjiqw?si=g9TyXGGUzo5BysJ6](https://youtu.be/WgH192Tjiqw?si=g9TyXGGUzo5BysJ6)\n1. Let `mask=(1<<maximumBit)-1` which has all its bits set from 0th bit -(maxmumBit-1)-th bit\n2. `ans` is the answering vector\n3. Set `ans[n-1]=nums[0]^mask`\n4. Use loop from i=1 to n-1 setp 1 do `ans[n-1-i]^=(ans[n-i]^nums[i])`\n5. return `ans`\n6. 2nd approach using partial_sum is a 2-passes solution\n7. Python 1-liner is made using accumulate which is like C++ std::partial_sum\n# Comparision among codes\n|Method|Elapsed time|beats|space| \n|---|---|---|---|\n|C++ prefix sum with xor 1-pass|0ms|100%|95.22MB|\n|C++ partial_sum with xor 2-pass|0ms|100%|95.48MB|\n|Python prefix sum with xor 1-pass|59ms|53.57%|32.60MB|\n|Python accumulate with xor 1-pass|16ms|100%|32.97MB|\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(1)$$\n# Code|| C++ prefix sum 1 pass\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n const unsigned mask=(1<<maximumBit)-1, n=nums.size();\n vector<int> ans(n, 0);\n ans[n-1]=nums[0]^mask;\n for(int i=1; i<n; i++)\n ans[n-1-i]^=(ans[n-i]^nums[i]);\n return ans;\n }\n};\n```\n# C++ using partial_sum 2 passes||0ms beats 100%\n```\nclass Solution {\npublic:\n static vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n nums[0]^=(1<<maximumBit)-1; \n partial_sum(nums.begin(), nums.end(), nums.begin(), bit_xor<>());\n reverse(nums.begin(), nums.end());\n return nums;\n }\n};\n```\n# Python done in video\n```\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n #bitwise XOR has the perfect algebaic properties, e.g. x^x=0, x^y=y^x, x^(~x)=0\n #x^k=y=> x^k^x=k=y^x\n mask, n=(1<<maximumBit)-1, len(nums)\n ans=[0]*n\n ans[-1]=nums[0]^mask\n for i in range(1, n):\n ans[~i]^=(ans[n-i]^nums[i])\n return ans\n \n```\nIf you have still diffculties to understand, just list the operations as follows (suppose n is large) then the pattern for prefix sum using xor is not hard to find\n```\nans[n-i-1]=nums[0]^mask\nans[n-i-2]=nums[0]^nums[1]^mask=ans[n-i-1]^nums[1]\nans[n-i-3]=nums[0]^nums[1]^nums[2]^mask=ans[n-i-2]^nums[2]\n.....\n.....\nans[1]=nums[0]^...^nums[n-2]^mask=ans[2]^nums[1]\nans[0]=nums[0]^...^nums[n-1]^mask=ans[1]^nums[0]\n```\n# Python 1-liner using accumulate|| 16ms beats 100%\n```\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n return list(accumulate(nums, operator.xor, initial=(1<<maximumBit)-1))[:0:-1]\n```\n | 20 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'C++', 'Python3'] | 6 |
maximum-xor-for-each-query | [Java] XOR with MAX value and All Left Element in Array | java-xor-with-max-value-and-all-left-ele-1vso | XOR Properties\nA ^ A = 0\n0 ^ A = A\n\n int MAX = (int) Math.pow(2, maximumBit) - 1; // (1 << maximumBit) - 1\n\t\t\nnums[0] XOR nums[1] XOR ... XOR nu | i18n | NORMAL | 2021-04-17T16:16:43.341800+00:00 | 2021-04-17T16:30:21.011122+00:00 | 501 | false | XOR Properties\nA ^ A = 0\n0 ^ A = A\n\n int MAX = (int) Math.pow(2, maximumBit) - 1; // (1 << maximumBit) - 1\n\t\t\nnums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k = MAX\n\n1. Take XOR both side with k\nnums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] = MAX ^ K\n\n2. Take XOR both side with MAX \nnums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR MAX = k\n\n\n```\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int xor = 0;\n for (int n : nums)\n xor ^= n;\n\n int[] answer = new int[nums.length];\n int MAX = (int) Math.pow(2, maximumBit) - 1; // (1 << maximumBit) - 1\n for (int i = 0; i < nums.length; i++) {\n answer[i] = xor ^ MAX;\n xor ^= nums[nums.length - 1 - i];\n }\n \n return answer;\n }\n}\n``` | 19 | 14 | [] | 2 |
maximum-xor-for-each-query | Very Simple | Beats 100% | Java, C++, Go, Python | Solution for LeetCode#1829 | very-simple-beats-100-java-c-go-python-s-o1ub | Approach-01\n\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires us to use cumulative XOR operations while consi | samir023041 | NORMAL | 2024-11-08T02:20:20.542160+00:00 | 2024-11-08T07:45:36.176011+00:00 | 1,291 | false | # Approach-01\n\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to use cumulative XOR operations while considering a constraint on the maximum possible value determined by maximumBit. The goal is to maximize the XOR results by leveraging the bit representation efficiently.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Calculate the Maximum Bit Value:\n\t- Compute the highest value that can be represented with maximumBit bits, which is \n(pow(2,maximumBit)\u22121). This value is used to maximize the XOR operation.\n2. Compute the Cumulative XOR:\n\t- Start by calculating the XOR of all elements in nums. This gives the initial cumulative XOR value for the entire array.\n3. Generate Results in Reverse:\n\t- Iterate through the array from the last element to the first, computing the XOR of xorVal with numBitVal to get the maximum possible XOR.\n\t- Update xorVal by XORing it with the current element to reflect the cumulative XOR for the remaining elements.\n4. Return the Answer:\n\t- The results are collected in ans and returned.\n\t\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```Java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n=nums.length;\n int[] ans=new int[n];\n int val=(int)Math.pow(2, maximumBit)-1;\n \n int k=n-1;\n int curr=0;\n for(int i=0; i<n; i++){\n curr ^=nums[i];\n ans[k--]=curr^val; \n }\n \n return ans;\n }\n}\n```\n```Cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size();\n vector<int> ans(n);\n int val = (1 << maximumBit) - 1; // Equivalent to pow(2, maximumBit) - 1\n \n int k = n - 1;\n int curr = 0;\n \n for (int i = 0; i < n; i++) {\n curr ^= nums[i];\n ans[k--] = curr ^ val;\n }\n \n return ans;\n }\n};\n```\n```Python []\nclass Solution:\n def getMaximumXor(self, nums, maximumBit):\n n = len(nums)\n ans = [0] * n\n val = (1 << maximumBit) - 1 # Equivalent to 2^maximumBit - 1\n \n curr = 0\n for num in nums:\n curr ^= num\n\n for i in range(n):\n ans[i] = curr ^ val\n curr ^= nums[n - 1 - i]\n\n return ans\n\n```\n```C# []\npublic class Solution {\n public int[] GetMaximumXor(int[] nums, int maximumBit) {\n int n = nums.Length;\n int[] ans = new int[n];\n int val = (1 << maximumBit) - 1; // Equivalent to 2^maximumBit - 1\n\n int curr = 0;\n foreach (int num in nums) {\n curr ^= num;\n }\n\n for (int i = 0; i < n; i++) {\n ans[i] = curr ^ val;\n curr ^= nums[n - 1 - i];\n }\n\n return ans;\n }\n}\n```\n```Go []\nfunc getMaximumXor(nums []int, maximumBit int) []int {\n n := len(nums)\n ans := make([]int, n)\n val := (1 << maximumBit) - 1 // Equivalent to 2^maximumBit - 1\n\n curr := 0\n for _, num := range nums {\n curr ^= num\n }\n\n for i := 0; i < n; i++ {\n ans[i] = curr ^ val\n curr ^= nums[n-1-i]\n }\n\n return ans\n}\n```\n# ------------------------------------------------------\n\n# Approach-02\n\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem involves finding the maximum possible XOR for each element in an array when all elements are XORed cumulatively from left to right. We then need to generate results in a reversed manner using a specified bit constraint, maximumBit.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Cumulative XOR Calculation:\n\t- Compute the cumulative XOR of all elements from left to right and store the results in an array arr. This will keep track of the XOR value for each prefix of the input array nums.\n2. Maximum XOR Value:\n\t- Calculate the maximum possible XOR value given maximumBit. This value is calculated as \n(pow(2,maximumBit)\u22121), which ensures all bits are set to 1 within the limit of maximumBit.\n3. Reverse and XOR:\n\t- Traverse arr in reverse and compute the XOR of each element with the maximum possible value obtained in the previous step. This yields the desired result in a reversed order.\n\t\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n=nums.length;\n int[] arr=new int[n];\n int[] ans=new int[n];\n\n arr[0]=nums[0];\n for(int i=1; i<n; i++){\n arr[i] = arr[i-1]^nums[i];\n }\n\n int val=(int)Math.pow(2, maximumBit)-1;\n int k=0;\n for(int i=n-1; i>=0; i--){\n ans[k++] = arr[i]^val;\n }\n\n return ans;\n }\n}\n``` | 13 | 0 | ['Python', 'C++', 'Java', 'Go', 'C#'] | 6 |
maximum-xor-for-each-query | ✅ One Line Solution | one-line-solution-by-mikposp-ix4z | Code #1\nTime complexity: O(n). Space complexity: O(1).\npython3\nclass Solution:\n def getMaximumXor(self, a: List[int], q: int) -> List[int]:\n retu | MikPosp | NORMAL | 2024-11-08T09:26:11.955967+00:00 | 2024-11-08T09:26:11.956006+00:00 | 267 | false | # Code #1\nTime complexity: $$O(n)$$. Space complexity: $$O(1)$$.\n```python3\nclass Solution:\n def getMaximumXor(self, a: List[int], q: int) -> List[int]:\n return [2**q-1^x for x in accumulate(a,xor)][::-1]\n```\n\n# Code #2\nTime complexity: $$O(n)$$. Space complexity: $$O(1)$$.\n```python3\nclass Solution:\n def getMaximumXor(self, a: List[int], q: int) -> List[int]:\n return [*accumulate([2**q-1]+a,xor)][:0:-1]\n``` | 7 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Python', 'Python3'] | 2 |
maximum-xor-for-each-query | Python | Cumulative XOR Pattern | python-cumulative-xor-pattern-by-khosiya-sb74 | see the Successfully Accepted Submission\n\n# Code\npython3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n | Khosiyat | NORMAL | 2024-11-08T03:03:32.842054+00:00 | 2024-11-08T03:03:32.842082+00:00 | 421 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446340716/?envType=daily-question&envId=2024-11-08)\n\n# Code\n```python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n # Calculate the maximum possible value for the given number of bits\n max_value = (1 << maximumBit) - 1 # Equivalent to 2^maximumBit - 1\n \n # Calculate the cumulative XOR of all elements in nums\n cumulative_xor = 0\n for num in nums:\n cumulative_xor ^= num\n \n # Prepare the result array to hold the answer for each query\n result = []\n \n # Process each query in reverse (removing the last element each time)\n for num in reversed(nums):\n # Find k by XORing cumulative_xor with max_value\n result.append(cumulative_xor ^ max_value)\n \n # Update cumulative_xor by removing the effect of the last element\n cumulative_xor ^= num\n \n return result\n\n```\n\n\n# Solution Approach\n\n### Approach\n1. **Cumulative XOR Calculation**: First, calculate the cumulative XOR of the entire array `nums`. This will be our initial XOR result. For each query, as we remove the last element from `nums`, we update this cumulative XOR.\n \n2. **Determine Maximum Possible Value**: The highest possible value we can achieve for each XOR is \\(2^{\\text{maximumBit}} - 1\\). This is essentially a binary number with all bits set to `1` for `maximumBit` bits.\n\n3. **Optimal `k` Calculation**: To maximize the XOR result, `k` should be the bitwise complement of the current cumulative XOR limited to `maximumBit` bits. This can be computed as:\n \\[\n k = \\text{max\\_value} \\, \\text{XOR} \\, \\text{current\\_xor}\n \\]\n where `max_value` is \\(2^{\\text{maximumBit}} - 1\\).\n\n4. **Reverse Query**: Instead of modifying `nums` repeatedly, simulate the reverse queries using a loop where you decrementally adjust the cumulative XOR by removing elements from the end.\n\n### Explanation of the Code\n1. **Calculate `max_value`**: This is the largest value we can achieve with `maximumBit` bits, which is \\(2^{\\text{maximumBit}} - 1\\).\n\n2. **Initial Cumulative XOR**: Compute the XOR of all elements in `nums`.\n\n3. **Reverse Query Processing**:\n - For each element in `nums` (processed from the last to the first), calculate the `k` value that, when XORed with the cumulative XOR, yields the maximum possible XOR result.\n - Append this value to `result`.\n - Update `cumulative_xor` by XORing it with the current element (removing the last element effect).\n\n4. **Return Result**: Since we processed the queries in reverse order, `result` will have answers in the desired order after appending each result.\n\n### Complexity Analysis\n- **Time Complexity**: \\(O(n)\\), where \\(n\\) is the number of elements in `nums`. We iterate over `nums` twice: once to compute the initial cumulative XOR and once for processing each query.\n- **Space Complexity**: \\(O(n)\\), since we store the result for each query in the `result` list.\n\n\n\n\n | 7 | 0 | ['Python3'] | 1 |
maximum-xor-for-each-query | Simple O(n) solution using XOR | simple-on-solution-using-xor-by-mandysin-hind | Let ans[i] = nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k\nWe are asked to maximize ans[i] such that k < 2^maximumBit.\n\nThe approach is to find t | mandysingh150 | NORMAL | 2021-06-23T06:48:29.867294+00:00 | 2021-06-30T06:17:01.865027+00:00 | 557 | false | Let **ans[i] = nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k**\nWe are asked to maximize **ans[i]** such that k < 2^maximumBit.\n\nThe approach is to find the XOR of all the elements. Each ans[i] can be calculated by taking the XOR of currentXOR and (2^maximumBit - 1). Then, we remove the last array element from the currentXOR by again taking XOR of currentXOR with the last array element. This property works because **a^a = 0**, where ^ is XOR operator.\n\nUsing the above mentioned technique, we can easily bulid the answer array.\n\nP.S. - If there was no constraint like **k < 2^maximumBit**, then the every ans[i] would be equal to INT_MAX, because we could take **k = ~(currentXOR)**, where ~ is Bitwise NOT operator.\n```\nvector<int> getMaximumXor(vector<int>& a, int maximumBit) {\n vector<int> ans;\n int xr = 0, n = a.size();\n \n\t\t// calculate XOR aof all elements\n for(int i=0 ; i<n ; ++i)\n xr ^= a[i];\n \n\t\t// calculate (2^maximumBit)-1 using left shift operator\n maximumBit = (1 << maximumBit)-1;\n\t\t\n for(int i=n-1 ; i>=0 ; --i) {\n ans.push_back(maximumBit^xr);\n xr ^= a[i];\n }\n return ans;\n }\n```\n**Any suggestions are welcome \uD83D\uDE00.\nPlease upvote the post if you like the solution approach.** | 7 | 1 | [] | 1 |
maximum-xor-for-each-query | ✅Beats 100% | Very Short & Easily Understandable | beats-100-very-short-easily-understandab-2ldr | \n\n\n# Complexity\n- Time complexity:O(N)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(N)\n Add your space complexity here, e.g. O(n) \n | dheerajcodes | NORMAL | 2024-11-08T22:30:55.973105+00:00 | 2024-11-08T22:42:58.977140+00:00 | 63 | false | \n\n\n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int[] arr = new int[nums.length];\n \n // Calculate the target value, which is the maximum XOR we want to achieve\n int target = (1 << maximumBit) - 1;//Or u can use Math.pow(2,maximumBit)-1\n \n // Calculatating xor of all elements\n int xor = 0;\n for (int i=0i<nums.length;i++) {\n xor ^= nums[i];\n }\n \n // Fill the result array\n for (int i = 0; i < nums.length; i++) {\n arr[i] = target ^ xor;\n xor ^= nums[nums.length - 1 - i]; // Update xor by removing the last element\n }\n \n return arr;\n }\n}\n\n```\n```python []\nclass Solution(object)\n def getMaximumXor(self, nums, maximumBit):\n arr = [0] * len(nums)\n \n # Calculate the target value, which is the maximum XOR we want to achieve\n target = (1 << maximumBit) - 1\n \n # Calculate the cumulative XOR of all elements in nums\n xor = 0\n for num in nums:\n xor ^= num\n \n # Fill the result array\n for i in range(len(nums)):\n arr[i] = target ^ xor\n xor ^= nums[len(nums) - 1 - i] # Update xor by removing the last element\n \n return arr\n```\n```c++ []\n#include <vector>\n#include <cmath>\nusing namespace std;\n\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n vector<int> arr(nums.size());\n \n // Calculate the target value, which is the maximum XOR we want to achieve\n int target = (1 << maximumBit) - 1;\n \n // Calculate the cumulative XOR of all elements in nums\n int xorVal = 0;\n for (int num : nums) {\n xorVal ^= num;\n }\n \n // Fill the result array\n for (int i = 0; i < nums.size(); i++) {\n arr[i] = target ^ xorVal;\n xorVal ^= nums[nums.size() - 1 - i]; // Update xor by removing the last element\n }\n \n return arr;\n }\n};\n```\n**THIS IS MY FIRST TIME SHARING A SOLUTION PLEASE DO UPVOTE**\n\n\n\n | 6 | 0 | ['Bit Manipulation', 'Python', 'C++', 'Java'] | 1 |
maximum-xor-for-each-query | Only for beginners of bitwise operations make a Crystel clear understanding, with added YouTube | only-for-beginners-of-bitwise-operations-1eiu | Youtube explanation https://youtu.be/Hvav8wtwEN8\n# Intuition and approach\nlets talk about XOR ==> \n1) xor of any numnber with Zero is number itself\n2) xor o | vinod_aka_veenu | NORMAL | 2024-11-08T13:38:16.099611+00:00 | 2024-11-08T15:26:44.089327+00:00 | 429 | false | # Youtube explanation https://youtu.be/Hvav8wtwEN8\n# Intuition and approach\nlets talk about XOR ==> \n1) **xor of any numnber with Zero** is number itself\n2) **xor of a number with itself** is always produce **Zero**.\n\n3) if we do xor few numbers, it would be like below.\n \nint **xorVal = x1^x2^x3^x4^x5**\n\nnow if **we remove any number lets say last number = x5**, what would be xor of remaing numbers?\n\nXOR of remaining numebrs\nx1^x2^x3^x4 = **xorVal ^ x5** (so we don\'t need to calculate again from starting)\n\ncome to this question, what we would do is :- \n\na) get the **xor** of all numbers\n for i = 0 to n-1 \n xorVal = xorVal ^ nums[i]; \n\nb) now we will have an value **val = pow(2, maxBit)-1**\n **whay this?**\n because we need this number so that we can pick a number \n**0 to val** (inclusive) to make a xorVal such that we would get the maximum xor value.\n\n**how ??**\nbecause this makes sure that we can pick any number from this range and do the XOR so that it would make largest value.\n\n**so now question is how to pick a value from this range, we don\'t know which one will make largest value if we do XOR??**\n\nlets explaing with example:- \nsuppose we have xor of all the array elements xorVal = 100101\n maxBit = 2\n val = pow(2, 2)-1 = 4-1 = 3 = 0011\n\nnow from 0000 to 0011 we need to pick a number that make largest xor\nnow \nexisting **xorVal = 100101 ==>** look at last **2 bits ==> 01** ==> means if make a xor with (**10**) this wpould produce both 1 ==> **means this value is the just reverse flipped value of last 2 bits**===> means if we do the xor of **val ^ xorVal** ==> **produce reversed value===> record it intp array , remove last elements** and keep on doing for all elemtns....return answer array.\n\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length;\n int xorVal = 0;\n int togglePoint = (1 << maximumBit) - 1; \n for (int x : nums) {\n xorVal ^= x;\n }\n\n int[] ans = new int[n];\n for (int i = 0; i < n; i++) {\n ans[i] = xorVal ^ togglePoint;\n xorVal ^= nums[n - 1 - i];\n }\n\n return ans;\n }\n}\n\n```\n```C# []\nclass Solution {\n public int[] GetMaximumXor(int[] nums, int maximumBit) {\n int n = nums.Length;\n int xorVal = 0;\n int togglePoint = (1 << maximumBit) - 1; \n for (int i=0; i<n; i++) {\n xorVal ^= nums[i];\n }\n\n int[] ans = new int[n];\n for (int i = 0; i < n; i++) {\n ans[i] = xorVal ^ togglePoint;\n xorVal ^= nums[n - 1 - i];\n }\n\n return ans;\n }\n}\n\n```\n | 6 | 0 | ['Array', 'Bit Manipulation', 'Bitmask', 'Java', 'C#'] | 2 |
maximum-xor-for-each-query | Easiest C++ solution || Beginner-friendly approach || With complete explanation !! | easiest-c-solution-beginner-friendly-app-t22d | Approach\n- Firstly, we find k = pow(2,maximumBit) -1 because our maximum value possible can be k. Example, maximum value possible with 3 bit is \npow(2,3) - 1 | prathams29 | NORMAL | 2023-06-27T06:40:14.782621+00:00 | 2023-06-27T08:24:48.725484+00:00 | 220 | false | # Approach\n- Firstly, we find `k = pow(2,maximumBit) -1` because our maximum value possible can be k. Example, maximum value possible with 3 bit is \n`pow(2,3) - 1 = 7` \n- The first for loop calculates the XOR for all the nums[i].\n- Before going to the second for loop, we must know that \n`If a XOR B = c then a XOR c = b` \n- Consider that **a is XOR of all nums[i]** and **b is the value that we need in order to get maximum possible value c**. We know that the **maximum possible value is k** as we found out so c=k and since we have a and c and we need to find b, hence we perform **a XOR c and push it in our ans vector**. \n- After that, we are told to delete the last element. Instead, we perform XOR of last element with a (XOR of all elements) because **XOR of an element with itself is 0**\n# Code\n```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size(), a=0, res=0, k=pow(2, maximumBit)-1;\n vector<int> ans; \n for(int i=0; i<n; i++)\n a^=nums[i];\n for(int i=n-1; i>=0; i--)\n {\n res = a;\n ans.push_back(res^k);\n a^=nums[i];\n }\n return ans;\n }\n};\n``` | 6 | 0 | ['C++'] | 1 |
maximum-xor-for-each-query | Just take bitwise negation of last maximumBit number of bits | just-take-bitwise-negation-of-last-maxim-cdx7 | If we know the cumulative XOR of the numbers upto the ith index, then we can just take the last maximumBit number of bits, invert them and that results in our k | rishabhsetiya7 | NORMAL | 2021-04-17T17:24:14.107204+00:00 | 2021-04-17T17:24:14.107277+00:00 | 332 | false | If we know the cumulative XOR of the numbers upto the ith index, then we can just take the last maximumBit number of bits, invert them and that results in our k for elements upto that index.\n\nIf we have a number and we want to invert the last x bits of that number then we can left shift the number by (32-x) bits, as 32 is the total number of bits in an integer, take the bitwise negation of the number and then right shift by (32-x) bits. \n\n```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int b) {\n vector<int> result(nums.size());\n b=32-b;\n unsigned int cumulativeXOR=nums[0];\n result[nums.size()-1]=~(cumulativeXOR<<b)>>b;\n for(int i=1; i<nums.size(); i++){\n cumulativeXOR^=nums[i];\n result[nums.size()-1-i]=~(cumulativeXOR<<b)>>b;\n }\n return result;\n }\n};\n``` | 6 | 2 | [] | 1 |
maximum-xor-for-each-query | Python [one loop + reverse] | python-one-loop-reverse-by-it_bilim-uro7 | \nclass Solution(object):\n def getMaximumXor(self, nums, maximumBit):\n k = 2**maximumBit - 1\n tmp = nums[0]\n res = [k ^ tmp]\n | it_bilim | NORMAL | 2021-04-17T17:07:09.321293+00:00 | 2021-04-17T17:07:09.321325+00:00 | 447 | false | ```\nclass Solution(object):\n def getMaximumXor(self, nums, maximumBit):\n k = 2**maximumBit - 1\n tmp = nums[0]\n res = [k ^ tmp]\n for i in range(1, len(nums)):\n tmp = tmp ^ nums[i]\n res.append(k ^ tmp)\n return reversed(res)\n``` | 6 | 1 | [] | 1 |
maximum-xor-for-each-query | Mastering XOR Magic: Maximizing Array Prefixes with Bitwise Tricks! | mastering-xor-magic-maximizing-array-pre-cn5y | Intuition\n\nThe main goal of this problem is to find the maximum XOR values for each prefix of the array nums in reverse order. By leveraging the XOR operation | suhas_sr7 | NORMAL | 2024-11-08T17:38:12.483133+00:00 | 2024-11-08T17:38:12.483172+00:00 | 48 | false | ### Intuition\n\nThe main goal of this problem is to find the maximum XOR values for each prefix of the array `nums` in reverse order. By leveraging the XOR operation\'s properties, we can maximize each prefix XOR by XORing it with the largest number possible given a specific number of bits (`maximumBit`). The maximum integer within `maximumBit` bits (all bits set to `1`) provides a way to "flip" all bits of the prefix XOR, giving the largest result possible.\n\n### Approach\n\n1. **Calculate the Maximum Possible XOR Mask**: \n - Compute `maxi` as `(2^maximumBit) - 1`, which represents the maximum integer value with all bits set to `1` within the `maximumBit` constraint. This `maxi` allows us to maximize the XOR of each prefix.\n\n2. **Build the Prefix XOR**:\n - Initialize `val` to `0`, which will store the cumulative XOR of `nums` from the start up to each position.\n - Traverse through `nums` and, for each element:\n - Update `val` to include the XOR of the current element.\n - Compute the maximum XOR for the current prefix by XORing `val` with `maxi` and store this result in `nums`.\n\n3. **Reverse the Result**:\n - Since the maximum XOR values need to be returned in reverse order, reverse `nums` before returning it.\n\n### Complexity\n\n- **Time Complexity**: \\(O(n)\\), where \\(n\\) is the length of `nums`. The code iterates through `nums` once to compute the prefix XORs and calculate the maximum XOR values, making it linear in time.\n\n- **Space Complexity**: \\(O(1)\\) additional space since the calculations are done in place in `nums`, except for reversing it at the end.\n\n### Code\n\n```python\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n # Initialize the cumulative XOR value and the maximum possible XOR mask\n val = 0\n maxi = (2 ** maximumBit) - 1\n \n # Traverse through `nums` and update each element with the maximum XOR for each prefix\n for i in range(len(nums)):\n val ^= nums[i] # Update cumulative XOR\n nums[i] = val ^ maxi # Calculate maximum XOR by XORing with `maxi`\n \n # Return the results in reverse order as required by the problem\n return nums[::-1]\n``` | 5 | 0 | ['Python3'] | 0 |
maximum-xor-for-each-query | simple py,JAVA code explained in detail using Bit masking!!! | simple-pyjava-code-explained-in-detail-u-iis9 | Problem understanding\n\n- Ah this month is going to be little tricky coz it\'s BIT MANUPULATION,but let me make it easy for you.\n- They have given an array nu | arjunprabhakar1910 | NORMAL | 2024-11-08T16:28:59.704098+00:00 | 2024-11-08T16:28:59.704140+00:00 | 65 | false | # Problem understanding\n\n- Ah this month is going to be little tricky coz it\'s *BIT MANUPULATION*,but let me make it easy for you.\n- They have given an array `nums` and a number `maximumBit`.\n- They have asked to find a `k` for each all the privious ***XOR-ed*** `prefixXOR` $XOR$ `k` should be maximised.This `k` is stored in `answer` arary and should be returned,where `answer[i]` is the $XOR$ after removing `i` elemts from the `nums`.\n```\nnums = [0,1,1,3], maximumBit = 2\nanswer=[0,3,2,3]\n1st query: nums = [0,1,1,3],since 0 XOR 1 XOR 1 XOR 3 XOR K \n2nd query: nums = [0,1,1], since 0 XOR 1 XOR 1 XOR K \n3rd query: nums = [0,1], since 0 XOR 1 XOR K \n4th query: nums = [0], since 0 XOR K\n``` \n- where `k` should be maximised in each query,whatever may be the `LHS` be maximum number for any `n` bit is $2^n-1$.\n```\n2bits--->11 \n3bits--->111\n```\n- so now we should find a `k` such that `k` $XOR$ `prefixXOR` leads to $2^n-1$.\n- we already know property of `XOR` when we have `odd` bits of `1` we get `1` as answer.\n```\n### Query 1: nums = [0, 1, 1, 3]\n| Step | Operation | Result (Decimal) | Result |\n|-------------|--------------|------------------|---------|\n| Initial | - | 0 | 0000 |\n| Step 1 | 0 XOR 1 | 1 | 0001 |\n| Step 2 | 1 XOR 1 | 0 | 0000 |\n| Step 3 | 0 XOR 3 | 3 | 0011 |\n| Final XOR | 3 XOR K | 3 XOR K | 0011 XOR K |\n\n### Query 2: nums = [0, 1, 1]\n| Step | Operation | Result (Decimal) | Result |\n|-------------|--------------|------------------|---------|\n| Initial | - | 0 | 0000 |\n| Step 1 | 0 XOR 1 | 1 | 0001 |\n| Step 2 | 1 XOR 1 | 0 | 0000 |\n| Final XOR | 0 XOR K | 0 XOR K | 0000 XOR K |\n\n### Query 3: nums = [0, 1]\n| Step | Operation | Result (Decimal) | Result |\n|-------------|--------------|------------------|---------|\n| Initial | - | 0 | 0000 |\n| Step 1 | 0 XOR 1 | 1 | 0001 |\n| Final XOR | 1 XOR K | 1 XOR K | 0001 XOR K |\n\n### Query 4: nums = [0]\n| Step | Operation | Result (Decimal) | Result |\n|-------------|--------------|------------------|---------|\n| Initial | - | 0 | 0000 |\n| Final XOR | 0 XOR K | 0 XOR K | 0000 XOR K |\n\n```\n- At the `finalXOR` in above tabulation we can see we can see `prefixXOR` or `LHS` of $XOR$.\n- now we want to maximise this `LHS` to $2^n-1$ so our `k` is preety clear right now which is *bits flipped* of `LHS`.Because when we bit flip for ever `0` in `LHS` we will maximise the answer,so our `k` is constructed by \n```\nbitmask=(1<<maximumBit)-1\n```\n - `1<<maximumBit` gives `100000000000000000.........` on `-1`,all bits before that are flipped to `0111111..........` the zer is `maximumbit+1` *position* so never mind!\n - now if we $XOR$ `bitmask` and `LHS` we get our `k`.\n```\n#Query-1\n0011 XOR 011 => 00(k=0)\n\n#Query-2\n0000 XOR 011 => 11(k=3)\n\n#Query-3\n0001 XOR 011 => 10(k=2)\n\n#Query-4\n0000 XOR 011 => 11(k=3)\n\n\n```\n---\n\n\n# Approach:Bit masking+prefix sum\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- We use `prefixXOR` to store all the prefix sum of $XOR$,answer to store `k` for each query.\n- Then we apply the above logic to find `k` by $XOR$ of `mask` and `prefixXOR[i]`.\n\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:$O(n)$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$O(n)$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```Python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n prefixXOR=[0]*len(nums)\n prefixXOR[0]=nums[0]\n answer=[]\n for i in range(1,len(nums)):\n prefixXOR[i]=prefixXOR[i-1] ^ nums[i]\n mask=(1<<maximumBit)-1\n for i in range(len(prefixXOR)-1,-1,-1):\n answer.append(prefixXOR[i] ^ mask)\n return answer\n\n \n \n \n \n\n```\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n ArrayList<Integer> prefixXOR=new ArrayList<>();\n prefixXOR.add(nums[0]);\n int[] answer=new int[nums.length];\n for(int i=1;i<nums.length;i++){\n prefixXOR.add(prefixXOR.get(i-1)^nums[i]);\n }\n int mask=(1<<maximumBit)-1;\n for(int i=prefixXOR.size()-1;i>-1;i--){\n answer[nums.length-1-i]=prefixXOR.get(i)^mask;\n }\n return answer;\n }\n}\n``` | 5 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Java', 'Python3'] | 0 |
maximum-xor-for-each-query | Python3 beats 100% ✅ 💯 Super easy to understand! 💡 | python3-beats-100-super-easy-to-understa-x4wy | \n# Approach\n1. Calculate Maximum Possible XOR: Compute (1 << maximumBit) - 1 to get the maximum XOR value within maximumBit bits. This sets all bits to 1 in b | rinsane | NORMAL | 2024-11-08T12:27:36.684068+00:00 | 2024-11-11T10:54:12.107906+00:00 | 24 | false | \n# Approach\n1. **Calculate Maximum Possible XOR**: Compute `(1 << maximumBit) - 1` to get the maximum XOR value within `maximumBit` bits. This sets all bits to 1 in binary, allowing us to achieve the highest possible XOR for any number with `maximumBit` bits.\n\n2. **Maintain Prefix XOR**: Initialize `curr` to store the cumulative XOR of elements in `nums` as we iterate through them.\n\n3. **Compute Result in Reverse Order**:\n - For each element in `nums`, update `curr` by XORing it with the current number. This accumulates the XOR for the prefix up to the current element.\n - Calculate the maximum XOR for the current prefix by using `maxm - curr`. Here\u2019s why:\n - **Explanation**: XORing `curr` with `maxm` (all bits set to 1) effectively inverts `curr`\'s bits. This "flips" each bit of `curr`, generating the highest possible XOR result by maximizing differences between `curr` and `maxm`. It ensures that each bit in the result is the opposite of `curr`, creating the maximum XOR value possible with `maximumBit` bits.\n - Insert this result at the beginning of our result list with `appendleft` to build the answer in reverse order.\n\n4. **Return the Result**: Convert the deque to a list and return.\n\n# Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n ans = deque([])\n curr = 0\n maxm = (1 << maximumBit) - 1\n for i in nums:\n curr ^= i\n ans.appendleft(maxm - curr)\n\n return list(ans)\n\n``` | 5 | 0 | ['Python3'] | 1 |
maximum-xor-for-each-query | Easy C++ Solution | Video Explanation | Full intuition explained | just 10 mins | easy-c-solution-video-explanation-full-i-nf4c | Video Solution\nhttps://youtu.be/v-MOHFW6dSs\n# Intuition\n Describe your first thoughts on how to solve this problem. \n1. Goal: For each index i (starting fro | Atharav_s | NORMAL | 2024-11-08T04:29:02.742364+00:00 | 2024-11-08T06:01:26.070211+00:00 | 497 | false | # Video Solution\nhttps://youtu.be/v-MOHFW6dSs\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. Goal: For each index i (starting from the last element and moving backward), we want to find the maximum XOR result by XORing a prefix XOR with a number that can be formed by maximumBit bits.\n2. Insight: The maximum number that can be represented with maximumBit bits is 2^maximumBit - 1. To achieve the maximum XOR result, we should XOR the prefix XOR with this number (let\u2019s call it mask).\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Calculate the Initial XOR:\n\n- Initialize xoor as the XOR of all elements in nums from the beginning.\n- This will store the cumulative XOR of the array, which we will modify as we move backward.\n2. Determine the Maximum Possible XOR:\n\n- Calculate mask = (1 << maximumBit) - 1, which is the maximum integer possible with maximumBit bits all set to 1 (binary 111...111). \n- This mask will give the highest possible XOR with xoor in each iteration.\n3. Iterate Backward and Compute Results:\n\n- Start from the last element in nums and move backward. For each index i:\nXOR xoor with mask to find the maximum XOR result for that step.\n- Append the result to result.\n- XOR xoor with nums[i] to remove the effect of nums[i] from xoor for the next iteration.\n4. Return the Result:\n\n- The result is collected in reverse order (from last element to first), so we can return it as it is.\n# Complexity\n- Time complexity:$$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n\n vector<int> result;\n\n int xoor = 0;\n\n for(auto&num: nums) xoor ^= num;\n\n int mask = pow(2,maximumBit) - 1; \n\n int n = nums.size();\n\n for(int i=n-1;i>=0;i--){\n\n result.push_back(xoor ^ mask);\n\n xoor ^= nums[i];\n }\n\n return result;\n \n \n }\n};\n``` | 5 | 0 | ['C++'] | 1 |
maximum-xor-for-each-query | Easy Solution | Beats 100% | step by step Explained | O(n) | easy-solution-beats-100-step-by-step-exp-5opk | Approach\n Describe your approach to solving the problem. \n\n### Approach Explanation\n\n1. Understanding the Mask:\n - We first calculate a mask. The mask i | iambandirakesh | NORMAL | 2024-11-08T00:26:10.200645+00:00 | 2024-11-08T00:26:10.200669+00:00 | 918 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n\n### Approach Explanation\n\n1. *Understanding the Mask*:\n - We first calculate a mask. The mask is created using the expression (1 << maximumBit) - 1. This constructs a binary number that has maximumBit bits all set to 1. For example:\n - If maximumBit is 3, the mask will be 0b111 (which is 7 in decimal).\n - This mask will help in flipping the bits of the numbers in the nums array.\n\n2. *XOR Operation*:\n - The XOR operation (^) is a bitwise operation that returns 1 for each bit position where the corresponding bits of its operands are different. \n - To maximize the XOR result for any number in nums, we can XOR it with the mask. This effectively flips the bits of the number, giving us the potential maximum value for that number within the constraints of maximumBit.\n\n3. *Cumulative XOR Update*:\n - As we iterate through each number x in nums, we perform two operations:\n - XOR x with the mask and store the result in the pre array. This gives us the maximum XOR possible for that number.\n - Update the mask by XORing it with x. This keeps track of the cumulative XOR of all previously seen numbers in nums, which ensures that the mask reflects the state of previous computations.\n\n4. *Reversing the Result*:\n - After populating the pre array with maximum XOR values, we reverse it. This is because the problem specifies that the results should be in descending order.\n\n5. *Return the Result*:\n - Finally, the function returns the reversed pre array, which contains the results in the required order.\n\n### Complexity\n- *Time Complexity*: O(n), where n is the length of the nums array. We perform a constant amount of work for each element in nums.\n- *Space Complexity*: O(n) for storing the results in the pre array.\n# Code\n```javascript []\n/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function(nums, maximumBit) {\n let n = nums.length;\n let pre = [];\n let mask = (1 << maximumBit) - 1; // Sets mask to the max possible value with \'maximumBit\' bits (e.g., 3 bits => 0b111 = 7)\n for (let x of nums) {\n pre.push(mask ^ x); // XOR each number in nums with the mask, effectively flipping bits to get maximum XOR\n mask ^= x; // Update the mask to reflect cumulative XOR of seen numbers\n }\n pre.reverse(); // Reverse the list to follow the descending order requirement\n return pre;\n};\n\n```\n``` python []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n n=len(nums)\n pre=[]\n mask=(1<<maximumBit)-1\n for x in nums:\n pre.append(mask^x)\n mask=mask^x\n pre.reverse()\n return pre\n```\n``` c++ []\n#include <vector>\nusing namespace std;\n\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size();\n vector<int> result;\n \n // Create the mask with \'maximumBit\' bits set to 1\n int mask = (1 << maximumBit) - 1; \n \n for (int x : nums) {\n result.push_back(mask ^ x); // XOR the current number with the mask\n mask ^= x; // Update the mask with the cumulative XOR\n }\n \n reverse(result.begin(), result.end()); // Reverse the array to meet descending order requirement\n return result; // Return the final result\n }\n};\n```\n``` java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length;\n int[] result = new int[n];\n \n // Create the mask with \'maximumBit\' bits set to 1\n int mask = (1 << maximumBit) - 1;\n int cumulativeXor = 0;\n \n // Calculate cumulative XOR for all elements in nums\n for (int num : nums) {\n cumulativeXor ^= num;\n }\n \n // Calculate the maximum XOR for each query\n for (int i = 0; i < n; i++) {\n result[i] = cumulativeXor ^ mask; // XOR cumulativeXor with mask to get the max XOR value\n cumulativeXor ^= nums[n - 1 - i]; // Update cumulative XOR for the next query\n }\n \n return result;\n }\n}\n```\n``` c# []\npublic class Solution {\n public int[] GetMaximumXor(int[] nums, int maximumBit) {\n int n = nums.Length;\n int[] result = new int[n];\n \n // Create a mask with \'maximumBit\' bits set to 1\n int mask = (1 << maximumBit) - 1;\n int cumulativeXor = 0;\n\n // Calculate the cumulative XOR for all elements in nums\n foreach (int num in nums) {\n cumulativeXor ^= num;\n }\n\n // Calculate the maximum XOR for each query\n for (int i = 0; i < n; i++) {\n result[i] = cumulativeXor ^ mask; // XOR cumulativeXor with mask to get max XOR value\n cumulativeXor ^= nums[n - 1 - i]; // Update cumulative XOR for the next query\n }\n\n return result;\n }\n}\n``` | 5 | 0 | ['Array', 'Bit Manipulation', 'Python', 'C++', 'Java', 'Python3', 'JavaScript', 'C#'] | 3 |
maximum-xor-for-each-query | [Go] O(n) solution with Explanation | go-on-solution-with-explanation-by-pushk-4xoh | So basically we need to find an integer whose binary representation when XOR with nums elements one by one yields the highest number that can be formed from max | pushkar_singh_katiyar | NORMAL | 2021-04-17T17:12:22.626415+00:00 | 2021-04-17T17:34:21.881604+00:00 | 200 | false | So basically we need to find an integer whose binary representation when XOR with nums elements one by one yields the highest number that can be formed from maximumBit.\n\nMax Integer that can be formed from n bits = 2^n - 1\nConsider the given example [0,1,1,3] with maximumBit = 2\n\nMaximum Integer (maxNum) that can be formed from 2 bits = 2^2-1 = 3\nNow Take 1st element i.e. 0\nBinary Representaton of 0 in 2 bits = 00\nBinary Representaton of maxNum in 2 bits = 11\nSo 00 XOR 11 = 11 so ans = 11 i.e. 3\n\nNow Take 1st and 2nd element i.e. 0, 1\nBinary Representaton of 0 XOR 1 in 2 bits = 00 XOR 01 = 01\nBinary Representaton of maxNum in 2 bits = 11\nSo 01 XOR 11 = 10 so ans = 10 i.e. 2\n\nNow Take 1st, 2nd and 3rd element i.e. 0, 1, 1\nBinary Representaton of 0 XOR 1 XOR 1 in 2 bits = 00 XOR 01 XOR 01 = 00\nBinary Representaton of maxNum in 2 bits = 11\nSo 00 XOR 11 = 11 so ans = 11 i.e. 3\n\n**Can you see the repetitive code??**\nWe are doing XOR for previous elements which we have already done. So better to store the previous XOR output at some place so that it can be directly used by proceeding elements\n\nNow Take firtst 4 elements i.e. 0, 1, 1, 3\nBinary Representaton of previous 3 elements XOR operation = 00\nBinary Representaton of previous 3 elements XOR operation XOR 3 = 00 XOR 11 = 11\nBinary Representaton of maxNum in 2 bits = 11\nSo 11 XOR 11 = 0 so ans = 00 i.e. 0\n\nNow store the result of each step in the array and reverse the array or insert the each step result in reverse order as mentioned in code.\n\n**BONUS**\nGo coders I got TLE for using the below syntax-\n```\nxorArr = append(xorArr, opr)\nresult = append([]int{maxNum-opr}, result...)\n```\nPlease note that append operation on slice are costly so better to declare the size in advance using make command and insert the value directly to the defined positions.\n\nHope this helps :)\nDo share and shower some up votes if you like it. Comment for any further explanation.\n\n```\nfunc getMaximumXor(nums []int, maximumBit int) []int {\n size := len(nums)\n maxNum := int(math.Pow(float64(2), float64(maximumBit)) - 1)\n xorArr := make([]int, size)\n result := make([]int, size)\n xorArr[0] = nums[0]\n result[size-1] = maxNum-nums[0]\n for i:=1;i<size;i++{\n opr := xorArr[i-1]^nums[i]\n xorArr[i] = opr\n result[size-i-1] = maxNum-opr\n }\n return result\n}\n``` | 5 | 1 | ['Go'] | 2 |
maximum-xor-for-each-query | [C++] - Simple Solution | Explained | c-simple-solution-explained-by-morning_c-cwxv | Store cumulative xor in extra array\n2. Calculate max possible value of k (2^maximumBit-1)\n3. Xor will be maximum when all 0\'s will be converted to 1s and vic | morning_coder | NORMAL | 2021-04-17T16:15:30.761602+00:00 | 2021-04-18T01:03:38.905408+00:00 | 812 | false | 1. Store cumulative xor in extra array\n2. Calculate max possible value of k (2^maximumBit-1)\n3. Xor will be maximum when all 0\'s will be converted to 1s and vice versa. So compute this value by inverting element arr[i] \n4. Add inverted element to ans vector\n\n```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n=nums.size();\n vector<int> arr(n,0);\n arr[0]=nums[0];\n for(int i=1;i<n;i++){\n arr[i]=arr[i-1]^nums[i];\n }\n int idx=0;\n int k = (1<<maximumBit) - 1;\n \n for(int i=n-1;i>=0;i--){\n int curr=k & ~(arr[i]); //invertBits(arr[i]);\n nums[idx++]=curr;\n }\n return nums;\n }\n};\n``` | 5 | 0 | ['Bit Manipulation', 'C', 'Bitmask', 'C++'] | 2 |
maximum-xor-for-each-query | Easy Solve.....🔥🔥🔥 | easy-solve-by-pritambanik-9qzo | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | PritamBanik | NORMAL | 2024-11-08T17:14:39.682708+00:00 | 2024-11-08T17:14:39.682730+00:00 | 44 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: 100%\u2705\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: 100%\u2705\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```c []\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* getMaximumXor(int* nums, int numsSize, int maximumBit, int* returnSize) {\n int k = (1 << maximumBit) - 1, *res = NULL, xora = 0;\n\n *returnSize = numsSize;\n res = malloc(sizeof(int) * numsSize);\n\n for(int i=0; i<numsSize; i++){\n xora ^= nums[i];\n res[numsSize - 1 - i] = xora ^ k;\n }\n\n return res;\n}\n``` | 4 | 0 | ['Array', 'Bit Manipulation', 'C', 'Prefix Sum'] | 0 |
maximum-xor-for-each-query | XOR Approach | Simple | Python | Beats 95% of the users | xor-approach-simple-python-beats-95-of-t-sq6a | Intuition\nThe problem involves finding the maximum XOR result for each query when removing the last element from a given array and considering all previous ele | i_god_d_sanjai | NORMAL | 2024-11-08T15:20:06.749265+00:00 | 2024-11-08T15:20:06.749296+00:00 | 36 | false | # Intuition\nThe problem involves finding the maximum XOR result for each query when removing the last element from a given array and considering all previous elements. The XOR operation has a unique property: if you know the XOR from the beginning up to some point, you can deduce the XOR up to the end of the array by XOR-ing with that known value. \n\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n pre = [nums[0]]\n for i in range(1,len(nums)):\n pre.append(pre[-1]^nums[i])\n ma = (1<<maximumBit) - 1\n out = []\n for i in range(len(pre)-1,-1,-1):\n out.append(pre[i]^ma)\n return out\n``` | 4 | 0 | ['Python3'] | 0 |
maximum-xor-for-each-query | 💡 C++ | O(n) | Simple Logic | With Full Explanation ✏ | c-on-simple-logic-with-full-explanation-xea4c | Intuition\nThe problem requires finding the maximum XOR value from the given array for specific operations. \nXOR is a binary operation where the result is 1 if | Tusharr2004 | NORMAL | 2024-11-08T09:53:47.548129+00:00 | 2024-11-08T09:53:47.548157+00:00 | 22 | false | # Intuition\nThe problem requires finding the maximum XOR value from the given array for specific operations. \nXOR is a binary operation where the result is `1` if the bits are different and `0` if they are the same.\n To maximize the XOR value with a given number `x`, one approach is to XOR it with a number that has the highest number of `1`s in its binary representation, up to a certain bit length. \nThis leads us to use `(2^maximumBit - 1)` as the value to XOR against because it represents the highest possible value with `maximumBit` bits all set to `1`.\n\n\n# Approach\n\n1. **Prefix XOR Calculation**:\n - We start by calculating the cumulative XOR of the elements in the input array `nums`. This helps in finding the XOR of subarrays efficiently.\n - We store these cumulative XOR values in reverse order in a vector `pre` to align with the output requirement.\n\n2. **XOR with Maximum Possible Value**:\n - For each precomputed cumulative XOR value, we XOR it with `(2^maximumBit - 1)`. This operation effectively flips all the bits up to `maximumBit`, helping to get the maximum XOR value for that prefix.\n - Use `static_cast<int>(pow(2, maximumBit) - 1)` to calculate `(2^maximumBit - 1)`.\n\n3. **Return the Result**:\n - Modify the `nums` array to hold the result of each XOR operation and return it.\n\n### Detailed Steps:\n- Initialize a prefix XOR array `pre` with the same size as `nums`.\n- Traverse `nums` to calculate the cumulative XOR in reverse order and store it in `pre`.\n- For each value in `pre`, XOR it with `(2^maximumBit - 1)` and store the result back in `nums`.\n- Return `nums` as the result.\n\n\n# Complexity\n- Time complexity:\n```O(n)```\n\n- Space complexity:\n```O(n)```\n\n# Code\n```cpp []\n#pragma GCC optimize("Ofast") \nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n cin.tie(0);\n cout.tie(0);\n ios::sync_with_stdio(false);\n\n int n=nums.size();\n vector<int>pre(n,0);\n pre[n-1]=nums.at(0);\n int xo=nums[0];\n\n // Compute the prefix XOR from left to right and store in reverse order\n for(int i=1;i<nums.size();i++){\n xo^=nums.at(i); // XOR current value with the cumulative XOR\n pre[n-i-1]=xo; // Store the result in reverse index\n }\n \n // Calculate the result using XOR with (2^maximumBit - 1)\n for(int j=0;j<pre.size();j++){\n nums[j]=(static_cast<int>(pow(2,maximumBit)-1))^pre[j];\n }\n return nums;\n }\n};\n```\n\n\n\n# Ask any doubt !!!\nReach out to me \uD83D\uDE0A\uD83D\uDC47\n\uD83D\uDD17 https://tushar-bhardwaj.vercel.app/ | 4 | 0 | ['Array', 'Prefix Sum', 'C++'] | 0 |
maximum-xor-for-each-query | JAVA SOLTUION || 100% FASTER SOLUTION | java-soltuion-100-faster-solution-by-vip-9aa1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | viper_01 | NORMAL | 2024-11-08T07:06:20.079891+00:00 | 2024-11-08T07:06:20.079918+00:00 | 10 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int xor = 0;\n int n = nums.length;\n int mask = (int)(Math.pow(2, maximumBit)) - 1;\n\n for(int i : nums) {\n xor = xor ^ i;\n }\n\n int[] ans = new int[n];\n\n for(int i = 0; i < n; i++) {\n ans[i] = mask & (~(xor));\n\n xor = xor ^ nums[n - 1 - i];\n }\n\n return ans;\n \n }\n}\n``` | 4 | 0 | ['Array', 'Bit Manipulation', 'Java'] | 1 |
maximum-xor-for-each-query | The Great Bit Flip: Chasing Maximum XOR Glory! | the-great-bit-flip-chasing-maximum-xor-g-fbdp | Intuition\n Describe your first thoughts on how to solve this problem. \nTo maximize the XOR for each prefix in nums, we can calculate the cumulative XOR up to | LalithSrinandan | NORMAL | 2024-11-08T02:01:05.451892+00:00 | 2024-11-08T02:01:05.451919+00:00 | 97 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo maximize the XOR for each prefix in nums, we can calculate the cumulative XOR up to each index. XORing this cumulative result with the maximum possible value for a given number of bits (maximumBit) gives the desired results.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nCalculate Cumulative XOR:\n\nInitialize an array ans to store cumulative XOR values.\nFor each index i in nums, calculate the cumulative XOR up to i by XORing the previous cumulative XOR (ans[i-1]) with nums[i].\nCompute Maximum XOR Value:\n\nThe maximum value achievable with maximumBit bits is calculated as n = (1 << maximumBit) - 1.\nXOR each element in ans with n to maximize the XOR for each prefix.\nReverse the Array:\n\nReverse the ans array in-place to return the results in the required order.\nReturn Result:\n\nReturn the modified ans array.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n), where \nn is the length of nums. We iterate through nums once to calculate the cumulative XOR, then once to apply the maximum XOR, and finally once to reverse the array.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(n), as we store the cumulative XOR results in the ans array.\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int[] ans= new int[nums.length];\n ans[0]= nums[0];\n for(int i=1; i<nums.length; i++){\n ans[i]= ans[i-1]^nums[i];\n }\n int n= (int)(Math.pow(2, maximumBit)-1);\n for(int i=0; i<nums.length; i++){\n ans[i]= ans[i]^n;\n }\n int l= 0, r= ans.length-1;\n while(l<r){\n int t= ans[l];\n ans[l]= ans[r];\n ans[r]=t;\n l+=1;\n r-=1;\n }\n return ans;\n }\n}\n``` | 4 | 0 | ['Java'] | 0 |
maximum-xor-for-each-query | simple c++ code 93%faster | simple-c-code-93faster-by-ujjwal_pratik-zu9x | vector x;\n int i,a,xo=0;\n for(i=0;i<nums.size();i++)\n {\n xo=xo^nums[i];x.push_back(xo);\n }\n int k=pow(2,maxi | ujjwal_pratik | NORMAL | 2021-12-01T14:30:43.009628+00:00 | 2021-12-01T14:30:43.009668+00:00 | 325 | false | vector<int> x;\n int i,a,xo=0;\n for(i=0;i<nums.size();i++)\n {\n xo=xo^nums[i];x.push_back(xo);\n }\n int k=pow(2,maximumBit);\n for(i=0;i<x.size();i++)\n {\n \n x[i]=k-x[i]-1;\n }\n reverse(x.begin(),x.end());\n \n return x; | 4 | 0 | ['Math'] | 0 |
maximum-xor-for-each-query | Python easy to understand code, Using XOR property | python-easy-to-understand-code-using-xor-5c0z | Hints Given\nNote that the maximum possible XOR result is always 2^(maximumBit) - 1\nSo the answer for a prefix is the XOR of that prefix XORed with 2^(maximumB | palashbajpai214 | NORMAL | 2021-07-13T03:33:07.412120+00:00 | 2021-07-13T03:33:07.412166+00:00 | 323 | false | **Hints Given**\nNote that the maximum possible XOR result is always 2^(maximumBit) - 1\nSo the answer for a prefix is the XOR of that prefix XORed with 2^(maximumBit)-1\n\n\n\n```\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n max_possible=2**maximumBit-1 #max possible ans\n array_xor=0 #current array xor\n n=len(nums)\n \n ans=[] #stores ans of queries\n \n #finding xor of nums\n for i in nums:\n array_xor^=i\n \n ans.append(max_possible^array_xor)\n \n for i in range(n-1,0,-1):\n array_xor^=nums[i] #removing largest values from ans\n ans.append(array_xor^max_possible)\n \n return ans\n \n``` | 4 | 0 | ['Bit Manipulation', 'Python', 'Python3'] | 0 |
maximum-xor-for-each-query | C++ SOLUTION | c-solution-by-shruti_mahajan-atoy | \nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int r=0;\n for(auto i:nums)\n r=r^i;\ | shruti_mahajan | NORMAL | 2021-04-26T16:24:23.417418+00:00 | 2021-06-19T20:38:37.926314+00:00 | 239 | false | ```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int r=0;\n for(auto i:nums)\n r=r^i;\n long long p=pow(2,maximumBit);\n vector<int> res;\n for(int i=nums.size()-1;i>=0;i--)\n {\n res.push_back(p-1-r);\n r=r^nums[i];\n }\n return(res);\n }\n};\n``` | 4 | 0 | [] | 1 |
maximum-xor-for-each-query | [Python 3] One-liner faster than 100% | python-3-one-liner-faster-than-100-by-ki-65jz | \nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]: \n return list(accumulate([nums[0] ^ 2 ** maximumBi | kingjonathan310 | NORMAL | 2021-04-19T22:08:15.571053+00:00 | 2021-04-19T22:08:15.571126+00:00 | 259 | false | ```\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]: \n return list(accumulate([nums[0] ^ 2 ** maximumBit - 1] + nums[1:], ixor))[::-1]\n```\n\nor\n\n```\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]: \n return list(accumulate([nums[0] ^ (1 << maximumBit) - 1] + nums[1:], ixor))[::-1]\n``` | 4 | 0 | ['Python', 'Python3'] | 1 |
maximum-xor-for-each-query | Python, XOR'em'all | python-xoremall-by-warmr0bot-tuc9 | Idea\nMust know: \n XOR of a number with 0 is the number itself. \n XOR of a number with itself is 0.\n\nThe rest is bit manipulation based on the above facts:\ | warmr0bot | NORMAL | 2021-04-17T17:13:37.352380+00:00 | 2021-04-17T17:13:37.352408+00:00 | 249 | false | # Idea\nMust know: \n* XOR of a number with 0 is the number itself. \n* XOR of a number with itself is 0.\n\nThe rest is bit manipulation based on the above facts:\n```\ndef getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n\trxor = 0\n\tfor n in nums:\n\t\trxor ^= n\n\n\tmaxnum = (1 << maximumBit) - 1\n\tresults = []\n\twhile nums:\n\t\tlw = (rxor & maxnum) ^ maxnum\n\t\tresults.append(lw)\n\t\trxor ^= nums.pop()\n\n\treturn results\n``` | 4 | 2 | ['Python', 'Python3'] | 2 |
maximum-xor-for-each-query | [C++] Easy to understand solution (100% time and 100% space) | c-easy-to-understand-solution-100-time-a-ykkz | \nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n vector<int> result(nums.size());\n \n int | bhaviksheth | NORMAL | 2021-04-17T16:11:42.887903+00:00 | 2021-04-17T16:12:55.570658+00:00 | 212 | false | ```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n vector<int> result(nums.size());\n \n int maxInt = pow(2, maximumBit) - 1;\n \n int bit = nums[0];\n \n for (int i = 1; i < nums.size(); ++i) bit ^= nums[i];\n \n for (int i = 0; i < nums.size(); ++i) {\n result[i] = bit ^ maxInt;\n bit ^= nums[nums.size() - i - 1];\n }\n \n return result;\n }\n};\n``` | 4 | 0 | [] | 0 |
maximum-xor-for-each-query | 💢☠💫Easiest👾Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅C#✅💥🔥💫Explained☠💥🔥 Beats 100 | easiestfaster-lesser-cpython3javacpython-jtx9 | \n\n# Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n- JavaScript Code | Edwards310 | NORMAL | 2024-11-08T11:18:31.237285+00:00 | 2024-11-08T11:18:31.237316+00:00 | 64 | false | \n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- ***JavaScript Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446669789\n- ***C++ Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446659383\n- ***Python3 Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446664401\n- ***Java Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446656714\n- ***C Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446666606\n- ***Python Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446664168\n- ***C# Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446668407\n- ***Go Code -->*** https://leetcode.com/problems/maximum-xor-for-each-query/submissions/1446673049\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n O(N)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n O(N)\n# Code\n\n | 3 | 0 | ['Array', 'Bit Manipulation', 'C', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3', 'JavaScript', 'C#'] | 0 |
maximum-xor-for-each-query | EASY || 100% ASSURED ANSWER || EASY PROCEDURE || BEST APPROACH | easy-100-assured-answer-easy-procedure-b-ly72 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | HarshvinderSingh | NORMAL | 2024-11-08T10:22:50.541540+00:00 | 2024-11-08T10:22:50.541578+00:00 | 11 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n final int n = nums.length;\n final int mx = (1 << maximumBit) - 1;\n int[] ans = new int[n];\n int xors = 0;\n\n for (int i = 0; i < n; ++i) {\n xors ^= nums[i];\n ans[n - 1 - i] = xors ^ mx;\n }\n\n return ans;\n }\n}\n``` | 3 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Java'] | 0 |
maximum-xor-for-each-query | Kotlin | Rust | kotlin-rust-by-samoylenkodmitry-z3gb | \n\nhttps://youtu.be/yoKp2xrbnk4\n\n#### Join me on Telegram\n\nhttps://t.me/leetcode_daily_unstoppable/794\n\n#### Problem TLDR\n\nRunning xor to make 2^k-1 #m | SamoylenkoDmitry | NORMAL | 2024-11-08T07:33:49.127448+00:00 | 2024-11-08T07:34:04.479500+00:00 | 63 | false | \n\nhttps://youtu.be/yoKp2xrbnk4\n\n#### Join me on Telegram\n\nhttps://t.me/leetcode_daily_unstoppable/794\n\n#### Problem TLDR\n\nRunning `xor` to make `2^k-1` #medium #bit_manipulation\n\n#### Intuition\n\nLet\'s observe what\'s happening:\n\n```j\n\n // n xor[..i] k < 2^mb(=b100)\n // b00 0 00 00^x = 11 11 = 3\n // b01 1 01 01^x = 11 10 = 2\n // b01 1 00 00^x = 11 11 = 3\n // b11 3 11 11^x = 11 00 = 0\n // 100 ans = [ 0 3 2 3 ]\n\n```\nFor `k=2` we have to make maximum `2^k-1 = b011`. Consider each column of bits independently: we can count them and `even` would give `0`, odd `1`. So, one way to solve is to count `k` bits and set all that happend to be `even` to `1`.\nOn a second thought, all this equalized into a `xor` operation: `xor[0..i] ^ res[i] = 0b11`.\n\n#### Approach\n\n* we don\'t have to do `xor 2^k-1` on each item, just start with it\n* let\'s use `scan` iterator in Kotlin\n* Rust also has a `scan` but it is more verbose\n\n#### Complexity\n\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n#### Code\n\n```kotlin []\n\n fun getMaximumXor(nums: IntArray, maximumBit: Int) = nums\n .scan((1 shl maximumBit) - 1) { r, t -> r xor t }\n .drop(1).reversed()\n\n```\n```rust []\n\n pub fn get_maximum_xor(nums: Vec<i32>, maximum_bit: i32) -> Vec<i32> {\n let mut r = (1 << maximum_bit) - 1;\n let mut res = nums.iter().map(|n| { r ^= n; r }).collect::<Vec<_>>();\n res.reverse(); res\n }\n\n```\n```c++ []\n\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int x = (1 << maximumBit) - 1, i = nums.size(); vector<int> res(i); \n for (;i;i--) res[i - 1] = x ^= nums[nums.size() - i];\n return res;\n }\n\n``` | 3 | 0 | ['Bit Manipulation', 'C++', 'Rust', 'Kotlin'] | 1 |
maximum-xor-for-each-query | Simple and Easy O(n) Explained code | simple-and-easy-on-explained-code-by-ahe-0091 | Maximum XOR Bound: We begin by calculating max_XOR = 2^maximumBit - 1. This value represents the highest possible XOR within the given maximumBit bit limit, as | ahen11 | NORMAL | 2024-11-08T06:14:05.205817+00:00 | 2024-11-08T06:14:05.205847+00:00 | 52 | false | 1. **Maximum XOR Bound:** We begin by calculating `max_XOR = 2^maximumBit - 1`. This value represents the highest possible XOR within the given `maximumBit` bit limit, **as it has all bits set to 1**. Using this value ensures that for each query, we maximize the XOR result.\n\n2. **Initializing `xor_till_now`:** We set `xor_till_now` to 0 before starting the loop. This initialization is important because 0 acts as the identity for XOR operations, meaning `0 ^ a = a` for any integer `a`. This allows us to progressively build `xor_till_now` by XORing it with each element in nums.\n\n3. **Iterating Through nums:** For each element `n` in `nums`, we update `xor_till_now` by XORing it with the current element (`xor_till_now ^= n`). This accumulates the XOR of all elements encountered so far, giving us the cumulative XOR at each step.\n\n5. **Results (k) in Reverse Order:** For each query, the required `k` is calculated by XORing `xor_till_now` with `max_XOR`. As the problem requires `k` for each query, by **removing elements from the end of `nums` for every next query**, we store each result(k) from the end of the `ans` array by decrementing an index variable `i` from -1 backwards.\n\n```python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n ans = [0]*len(nums)\n max_XOR = pow(2,maximumBit)-1\n xor_till_now = 0 \n i = -1\n for n in nums:\n xor_till_now ^= n\n ans[i]=max_XOR ^ xor_till_now\n i-=1\n return ans\n\n``` | 3 | 0 | ['Python3'] | 0 |
maximum-xor-for-each-query | Very intuitive one pass solution | Beats 100% ✅| Java | C++ | Python3 | very-intuitive-one-pass-solution-beats-1-r2ko | Intuition\nTo maximize the XOR for each query, we need to find a number k such that, when XORed with the cumulative XOR of all numbers in nums (let\u2019s call | prakharpandey1198 | NORMAL | 2024-11-08T05:58:28.651372+00:00 | 2024-11-08T05:58:28.651399+00:00 | 158 | false | # Intuition\nTo maximize the XOR for each query, we need to find a number k such that, when XORed with the cumulative XOR of all numbers in nums (let\u2019s call it currXor), it yields the maximum possible result.\n\nThe maximum possible result given maximumBit is 2^maximumBit - 1 (which has all bits up to maximumBit set to 1), referred to here as maxNum. This leads us to the equation:\n\n**currXor \u2295 \uD835\uDC58 = maxNum**\n\nFrom this, by XORing both sides with currXor, we find:\n\n**\uD835\uDC58 = maxNum \u2295 currXor**\n\nThus, for each query, the optimal value of k can be obtained simply as maxNum ^ currXor.\n\n# Approach\n\nInitialize maxNum: Calculate maxNum as (1 << maximumBit) - 1, which represents the maximum value within the bit limit, with all bits up to maximumBit set to 1.\n\nCumulative XOR (currXor): Iterate over nums and maintain a cumulative XOR of all elements encountered so far.\n\nCompute k for Each Query: For each query, k is computed as maxNum ^ currXor, yielding the maximum achievable XOR value for that query. We store the result in the answer array in reverse order to simulate removing the last element in each query.\n\nReturn the Result: The answer array stores the results for each query in the required order.\n\n# Complexity\nTime complexity: O(n)\n\nSpace complexity: O(1) we are not using any space for algorithmic purpose so O(1) is actual space complexity.\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int [] answer= new int[nums.length];\n int currXor = 0;\n int maxNum = (1 << maximumBit) - 1; // Set all bits up to maximumBit to 1\n\n for(int i = 0; i < nums.length; i++) {\n currXor ^= nums[i]; // Update cumulative XOR\n answer[nums.length - i - 1] = currXor ^ maxNum; // Find k by flipping bits in totalXor\n }\n\n return answer;\n }\n}\n\n\n```\n\n\n```python []\nclass Solution:\n def getMaximumXor(self, nums, maximumBit):\n answer = [0] * len(nums)\n totalXor = 0\n maxNum = (1 << maximumBit) - 1 # maxNum with all bits set up to maximumBit\n \n for i in range(len(nums)):\n totalXor ^= nums[i] # Update cumulative XOR\n answer[len(nums) - i - 1] = totalXor ^ maxNum # Calculate k and store in reverse order\n \n return answer\n```\n```c++ []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n vector<int> answer(nums.size());\n int totalXor = 0;\n int maxNum = (1 << maximumBit) - 1; // maxNum with all bits set up to maximumBit\n \n for (int i = 0; i < nums.size(); ++i) {\n totalXor ^= nums[i]; // Update cumulative XOR\n answer[nums.size() - i - 1] = totalXor ^ maxNum; // Calculate k and store in reverse order\n }\n \n return answer;\n }\n};\n\n```\n | 3 | 0 | ['C++', 'Java', 'Python3'] | 0 |
maximum-xor-for-each-query | Easiest Solution🔥| Beats 100% ✅|Optimal Approach | C++ | Prefix XOR | easiest-solution-beats-100-optimal-appro-sa6d | Intuition\nTo solve this problem, we need to leverage the XOR properties, which allow us to toggle bits between 1 and 0. We can utilize a prefix XOR array to st | ishanbagra | NORMAL | 2024-11-08T04:51:20.853653+00:00 | 2024-11-08T04:51:20.853693+00:00 | 18 | false | Intuition\nTo solve this problem, we need to leverage the XOR properties, which allow us to toggle bits between 1 and 0. We can utilize a prefix XOR array to store cumulative XOR results up to each element in nums, making it easier to calculate the desired maximum XOR values efficiently.\n\nApproach\nCalculate the Maximum XOR Value: Since maximumBit defines the bit length, the maximum possible XOR value that can be achieved with maximumBit bits is (1 << maximumBit) - 1. This value will help in toggling all bits up to maximumBit and allows us to maximize XOR for each prefix result.\n\nCreate a Prefix XOR Array:\n\nInitialize a prefix XOR array (prefixor) where each entry at index i contains the cumulative XOR of elements from the start up to i in nums.\nPopulate prefixor such that each prefixor[i] equals prefixor[i-1] ^ nums[i], with prefixor[0] = nums[0].\nCompute the Maximum XOR Values in Reverse Order:\n\nTo build the ans array, traverse the prefixor array from the end to the beginning.\nFor each element, XOR it with max (the maximum achievable XOR value with maximumBit bits) and store the result in ans.\nReturn ans:\n\nSince we computed elements in reverse, we have stored them in the desired order directly in ans.\nComplexity\nTime Complexity:O(n), where n is the number of elements in nums. This is because we traverse nums twice: once to build the prefix XOR array and once to compute the ans values.\n\nSpace Complexity:O(n), as we use an additional array (prefixor)ofsize n and another vector (ans) to store the results.\n\n# Code\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size();\n int max = (1 << maximumBit) - 1; \n vector<int> prefixor(n);\n vector<int> ans;\n\n prefixor[0] = nums[0];\n \n for (int i = 1; i < n; i++) {\n prefixor[i] = prefixor[i - 1] ^ nums[i];\n }\n\n for (int i = 0; i < n; i++) {\n ans.push_back(prefixor[n - i - 1] ^ max);\n }\n \n return ans;\n }\n};\n\n```\n\n | 3 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'C++'] | 1 |
maximum-xor-for-each-query | Bit Manipulation || Bit Mask || Proper Explanation | bit-manipulation-bit-mask-proper-explana-183e | Intuition\nThe problem requires finding a way to get the maximum XOR of elements up to each index in a specific way. The first thought is to use bitwise manipul | Anurag_Basuri | NORMAL | 2024-11-08T04:14:31.610414+00:00 | 2024-11-08T04:14:31.610439+00:00 | 102 | false | # Intuition\nThe problem requires finding a way to get the maximum XOR of elements up to each index in a specific way. The first thought is to use bitwise manipulation to keep track of the XOR of the entire array, updating it as elements are "removed" in the specified order, and then to obtain the maximum XOR for each step.\n\n# Approach\n1. Start by calculating the XOR of all elements in the input list `nums`.\n2. Compute a bitmask based on `maximumBit`. This mask ensures that we only consider the lower bits up to `maximumBit`, thus ignoring any higher bits.\n3. As we traverse `nums` in reverse order:\n - Calculate the XOR with the mask to get the maximum XOR value at that step.\n - Add this maximum XOR value to the result list.\n - Update the cumulative XOR by removing the effect of the current element.\n4. Return the result list at the end, which holds the maximum XOR values for each step in reverse order.\n\n# Complexity\n- **Time complexity:** \\(O(n)\\), where \\(n\\) is the number of elements in `nums`, since we traverse `nums` only once.\n- **Space complexity:** \\(O(n)\\), as we store the results in an output list.\n\n\n# Code\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n vector<int> answer;\n int XOR = 0;\n for(int i:nums) XOR ^= i;\n\n int mask = (1 << maximumBit) - 1;\n for(int i = nums.size() - 1; i >= 0; i--){\n answer.push_back(XOR ^ mask);\n XOR ^= nums[i];\n }\n\n return answer;\n }\n};\n```\n``` python []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n answer = []\n XOR = 0\n for num in nums:\n XOR ^= num\n\n mask = (1 << maximumBit) - 1\n for i in range(len(nums) - 1, -1, -1):\n answer.append(XOR ^ mask)\n XOR ^= nums[i]\n\n return answer\n``` | 3 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'C++', 'Python3'] | 2 |
maximum-xor-for-each-query | Easiest Solution | Beats 100% ✅| Optimal Approach | C++ | Java | Python3 | Prefix XOR | easiest-solution-beats-100-optimal-appro-hwas | Upvote if it helps \n\n{:height="30px" width="30px"} \n\n---\n\n### C++ Code\ncpp\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, | BijoySingh7 | NORMAL | 2024-11-08T01:49:41.017530+00:00 | 2024-11-08T02:20:56.021688+00:00 | 310 | false | # Upvote if it helps \n\n{:height="30px" width="30px"} \n\n---\n\n### C++ Code\n```cpp\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n vector<int> ans;\n int n = nums.size();\n int max_possible = (1 << maximumBit) - 1; \n int total_XOR = 0;\n\n // Calculate the cumulative XOR of the entire array\n for (int num : nums) {\n total_XOR ^= num;\n }\n\n // Generate the result for each query\n for (int i = 0; i < n; i++) {\n ans.push_back(total_XOR ^ max_possible);\n total_XOR ^= nums[n - 1 - i];\n }\n\n return ans;\n }\n};\n```\n\n### Python Code\n```python\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n ans = []\n max_possible = (1 << maximumBit) - 1\n total_XOR = 0\n\n # Calculate the cumulative XOR of the entire array\n for num in nums:\n total_XOR ^= num\n\n # Generate the result for each query\n for i in range(len(nums)):\n ans.append(total_XOR ^ max_possible)\n total_XOR ^= nums[-1 - i]\n\n return ans\n```\n\n### Java Code\n```java\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length;\n int[] ans = new int[n];\n int max_possible = (1 << maximumBit) - 1;\n int total_XOR = 0;\n\n // Calculate the cumulative XOR of the entire array\n for (int num : nums) {\n total_XOR ^= num;\n }\n\n // Generate the result for each query\n for (int i = 0; i < n; i++) {\n ans[i] = total_XOR ^ max_possible;\n total_XOR ^= nums[n - 1 - i];\n }\n\n return ans;\n }\n}\n```\n\n---\n\n### Explanation\n\n1. **Initial Observations**:\n - We need to calculate the cumulative XOR of all numbers in the `nums` array.\n - For each query, we are required to find an integer `k` such that the XOR of all elements plus `k` gives the maximum possible value with `maximumBit` bits set.\n\n2. **Approach**:\n - **Calculate Maximum Possible Value**: The maximum value achievable with `maximumBit` bits is `(1 << maximumBit) - 1`. For example, if `maximumBit` is 3, then the maximum is `7` (binary `111`).\n - **Cumulative XOR**: First, compute the cumulative XOR of all elements in the array to get the initial XOR result.\n - **Looping through Queries**:\n - For each query, calculate `k` by XORing `total_XOR` with `max_possible`.\n - This gives the optimal `k` to maximize the XOR operation.\n - After calculating `k`, remove the last element of `nums` from the cumulative XOR by XORing it out. This simulates removing elements from the end of `nums` in each query.\n\n3. **Complexity Analysis**:\n - **Time Complexity**: \\(O(n)\\) since we are iterating through `nums` twice (once to calculate the cumulative XOR and once to generate the answers).\n - **Space Complexity**: \\(O(n)\\) for storing the results in `ans`.\n\n\n | 3 | 1 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Python', 'C++', 'Java', 'Python3'] | 4 |
maximum-xor-for-each-query | Maximum xor for each query - Easy Solution - Nice Explanation - 🌟Beats 100%🌟 | maximum-xor-for-each-query-easy-solution-q8r4 | Intuition\n1. Total prefix XOR for the entire nums array is calculated and using that , we\'ll find the value of k.\n\n1. The value of k can be find by reverse | RAJESWARI_P | NORMAL | 2024-11-08T01:22:03.819666+00:00 | 2024-11-08T01:22:03.819689+00:00 | 84 | false | # Intuition\n1. Total prefix XOR for the entire nums array is calculated and using that , we\'ll find the value of k.\n\n1. The value of k can be find by reverse processing of XOR-ing the maximum XOR with the available array and we find the k-value and is updated.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n**Initialization**:\n\n no of query to be processed is equal to the no of elements in the array.Since at the end of each query, last element is removed from the query.\n \n maxVal is initialized with 2 power of maximumBit - 1 in which all the bits are of 1.\n\n Declare the query_answer array of length no_query.\n\n Initialize curr_xor as 0 to perform prefix curr_xor of all the elements in the entire array.\n\n int no_query = nums.length;\n int maxVal = (1 << maximumBit) - 1; // Maximum possible value with `maximumBit` bits\n int[] query_answer = new int[no_query];\n int curr_xor = 0;\n\n**Prefix xor**:\n XOR of the entire array is calculated by xor-ing curr_xor with each elements of the nums array.\n\n // Compute the prefix XOR for the entire array\n for (int num : nums) {\n curr_xor ^= num;\n }\n\n**Process query**:\n Reverse process is followed to find k for each query by xor-ing the current xor (curr_xor) with the maxVal and storing it in the query_answer array.\n\n Since after each query processing, the last element is removed , so to reduce its effect -> xor-ing the prefix xor curr_xor with the last element of the nums array.\n\n for (int i = 0; i < no_query; i++) {\n query_answer[i] = curr_xor ^ maxVal; // Maximum XOR for the current prefix\n curr_xor ^= nums[no_query - 1 - i]; // Remove the last element\'s effect for the next query\n }\n\n**Returning a value**:\n Finally, the query_answer is returned which contains the k value in each query processing steps.\n\n return query_answer;\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- ***Time complexity***: **O(n)**.\n n -> number of elements in the nums array.\n\n**Prefix xor calculation(First loop)**: **O(n)**.\n This loop iterates through elements in the array atleast once.\n At each iteration , xor is calculated which may take O(1) time.Hence the overall complexity is **O(n) time**.\n\n**Processing each query in reverse order(Second loop)**: **O(n)**.\n This loop also iterates through each element in nums once to answer each query in reverse order.\n The XOR operation and accessing elements from the array each take constant time, so this loop also takes **O(n) time**.\n\n**Overall Time Complexity**: **O(n)**.\n Since in both the loop, the time complexity is O(n).\n Hence the overall time complexity is **O(n)+O(n) = O(n)**.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- ***Space complexity***: **O(n)**.\n n -> number of elements in the nums array.\n\n**Storage for query_answer array**: **O(n)**.\n query_answer is an array of size n to store the result for each query.This requires **O(n) space**.\n\n**Other variables**: **O(1)**.\n no_query, maxVal, and curr_xor are single integer variables, requiring **O(1) space each**.\n\n**Overall Space Complexity**: **O(n)**.\n Since the primary space usage comes from the query_answer array, the overall space complexity of the program is: **O(n)**.\n\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\npublic class Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int no_query = nums.length;\n int maxVal = (1 << maximumBit) - 1; // Maximum possible value with `maximumBit` bits\n int[] query_answer = new int[no_query];\n int curr_xor = 0;\n \n // Compute the prefix XOR for the entire array\n for (int num : nums) {\n curr_xor ^= num;\n }\n \n // Process each query in reverse order\n for (int i = 0; i < no_query; i++) {\n query_answer[i] = curr_xor ^ maxVal; // Maximum XOR for the current prefix\n curr_xor ^= nums[no_query - 1 - i]; // Remove the last element\'s effect for the next query\n }\n \n return query_answer;\n }\n}\n``` | 3 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Java'] | 0 |
maximum-xor-for-each-query | Straight forward approach ✅✅✅| beats 100% 💯🫡| easy to understand C++🎯 | straight-forward-approach-beats-100-easy-e9j2 | \n\n\n# Code\ncpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int x = 0, mask = (1 << maximumBit) | ritik6g | NORMAL | 2024-11-08T00:39:18.827826+00:00 | 2024-11-08T00:39:18.827856+00:00 | 198 | false | \n\n\n# Code\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int x = 0, mask = (1 << maximumBit) - 1;\n vector<int> res(nums.size());\n for (int num : nums)\n x ^= num;\n for (int i = nums.size() - 1; i >= 0; --i) {\n res[nums.size() - 1 - i] = x ^ mask;\n x ^= nums[i];\n }\n return res;\n }\n};\n\n``` | 3 | 0 | ['C++'] | 1 |
maximum-xor-for-each-query | SIMPLE PREFIX SUM C++ SOLUTION | simple-prefix-sum-c-solution-by-jeffrin2-2krl | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Jeffrin2005 | NORMAL | 2024-07-31T09:14:37.429112+00:00 | 2024-07-31T09:14:37.429137+00:00 | 229 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:o(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:o(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size();\n vector<int> answer(n);\n int maxXor = (1 << maximumBit) - 1; \n vector<int>prefixXor(n + 1);\n prefixXor[0] = nums[0];\n for(int i = 1; i < n;i++){\n prefixXor[i] = prefixXor[i - 1] ^ nums[i];\n }\n for(int i = 0; i < n;i++){\n answer[i] = maxXor ^ prefixXor[n - 1 - i]; \n }\n\n return answer;\n }\n};\n``` | 3 | 0 | ['C++'] | 1 |
maximum-xor-for-each-query | ✅ C++ | Intuition | Clean | 2 pass | c-intuition-clean-2-pass-by-lastropy-sj6r | Please Upvote if it helps\n\nIntuition - \nk -> [0 , pow(2,maximumBit) - 1]\n\nnow, let\'s say,\nRight now (j) elements are present in array, and their xor is ( | lastropy | NORMAL | 2022-07-28T13:23:28.895867+00:00 | 2022-08-05T07:03:03.573504+00:00 | 368 | false | **Please Upvote if it helps**\n\nIntuition - \nk -> [0 , pow(2,maximumBit) - 1]\n\nnow, let\'s say,\nRight now (j) elements are present in array, and their xor is (xr).\n\nWe want : xr ^ k -> maximum.\nNow, what is maximum xor possible ? \n*pow(2 , maximumBit) -1.*\n**Actually this is because " 0 <= nums[i] <= pow(2, maximumBit) - 1" as given in constraints**\n\n\nNow, let\'s say we want :\nxr ^ k = pow(2 , maximumBit) -1.\nNow, xor\'ing with (xr) on both sides -\nxr ^ xr ^ k = xr ^ (pow(2 , maximumBit) -1)\n\n## using property of xor, if, we xor a number 2 times to a value, we get the result as the value itself. \n\ni.e. x ^ x ^ y = y\nUsing xor property, we can simplify it as: *k = xr ^ (pow(2 , maximumBit) -1)*\n\nNow, all we need to do is to calculate (xr) for all elements and then keep moving pointer back by one and then, re-calculate (xr). Repeat until pointer is not -1. ( Thanks to @striker000 for the suggestion to not modify the input array. )\n\nAgain, using property of xor, if, we xor a number 2 times to a value, we get the result as the value itself.\n\ni.e. x ^ x ^ y = y.\n\nThis is shown in code.\n```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int xr = 0;\n for(int i: nums)\n xr = xr ^ i;\n vector<int> ans;\n int mx = pow(2, maximumBit)-1;\n\t\tfor( int pointer = nums.size() - 1; pointer >= 0; pointer--){\n int val = mx ^ xr;\n ans.push_back(val);\n xr = xr ^ nums[pointer];\n }\n return ans;\n }\n};\n```\nTC -> O(n) where n is size of nums (2 pass)\nSC -> O(n) because of ans array\n\n**Please Upvote if it helps** | 3 | 0 | ['Bit Manipulation', 'C'] | 1 |
maximum-xor-for-each-query | java easy to understand | beginner friendly | java-easy-to-understand-beginner-friendl-yno9 | \nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int[] xors = new int[nums.length];\n xors[0] = nums[0];\n | rmanish0308 | NORMAL | 2022-04-26T06:53:47.825980+00:00 | 2022-04-26T06:53:47.826023+00:00 | 238 | false | ```\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int[] xors = new int[nums.length];\n xors[0] = nums[0];\n for(int i=1;i<nums.length;i++)\n xors[i] = xors[i-1]^nums[i];\n \n int[] ans = new int[nums.length];\n int max = (int)Math.pow(2,maximumBit)-1;\n for(int i=0;i<nums.length;i++)\n ans[nums.length-1-i] = max^xors[i];\n return ans;\n }\n}\n```\nPlease upvote if u find my code easy to understand | 3 | 0 | ['Java'] | 1 |
maximum-xor-for-each-query | Beginner friendly JavaScript Solution | beginner-friendly-javascript-solution-by-r9wc | Time Complexity : O(n)\n\n/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function(nums, maximumB | HimanshuBhoir | NORMAL | 2022-02-15T04:06:40.619434+00:00 | 2022-02-15T04:06:40.619477+00:00 | 175 | false | **Time Complexity : O(n)**\n```\n/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function(nums, maximumBit) {\n let xor = (1 << maximumBit) - 1\n for(let i=0; i<nums.length; i++){\n xor ^= nums[i]\n nums[i] = xor\n }\n return nums.reverse()\n};\n``` | 3 | 0 | ['JavaScript'] | 0 |
maximum-xor-for-each-query | Python3 solution using single for loop | python3-solution-using-single-for-loop-b-qd8k | \nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n res = []\n for i in range(1,len(nums)):\n | EklavyaJoshi | NORMAL | 2021-06-18T18:45:59.687121+00:00 | 2021-06-18T18:45:59.687165+00:00 | 254 | false | ```\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n res = []\n for i in range(1,len(nums)):\n res.append(2**maximumBit - 1 - nums[i-1])\n nums[i] = nums[i-1]^nums[i]\n res.append(2**maximumBit - 1 - nums[-1])\n return res[::-1]\n```\n**If you like this solution, please upvote for this** | 3 | 0 | ['Python3'] | 1 |
maximum-xor-for-each-query | JAVA O(n) time | java-on-time-by-adarsh_goswami-o86u | 1.To get the max xor value we can toggle every bit of xor_prefix to get k \n2.But we have a constraint that k should be smaller than 2^maxBit so all we need to | adarsh_goswami | NORMAL | 2021-04-17T16:21:30.648494+00:00 | 2021-04-18T01:12:50.964210+00:00 | 158 | false | 1.To get the max xor value we can toggle every bit of xor_prefix to get k \n2.But we have a constraint that k should be smaller than 2^maxBit so all we need to do is consider only the last maximumBits of the value k which we calc in step 1. \n\n\n\tclass Solution {\n private int max= (int)((1l<<31)-1);\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n= nums.length;\n int[] ans= new int[n];\n \n int xor_prefix= 0;\n for(int i= 0;i<n;i++) {\n xor_prefix^= nums[i];\n int k= xor_prefix^max;\n k= k& (1<<maximumBit)- 1;\n ans[n-i-1]= k;\n }\n \n return ans;\n }\n} | 3 | 1 | [] | 0 |
maximum-xor-for-each-query | [C++/Java] Get the max K by answer.push_back(lim ^ total_XOR); | cjava-get-the-max-k-by-answerpush_backli-i79o | C++\n\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n // get the max limit, in order to get the max K\n | guanwenw | NORMAL | 2021-04-17T16:12:37.280678+00:00 | 2021-04-17T16:12:37.280709+00:00 | 159 | false | C++\n```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n // get the max limit, in order to get the max K\n int lim = (int)pow(2, maximumBit) - 1;\n int tot = 0;\n vector<int> answer;\n \n // get the XOR result for all elements\n for(int i : nums){\n tot = tot ^ i;\n }\n \n // get the first max k, all 1s is the max\n answer.push_back(lim ^ tot);\n for(int i = nums.size() - 1; i >= 1; i--){\n // remove the last element by XOR\n tot = tot ^ nums[i]; \n answer.push_back(lim ^ tot);\n }\n return answer; \n }\n};\n```\nJava\n```\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n // get the max limit, in order to get the max K\n int lim = (int)Math.pow(2, maximumBit) - 1;\n int tot = 0;\n int[] answer = new int[nums.length];\n \n // get the XOR result for all elements\n for(int i : nums){\n tot = tot ^ i;\n }\n \n // get the first max k, all 1s is the max\n answer[0] = lim ^ tot;\n for(int i = nums.length - 1; i >= 1; i--){\n // remove the last element by XOR\n tot = tot ^ nums[i]; \n answer[nums.length - 1 - i+1] = lim ^ tot;\n }\n return answer;\n }\n}\n``` | 3 | 1 | [] | 0 |
maximum-xor-for-each-query | Java Easy to understand O(n) | java-easy-to-understand-on-by-manjumalle-n5oe | If A xor B = C Then A xor C = B\n\n\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n \xA0 \xA0 \xA0 \xA0int max = (int)(Math.po | manjumallesh | NORMAL | 2021-04-17T16:09:54.914892+00:00 | 2021-04-17T16:31:29.878970+00:00 | 177 | false | If A xor B = C Then A xor C = B\n\n```\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n \xA0 \xA0 \xA0 \xA0int max = (int)(Math.pow(2,maximumBit));\n int[] res = new int[nums.length];\n int xorSum = 0;\n for(int x : nums){\n xorSum ^= x;\n }\n for(int i = 0; i < nums.length; i++){\n \xA0 \xA0 \xA0 \xA0 \xA0 \xA0res[i] = (xorSum ^ max - 1);\n xorSum ^= nums[nums.length - i - 1];\n }\n return res;\n }\n}\n``` | 3 | 1 | [] | 1 |
maximum-xor-for-each-query | O(n) solution || Prefix XOR || Maximum XOR for Each Query | on-solution-prefix-xor-maximum-xor-for-e-nbzs | The solution is very simple. We have to just calculate the xor value of every subarray required.\nWe can do this in O(n) with prefix array properties.\n\nAfter | tannatsri | NORMAL | 2021-04-17T16:01:06.498309+00:00 | 2021-04-17T16:01:06.498341+00:00 | 276 | false | The solution is very simple. We have to just calculate the xor value of every subarray required.\nWe can do this in O(n) with prefix array properties.\n\nAfter calculating the prefix array, we have to find the maximum xor value which we can get. \nThe maximum value which we can get is `2 ^ maximumBits - 1` (**inverse of XOR is XOR**).\n\nSo now we have to just calculate the value of `k` for every subarray.\n\nLet `x = 2 ^ maximumBits - 1`.\nTherefore, `k = a[i] ^ x`\n\nNow we have to just calulatte for each Prefix XOR array in `reverse` order.\n\nTotal time complexity `O(n)`\nTotal space complexity `O(n)`\n\n\n```\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size();\n int prefixXOR[100004] = {0};\n prefixXOR[0] = nums[0];\n for(int i = 1; i < n; ++i) {\n prefixXOR[i] = prefixXOR[i - 1] ^ nums[i];\n \n }\n int x = pow(2,maximumBit) - 1;\n vector<int> ans;\n for(int i = n - 1; i >= 0; --i) {\n ans.push_back(prefixXOR[i] ^ x);\n }\n return ans;\n \n }\n};\n``` | 3 | 0 | [] | 0 |
maximum-xor-for-each-query | Beats 100% 🥇 || Easy Java Solution || Beginner Friendly ✅ | beats-100-easy-java-solution-beginner-fr-mx2z | \n\n# Intuition\n1.\tThe problem requires finding a value k such that the XOR of all elements in nums with k is maximized.\n2.\tSince each k must be less | KallemSahana | NORMAL | 2024-11-08T19:59:02.564570+00:00 | 2024-11-08T19:59:02.564611+00:00 | 7 | false | \n\n# Intuition\n1.\tThe problem requires finding a value k such that the XOR of all elements in nums with k is maximized.\n2.\tSince each k must be less than 2^{maximumBit} , the maximum value k can take is 2^{maximumBit} - 1 , which is essentially a bitmask with all bits set to 1 up to the given maximumBit.\n3.\tThe XOR operation between a number and a bitmask with all bits set to 1 yields the maximum possible XOR for that number, effectively \u201Cflipping\u201D all bits within the constraints of maximumBit.\n4.\tWe maintain a cumulative XOR of the elements in nums, and with each query, we calculate the XOR with the maximum possible k (bitmask) to get the maximum XOR for the current array state.\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1.\tCalculate the maximum possible k as {max} = 2^{maximumBit} - 1 .\n2.\tCompute the cumulative XOR of the entire array, stored in s. This XOR represents the result of XOR-ing all elements in nums at the start.\n3.\tFor each query, the answer for the current state is s ^ max, which flips the bits of s to give the maximum XOR.\n4.\tAfter calculating the answer for the current state, remove the last element of nums from s by XOR-ing s with nums[len - i - 1] (starting from the last element and moving backward).\n5.\tRepeat the process for each query, storing the result in an array ans and decrementing the effective size of nums by removing the last element.\n\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: **O(n)** , where n is the length of nums.\n\u2022\tCalculating the initial cumulative XOR takes O(n).\n\u2022\tEach query requires a constant time operation, as we are just performing XOR operations and updating s, so processing all queries also takes O(n).\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: **O(n).**\n\u2022\tWe use an additional array ans of size n to store the results, so the space complexity is O(n) . The space for s and max is O(1) , which is negligible in this context.\n\n\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int len = nums.length;\n int[] ans = new int[len];\n int s = 0, max = (int)Math.pow(2, maximumBit) -1;\n // Calculate cumulative XOR of all elements in nums\n for(int i=0; i<len; i++){\n s ^= nums[i];\n }\n // Process each query in reverse\n for(int i =0; i<len; i++){\n ans[i] = s^max; // Maximum XOR for current state\n s = s^nums[len - i - 1]; // Remove last element of current state from s\n }\n return ans;\n }\n}\n``` | 2 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Java'] | 0 |
maximum-xor-for-each-query | Solution | solution-by-vijay_sathappan-5ypx | \n\n# Code\npython []\nclass Solution(object):\n def getMaximumXor(self, nums, maximumBit):\n max_k = (1<<maximumBit)-1 \n n=len(nums) \n | vijay_sathappan | NORMAL | 2024-11-08T17:50:48.053459+00:00 | 2024-11-08T17:50:48.053488+00:00 | 12 | false | \n\n# Code\n```python []\nclass Solution(object):\n def getMaximumXor(self, nums, maximumBit):\n max_k = (1<<maximumBit)-1 \n n=len(nums) \n dp=[0]*n \n curr=0 \n for i in range(n):\n curr^=nums[i] \n dp[n-i-1]=curr^max_k \n return dp\n\n \n \n``` | 2 | 0 | ['Python'] | 0 |
maximum-xor-for-each-query | Simple and easy java solution | simple-and-easy-java-solution-by-apoorvm-rqxc | Intuition\nWe have to calculate the cumulative XOR of all elements in the array nums. This will allow us to quickly determine the XOR of the remaining elements | Apoorvmittal | NORMAL | 2024-11-08T17:05:25.107320+00:00 | 2024-11-08T17:05:25.107350+00:00 | 11 | false | # Intuition\nWe have to calculate the cumulative XOR of all elements in the array nums. This will allow us to quickly determine the XOR of the remaining elements after removing the last element.\n\n# Approach\nInitialize an array ans which stores the final output elements of size n.\nWe calculate the cumulative XOR as c of all the elements in nums.\nWe calculate the maximum possible value as max of k as 2^maximumBit - 1, which is the largest number that can be formed with maximumBit bits.\nWe iterate through array nums and calculate xor value of cumulative XOR and max and store it in ans[i]\nUpdate c by value of taking its with xor of nums[n-i-1] basically removing the effect of last element in each query\nreturn the final ans array.\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(N)\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length;\n int[] ans = new int[n];\n int c = 0;\n for (int num : nums) {\n c ^= num;\n }\n int max = (1 << maximumBit) - 1;\n for (int i = 0; i <= n - 1; i++) {\n ans[i] = c ^ max;\n c ^= nums[n - i - 1];\n }\n return ans;\n }\n}\n``` | 2 | 0 | ['Java'] | 0 |
maximum-xor-for-each-query | 0 ms || beat 100% 🔥🔥 | 0-ms-beat-100-by-saurabh_bhandariii-aeqw | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | saurabh_bhandariii | NORMAL | 2024-11-08T15:40:11.469012+00:00 | 2024-11-08T15:40:11.469042+00:00 | 7 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution{\npublic:\n vector<int>getMaximumXor(vector<int>&nums,int maximumBit){\n vector<int>res(nums.size());\n int xorsum=0;\n int maxVal=(1<<maximumBit)-1;\n for(int i=0;i<nums.size();++i){\n xorsum^=nums[i];\n res[nums.size()-i-1]=xorsum^maxVal;\n }\n return res;\n }\n};\n``` | 2 | 0 | ['C++'] | 0 |
maximum-xor-for-each-query | Easiest solution | FEW LINES | JavaScript | C++ | Python | easiest-solution-few-lines-javascript-c-g3exn | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Nurliaidin | NORMAL | 2024-11-08T12:42:19.839701+00:00 | 2024-11-08T12:42:19.839740+00:00 | 49 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n# Code\n```javascript []\n/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function(nums, maximumBit) {\n // Initialize an array to store the result, with the same length as nums\n let res = new Array(nums.length).fill(0);\n // temp will accumulate the XOR of elements in nums\n let temp = 0;\n \n for (let i = 0; i < nums.length; i++) {\n // Update temp to be the XOR of itself and the current element in nums\n temp ^= nums[i];\n // Calculate the maximum XOR value for the current position by XORing temp with (2^maximumBit - 1)\n // (2^maximumBit - 1) creates a bitmask with maximumBit bits set to 1\n res[nums.length - i - 1] = Math.pow(2, maximumBit) - 1 ^ temp;\n }\n \n // Return the resulting array with XOR values in reverse order\n return res;\n};\n```\n``` cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n // Initialize an array to store the result, with the same length as nums\n vector<int> res(nums.size(), 0);\n // temp will accumulate the XOR of elements in nums\n int temp = 0;\n for (int i=0; i<nums.size(); i++) {\n // Update temp to be the XOR of itself and the current element in nums\n temp ^= nums[i];\n int k = pow(2, maximumBit)-1;\n // Calculate the maximum XOR value for the current position by XORing temp with (2^maximumBit - 1)\n // (2^maximumBit - 1) creates a bitmask with maximumBit bits set to 1\n res[nums.size()-i-1] = k ^ temp;\n }\n // Return the resulting array with XOR values in reverse order\n return res;\n }\n};\n```\n``` python []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n # Initialize an array to store the result, with the same length as nums\n res = [0]*len(nums)\n # temp will accumulate the XOR of elements in nums\n temp = 0\n for i in range(0, len(nums)):\n # Update temp to be the XOR of itself and the current element in nums\n temp ^= nums[i]\n # Calculate the maximum XOR value for the current position by XORing temp with (2^maximumBit - 1)\n # (2^maximumBit - 1) creates a bitmask with maximumBit bits set to 1\n k = int (math.pow(2, maximumBit)-1)\n res[len(nums)-i-1] = k ^ temp\n # Return the resulting array with XOR values in reverse order\n return res\n``` | 2 | 0 | ['C++', 'Python3', 'JavaScript'] | 0 |
maximum-xor-for-each-query | ✅✅ Beats 100% || Clean Solution | beats-100-clean-solution-by-karan_aggarw-ue8u | Code\ncpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n=nums.size();\n vector<int>ans( | Karan_Aggarwal | NORMAL | 2024-11-08T11:17:57.757578+00:00 | 2024-11-08T11:17:57.757610+00:00 | 2 | false | # Code\n```cpp []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n=nums.size();\n vector<int>ans(n);\n int XOR=0;\n // Find all XORs\n for(int num:nums){\n XOR ^= num;\n }\n // Create a mask\n int mask=(1<<maximumBit)-1;\n for(int i=0;i<n;i++){\n // push value by XORing the res by mask\n ans[i]=XOR^mask;\n // remove last element from array\n XOR ^= nums[n-i-1];\n }\n return ans;\n }\n};\n```\n# Have a Good Day \uD83D\uDE0A | 2 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'C++'] | 0 |
maximum-xor-for-each-query | Rust | 0ms | 100% | O(n) | rust-0ms-100-on-by-dmitry-shulaykin-ggol | Intuition\nApplying XOR with the same operands will result in 0.\n\n# Approach\nWe can built XOR of all nums and then reverse XOR by applying nums again one by | dmitry-shulaykin | NORMAL | 2024-11-08T10:35:32.271325+00:00 | 2024-11-08T10:35:32.271356+00:00 | 13 | false | # Intuition\nApplying XOR with the same operands will result in 0.\n\n# Approach\nWe can built XOR of all nums and then reverse XOR by applying nums again one by one from the back of the array.\n\nWe can maximize XOR of sum with k by choosing k as negative of the sum.\n\nUse bitwise AND to take only bits that matter (up to maximum bit) \n\n# Complexity\n- Time complexity: O(n) - looping nums array only twice (forward and backward)\n\n- Space complexity: O(1) - we don\'t store anything, except input and output\n\n# Code\n```rust []\nimpl Solution {\n pub fn get_maximum_xor(nums: Vec<i32>, maximum_bit: i32) -> Vec<i32> {\n let mut xor = 0;\n\n for num in &nums {\n xor ^= *num;\n }\n\n let mut ans = vec!();\n for num in nums.iter().rev() {\n ans.push((!xor) & ((1 << maximum_bit) - 1));\n xor ^= *num;\n }\n\n ans\n }\n}\n``` | 2 | 0 | ['Rust'] | 0 |
maximum-xor-for-each-query | Beats 100%|| Easy solution in Java, Python & Cpp | beats-100-easy-solution-in-java-python-c-kvi9 | Intuition\n1. Understanding the Goal:\n- We are given a sorted array nums and an integer maximumBit.\n- We need to find an integer k less than 2^maximumBit such | yashringe | NORMAL | 2024-11-08T07:33:18.139000+00:00 | 2024-11-08T07:33:18.139053+00:00 | 68 | false | # Intuition\n**1. Understanding the Goal:**\n- We are given a sorted array nums and an integer maximumBit.\n- We need to find an integer k less than 2^maximumBit such that nums[0] XOR nums[1] XOR ... XOR nums[nums.length-1] XOR k is maximized for each query.\n- After each query, we remove the last element of nums and repeat the process on the remaining array.\n\n**2. Maximizing XOR:**\n- To maximize XOR, we need to find a k that when XORed with the cumulative XOR of the array (xor_all) gives the largest possible result.\n- The largest value that can be formed with maximumBit bits is 2^maximumBit - 1. This number has all bits set to 1 up to maximumBit (e.g., for maximumBit = 3, the number is 111 in binary or 7 in decimal).\n- Thus, to maximize the XOR with xor_all, we can choose k to be xor_all XOR maxXor, where maxXor = 2^maximumBit - 1.\n\n**3. Reversing from the End:**\n- We\u2019re asked to start from the entire array and progressively "remove" the last element for each query.\n- For each query, we adjust xor_all by XORing out the last element in the current nums array, giving us an updated cumulative XOR for the remaining array.\n\n# Approach\n\n**1. Initialize Cumulative XOR and Maximum XOR Mask:**\n- First, compute the cumulative XOR of all elements in nums (xor_all). This gives us the XOR of the full array.\n- Calculate maxXor = 2^maximumBit - 1, which will be used to find k in each query.\n**2. Compute Each Answer Iteratively:**\n- For each query, find k by computing xor_all XOR maxXor, which maximizes the XOR value for the current cumulative XOR.\n- Append this result to ans.\n- Update xor_all by XORing out the last element of the current nums, effectively removing it from consideration for the next query.\n**3. Return the Result in Reverse Order:**\n- Since we processed the array from end to beginning, ans will contain the answers in reverse order, so we return it as-is.\n\n# Complexity\n- Time complexity:\nO(n) to traverse the array\n\n- Space complexity:\nO(1) but O(n) if we consider space used for answer.\n\n\n\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length;\n int xorr = nums[0];\n int maxXor = (1 << maximumBit) - 1;\n\n for (int i = 1; i < n; i++) {\n xorr ^= nums[i];\n }\n\n int[] ans = new int[n];\n for (int i = 0; i < n; i++) {\n ans[i] = xorr ^ maxXor;\n xorr ^= nums[n - 1 - i];\n }\n\n return ans;\n }\n}\n```\n\n```Python []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n n = len(nums)\n xorr = nums[0]\n max_xor = (1 << maximumBit) - 1\n \n for i in range(1, n):\n xorr ^= nums[i]\n \n ans = []\n for i in range(n):\n ans.append(xorr ^ max_xor)\n xorr ^= nums[n - 1 - i]\n \n return ans\n\n \n```\n\n```C++ []\nclass Solution {\npublic:\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size(),xorr = nums[0],maxxorr = pow(2,maximumBit)-1;\n for(int i=1;i<n;i++)xorr ^= nums[i];\n vector<int>ans(n);\n for(int i=0;i<n;i++){\n ans[i] = xorr^maxxorr;\n xorr ^= nums[n-1-i];\n }\n return ans;\n }\n};\n```\n\n\n | 2 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'C++', 'Java', 'Python3'] | 0 |
maximum-xor-for-each-query | Beats 78%||Java||Bit manipulation||Easy solution | beats-78javabit-manipulationeasy-solutio-sal8 | Intuition\n\n1.The key idea here seems to be using XOR as a way to accumulate or "cancel out" the effects of certain elements in the input array nums[] and then | abhay__solanki | NORMAL | 2024-11-08T06:53:00.480827+00:00 | 2024-11-08T06:53:00.480848+00:00 | 32 | false | # Intuition\n\n1.The key idea here seems to be using XOR as a way to accumulate or "cancel out" the effects of certain elements in the input array nums[] and then manipulate these accumulated values to create a new output array arr[].\n\n2.The value of b (which is (1 << maximumBit) - 1) is used as a mask to potentially modify the XOR result in a controlled way.\n\n3.By iterating backward over the array and updating a as the XOR of previous elements, the code is ensuring that each element in arr[] is computed based on all the previous elements in nums[] but with the current element\'s influence removed.\n# Approach\n\n\n1.XOR all elements: We first compute the XOR of all elements in the input array nums[]. This is stored in the variable a.\n\n2.Apply the mask (b): For each element, we compute the result using the XOR of the current accumulated value a and the mask b. This result is stored in the output array arr[].\n\n3.Iterate backward: After each computation, we update a by XORing it with the current element to "remove" its influence for subsequent calculations.\n\n\n# Complexity\n- Time complexity:\nO(2n)\n\n- Space complexity:\nO(1)\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n=nums.length;\n int arr[]=new int[n];\n int b=(1<<maximumBit)-1;//max value of k\n int a=0;\n for(int i=0;i<n;i++){\n a=a^nums[i];\n }\n for(int i=n-1;i>=0;i--){\n arr[n-1-i]=a^b;//it calculate the value of k for\n //which the xor of elements is max.\n a=a^nums[i];\n }\n return arr;\n \n }\n}\n``` | 2 | 0 | ['Java'] | 0 |
maximum-xor-for-each-query | Easiest Solution || Beats 100% ✅|| C++|| Beginer friendly | easiest-solution-beats-100-c-beginer-fri-4lmf | Intuition\n Describe your first thoughts on how to solve this problem. \nThink what no. can be produced which is maximum ?\n\nAfter seeing constarints nums[i]<2 | ghanshyamgcs22 | NORMAL | 2024-11-08T06:15:24.238977+00:00 | 2024-11-08T06:15:24.239016+00:00 | 85 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThink what no. can be produced which is maximum ?\n\nAfter seeing constarints `nums[i]<2^maximumBit`, so maximum no. can be produced `2^maximumBit-1`, now take xor to find value of k for which both produced `2^maximumBit-1`\n\n\n# Example \n`nums = [0,1,1,3], maximumBit = 2`\n\nmaximum no. that can be produced is =2^2-1=3\n\nNow take xor of element one by one, for value of k take again xor to 3\n\nFor example x^y=c, where we know x and c, how we can find y?\ny--> x^c\n\nnums =[0,1,1,3]\n\nxor =[0,1,0,3]\n\nxor to 3(to find corrosponding value of k)= [3,2,3,0]\n\nnow reverse the answer [0,3,2,3]\n\n\n# Complexity\n- Time complexity:0(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n\n\n\n\n vector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n \n int x = pow(2, maximumBit) - 1; \n int currentXor = 0;\n vector<int> v(nums.size());\n\n \n for (int i = 0; i < nums.size(); i++) {\n currentXor ^= nums[i];\n v[nums.size() - 1 - i] = currentXor ^ x;\n }\n\n return v;\n }\n\n};\n``` | 2 | 0 | ['C++'] | 1 |
maximum-xor-for-each-query | 🔥BEATS 💯 % 🎯 |✨SUPER EASY BEGINNERS 👏 | beats-super-easy-beginners-by-codewithsp-njrg | \n\n\n\n---\n\n### Intuition\nThe goal is to return the maximum XOR value for a given array nums in a sequence based on maximumBit. Each time, we calculate the | CodeWithSparsh | NORMAL | 2024-11-08T05:53:57.201453+00:00 | 2024-11-08T05:53:57.201486+00:00 | 9 | false | \n\n\n\n---\n\n### Intuition\nThe goal is to return the maximum XOR value for a given array `nums` in a sequence based on `maximumBit`. Each time, we calculate the maximum XOR value by finding the XOR result that gives the highest bits, limited by `maximumBit`. \n\nGiven that XOR is reversible, we can leverage the cumulative XOR of all elements to avoid recalculating it for each new configuration.\n\n---\n\n### Approach\n1. **Calculate Cumulative XOR:** First, compute the XOR of the entire array (`totalXor`). This represents the cumulative XOR when all elements are present.\n2. **Determine Maximum XOR for Each State:** Using `totalXor` at each step, compute the highest `k` that can be achieved. For a given `maximumBit`, this maximum `k` is determined as `maxBit = (1 << maximumBit) - 1`, which sets all bits within the `maximumBit` constraint.\n3. **Iterate in Reverse Order:** For each state, remove the last element of `nums` (by updating `totalXor`) and compute the new XOR with `maxBit`.\n4. **Return Results in Reverse Order:** We calculate in reverse, as the problem demands results in decreasing length order.\n\n---\n\n### Complexity\n- **Time Complexity:** \\(O(n)\\), where \\(n\\) is the length of `nums`, as we only need one pass to calculate the XOR and one pass for the results.\n- **Space Complexity:** \\(O(n)\\), for storing the result array.\n\n---\n\n### Code Implementations\n\n```dart []\nimport \'dart:math\';\n\nclass Solution {\n List<int> getMaximumXor(List<int> nums, int maximumBit) {\n int totalXor = nums.reduce((a, b) => a ^ b);\n int maxBit = pow(2, maximumBit).toInt() - 1;\n List<int> answer = List.filled(nums.length, 0);\n\n for (int i = nums.length - 1; i >= 0; i--) {\n answer[nums.length - 1 - i] = totalXor ^ maxBit;\n totalXor ^= nums[i];\n }\n\n return answer;\n }\n}\n```\n\n```java []\nimport java.util.*;\n\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int totalXor = 0;\n for (int num : nums) {\n totalXor ^= num;\n }\n \n int maxBit = (1 << maximumBit) - 1;\n int[] answer = new int[nums.length];\n\n for (int i = nums.length - 1; i >= 0; i--) {\n answer[nums.length - 1 - i] = totalXor ^ maxBit;\n totalXor ^= nums[i];\n }\n\n return answer;\n }\n}\n```\n\n```python []\nfrom typing import List\n\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n totalXor = 0\n for num in nums:\n totalXor ^= num\n \n maxBit = (1 << maximumBit) - 1\n answer = [0] * len(nums)\n \n for i in range(len(nums) - 1, -1, -1):\n answer[len(nums) - 1 - i] = totalXor ^ maxBit\n totalXor ^= nums[i]\n \n return answer\n```\n\n```javascript []\nvar getMaximumXor = function(nums, maximumBit) {\n let totalXor = nums.reduce((acc, num) => acc ^ num, 0);\n const maxBit = (1 << maximumBit) - 1;\n const answer = Array(nums.length);\n\n for (let i = nums.length - 1; i >= 0; i--) {\n answer[nums.length - 1 - i] = totalXor ^ maxBit;\n totalXor ^= nums[i];\n }\n\n return answer;\n};\n```\n\n```cpp []\n#include <vector>\n#include <cmath>\n\nclass Solution {\npublic:\n std::vector<int> getMaximumXor(std::vector<int>& nums, int maximumBit) {\n int totalXor = 0;\n for (int num : nums) {\n totalXor ^= num;\n }\n \n int maxBit = (1 << maximumBit) - 1;\n std::vector<int> answer(nums.size());\n\n for (int i = nums.size() - 1; i >= 0; i--) {\n answer[nums.size() - 1 - i] = totalXor ^ maxBit;\n totalXor ^= nums[i];\n }\n\n return answer;\n }\n};\n```\n\n```go []\nfunc getMaximumXor(nums []int, maximumBit int) []int {\n totalXor := 0\n for _, num := range nums {\n totalXor ^= num\n }\n \n maxBit := (1 << maximumBit) - 1\n answer := make([]int, len(nums))\n\n for i := len(nums) - 1; i >= 0; i-- {\n answer[len(nums) - 1 - i] = totalXor ^ maxBit\n totalXor ^= nums[i]\n }\n\n return answer\n}\n```\n\nThis consistent approach across different languages will help produce correct results for the maximum XOR values given a dynamic array configuration. \n\n---\n {:style=\'width:250px\'} | 2 | 0 | ['Array', 'Bit Manipulation', 'C', 'Prefix Sum', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'Dart'] | 0 |
maximum-xor-for-each-query | Easiest Solution.Beats 100.00%.0 ms Solution🔥 | easiest-solutionbeats-100000-ms-solution-wv1s | Intuition\n Describe your first thoughts on how to solve this problem. \nThe maximum possible XOR result is always 2^(maximumBit)-1.So try to make the prefix-su | yashvars15114 | NORMAL | 2024-11-08T05:43:19.242902+00:00 | 2024-11-08T05:43:19.242937+00:00 | 59 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe maximum possible XOR result is always 2^(maximumBit)-1.So try to make the prefix-sum equal to maximum XOR.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nInstead of storing prefix-sum in another array/vector.Intialize a vector of size n and a variable xr to store XOR.Then traverse through the nums vector and compute the XOR and store it as \nans[n-i-1] = (mx)^(xr&mx) by this we avoid reversing the vector again.\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\nvector<int> getMaximumXor(vector<int>& nums, int maximumBit) {\n int n = nums.size();\n int mx = pow(2,maximumBit)-1;\n vector<int> ans(n,0);\n int xr = 0;\n for(int i=0;i<n;i++)\n {\n xr ^= nums[i]; \n ans[n-i-1] = (mx)^(xr&mx);\n }\n return ans; \n}\n};\n``` | 2 | 0 | ['C++'] | 0 |
maximum-xor-for-each-query | Basic XOR solution | basic-xor-solution-by-nurzhansultanov-eeyp | Code\npython3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n allx = 0\n for num in nums:\n | NurzhanSultanov | NORMAL | 2024-11-08T04:03:44.265200+00:00 | 2024-11-08T04:03:44.265230+00:00 | 9 | false | # Code\n```python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n allx = 0\n for num in nums:\n allx ^= num\n result = []\n n = (2**maximumBit) - 1 \n for num in reversed(nums):\n result.append(allx ^ n)\n allx ^= num \n return result\n\n``` | 2 | 0 | ['Python3'] | 0 |
maximum-xor-for-each-query | Java Clean Solution | java-clean-solution-by-shree_govind_jee-zx6b | Complexity\n- Time complexity:O(n)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity:O(n)\n Add your space complexity here, e.g. O(n) \n\n# Code | Shree_Govind_Jee | NORMAL | 2024-11-08T03:57:37.048347+00:00 | 2024-11-08T03:57:37.048381+00:00 | 81 | false | # Complexity\n- Time complexity:$$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int xor = 0;\n for(int i=0; i<nums.length; i++){\n xor ^= nums[i];\n }\n\n int[] res = new int[nums.length];\n int maxor = (int)Math.pow(2, maximumBit)-1;\n for(int i=0; i<nums.length; i++){\n res[i] = xor^maxor;\n xor ^= nums[nums.length-i-1];\n }\n return res;\n }\n}\n``` | 2 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Java'] | 0 |
maximum-xor-for-each-query | Bit Manipulation | Super simple code | 2ms Beats 100% | bit-manipulation-super-simple-code-2ms-b-eszm | Intuition\n Describe your first thoughts on how to solve this problem. \n1 << maximumBit - 1 will give us the max possible value of xor, for eg. taking maximumB | Rishab_Mandal | NORMAL | 2024-11-08T03:04:31.927671+00:00 | 2024-11-08T03:04:31.927691+00:00 | 4 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1 << maximumBit - 1 will give us the max possible value of xor, for eg. taking maximumBit = 2, max value will be 1 << 2 - 1 = 4 - 1 = 3.\n\nNow, since we know the max value, we can use it to calculate the value which would be xored with remaining elements to get that max value.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n\n# Code\n```java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int max = (1 << maximumBit) - 1, n = nums.length;\n int[] ans = new int[n];\n int curr = 0;\n for(int i = 0; i<n; i++){\n curr ^= nums[i];\n ans[n-i-1] = curr ^ max;\n }\n return ans;\n }\n}\n``` | 2 | 0 | ['Array', 'Bit Manipulation', 'Java'] | 0 |
maximum-xor-for-each-query | Very Simple PrefixXor Solution Java/C (100%) | very-simple-prefixxor-solution-javac-100-z2ig | Approach\n Describe your approach to solving the problem. \n1. max is calculated as (1 << maximumBit) - 1. This creates a number with maximumBit bits all set to | rajnarayansharma110 | NORMAL | 2024-11-08T02:51:55.370950+00:00 | 2024-11-08T02:51:55.370985+00:00 | 69 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. max is calculated as (1 << maximumBit) - 1. This creates a number with maximumBit bits all set to 1 (e.g., if maximumBit = 3, max would be 111 in binary, which is 7 in decimal).\n2. Calucate prefix XOR\n3. Maximum XOR Calculation:\n - The XOR of nums[size - i - 1] with max gives the maximum possible XOR value that can be achieved for each prefix of nums.\n\n# Example\n1. For nums = [0, 1, 1, 3] and maximumBit = 2:\n2. Calculate max:\n - max = (1 << 2) - 1 = 3 (binary 11).\n3. Prefix XOR Calculation:\n - After the first loop, nums becomes [0, 1, 0, 3] because:\n - nums[1] ^= nums[0] \u2192 nums[1] = 1\n - nums[2] ^= nums[1] \u2192 nums[2] = 0\n - nums[3] ^= nums[2] \u2192 nums[3] = 3\n4. Maximum XOR Calculation:\n - The second loop computes res as follows:\n - res[0] = nums[3] ^ max = 3 ^ 3 = 0\n - res[1] = nums[2] ^ max = 0 ^ 3 = 3\n - res[2] = nums[1] ^ max = 1 ^ 3 = 2\n - res[3] = nums[0] ^ max = 0 ^ 3 = 3\n5. Thus, res = [0, 3, 2, 3].\n\n# Complexity\n- Time complexity:O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```Java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int size = nums.length;\n int[] res = new int[size];\n int max = (1 << maximumBit) - 1;\n for (int i = 1; i < size; i++) {\n nums[i] ^= nums[i - 1];\n }\n for (int i = 0; i < size; i++) {\n res[i] = nums[size - i - 1] ^ max;\n }\n return res;\n }\n}\n```\n```c []\nint* getMaximumXor(int* nums, int size, int maximumBit, int* rSize) {\n int* res = (int*)malloc(sizeof(int) * size);\n *rSize = size;\n int max = pow(2, maximumBit) - 1;\n for (int i = 1; i < size; i++) {\n nums[i] ^= nums[i - 1];\n }\n for (int i = 0; i < size; i++) {\n res[i] = nums[size - i - 1] ^ max;\n }\n return res;\n}\n``` | 2 | 0 | ['C', 'Prefix Sum', 'Java'] | 0 |
maximum-xor-for-each-query | Python | Beats 100% | Bit Manipulation | python-beats-100-bit-manipulation-by-shi-qdlh | Intuition\n Describe your first thoughts on how to solve this problem. \nMy initial thought is to utilize the properties of XOR and how it interacts with bits. | shivamtld | NORMAL | 2024-11-08T02:49:26.976326+00:00 | 2024-11-08T02:49:26.976361+00:00 | 87 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nMy initial thought is to utilize the properties of XOR and how it interacts with bits. The key observation is that XORing a number with its bitwise complement yields a number where all bits are set to 1 up to the most significant bit of the original number. The challenge is to find the maximum XOR value possible with a given number after XORing it with some number from a given list, considering a specific number of bits.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- **Calculate Maximum Value:** First, calculate the maximum possible value that can be formed with the given number of bits, maximumBit. This value is (2**maximumBit) - 1. This value essentially sets all the bits to 1 up to the specified number of bits.\n\n- **Initialize Variables**: Start with an empty list output to store the results, and a variable xor_sum initialized to the first element of the list nums. This xor_sum will be used to store the cumulative XOR of elements as we iterate through the list.\n\n- **Iterate and Compute XOR**: For each number in the list (starting from the first), update the xor_sum with the XOR of the current number. After updating, compute the maximum XOR for this cumulative sum by XORing it with max_value and subtracting the result from max_value.\n\n- **Store Result in Reverse Order**: Since the problem specifies that the results should be returned in reverse order of the XOR operations, append each computed result to the output list. After completing the iterations, return the reversed list.\n\n- **Return the Result**: Finally, the list output is reversed to provide the results in the required order and returned.\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n # Calculate the maximum value possible with the given number of bits\n max_value = (2 ** maximumBit) - 1\n \n # Initialize the result list\n output = []\n \n # Compute the XOR of all numbers in the list\n xor_sum = nums[0]\n \n # Calculate the maximum XOR for the first element\n output.append(max_value - xor_sum)\n \n # Iterate over the rest of the numbers\n for num in nums[1:]:\n # Update the cumulative XOR sum\n xor_sum ^= num\n \n # Calculate the maximum XOR for the current cumulative XOR sum\n output.append(max_value - xor_sum)\n \n # Return the result in reverse order as required\n return output[::-1] \n```\n\n\n | 2 | 0 | ['Bit Manipulation', 'Python3'] | 1 |
maximum-xor-for-each-query | 🌟 Beats 100.00 % 👏 || Step-by-Step Breakdown 🔥💯 | beats-10000-step-by-step-breakdown-by-wi-2uxb | \n\n## \uD83C\uDF1F Access Daily LeetCode Solutions Repo : click here\n\n---\n\n\n\n---\n\n# Intuition\nTo solve the problem, we need to maximize the XOR result | withaarzoo | NORMAL | 2024-11-08T02:12:37.360867+00:00 | 2024-11-08T02:12:37.360894+00:00 | 88 | false | \n\n## **\uD83C\uDF1F Access Daily LeetCode Solutions Repo :** [click here](https://github.com/withaarzoo/LeetCode-Solutions)\n\n---\n\n\n\n---\n\n# Intuition\nTo solve the problem, we need to maximize the XOR result for each query by choosing an integer `k` that, when XORed with the cumulative XOR of `nums`, gives the highest possible value within a given range. A key insight here is to use the bitwise complement of the cumulative XOR to obtain this maximum XOR result. By keeping track of the XOR for all elements in `nums` and updating it each time we remove an element, we can efficiently determine the optimal `k` for each query.\n\n---\n\n# Approach\n1. **Initial Cumulative XOR**: First, we calculate the XOR for all elements in `nums`. This gives us the initial XOR for the entire array, which we can adjust as we process each query by removing elements from the back.\n \n2. **Determine Max XOR Value (`max_k`)**: The maximum value we can achieve with a given number of bits (`maximumBit`) is `2^maximumBit - 1`, which represents a number where all bits are set to `1`. This allows us to use `XORed ^ max_k` to get the highest possible XOR value for each query.\n\n3. **Iterate Backwards Through Queries**: For each query, calculate `k = XORed ^ max_k` and store it. Then, update `XORed` by removing the effect of the last element in `nums` (in reverse order).\n\n4. **Result Array**: Collect all values of `k` from each query in reverse order, producing the final answer array.\n\n---\n\n# Complexity\n- **Time Complexity**: \\(O(n)\\), where \\(n\\) is the length of `nums`, since we only traverse `nums` once to compute the XOR and once for each query.\n- **Space Complexity**: \\(O(n)\\), as we store the results in an array of size \\(n\\).\n\n---\n\n# Code\n```C++ []\nclass Solution\n{\npublic:\n vector<int> getMaximumXor(vector<int> &nums, int maximumBit)\n {\n int n = nums.size();\n vector<int> answer(n);\n int XORed = 0;\n\n // Calculate the cumulative XOR of the entire nums array\n for (int num : nums)\n {\n XORed ^= num;\n }\n\n // max_k is 2^maximumBit - 1\n int max_k = (1 << maximumBit) - 1;\n\n // Process each query in reverse\n for (int i = 0; i < n; ++i)\n {\n // Calculate the k that maximizes XOR\n answer[i] = XORed ^ max_k;\n\n // Update XORed by removing the effect of the last element\n XORed ^= nums[n - 1 - i];\n }\n\n return answer;\n }\n};\n\n```\n```Java []\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int n = nums.length;\n int[] answer = new int[n];\n int XORed = 0;\n\n // Calculate the cumulative XOR of the entire nums array\n for (int num : nums) {\n XORed ^= num;\n }\n\n // max_k is 2^maximumBit - 1\n int max_k = (1 << maximumBit) - 1;\n\n // Process each query in reverse\n for (int i = 0; i < n; i++) {\n // Calculate the k that maximizes XOR\n answer[i] = XORed ^ max_k;\n\n // Update XORed by removing the effect of the last element\n XORed ^= nums[n - 1 - i];\n }\n\n return answer;\n }\n}\n\n```\n```JavaScript []\n/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function (nums, maximumBit) {\n let n = nums.length;\n let answer = new Array(n);\n let XORed = 0;\n\n // Calculate the cumulative XOR of the entire nums array\n for (let num of nums) {\n XORed ^= num;\n }\n\n // max_k is 2^maximumBit - 1\n let max_k = (1 << maximumBit) - 1;\n\n // Process each query in reverse\n for (let i = 0; i < n; i++) {\n // Calculate the k that maximizes XOR\n answer[i] = XORed ^ max_k;\n\n // Update XORed by removing the effect of the last element\n XORed ^= nums[n - 1 - i];\n }\n\n return answer;\n};\n\n```\n```Python []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n n = len(nums)\n answer = [0] * n\n XORed = 0\n \n # Calculate the cumulative XOR of the entire nums array\n for num in nums:\n XORed ^= num\n \n # max_k is 2^maximumBit - 1\n max_k = (1 << maximumBit) - 1\n \n # Process each query in reverse\n for i in range(n):\n # Calculate the k that maximizes XOR\n answer[i] = XORed ^ max_k\n \n # Update XORed by removing the effect of the last element\n XORed ^= nums[n - 1 - i]\n \n return answer\n\n```\n```Go []\nfunc getMaximumXor(nums []int, maximumBit int) []int {\n n := len(nums)\n answer := make([]int, n)\n XORed := 0\n \n // Calculate the cumulative XOR of the entire nums array\n for _, num := range nums {\n XORed ^= num\n }\n \n // max_k is 2^maximumBit - 1\n max_k := (1 << maximumBit) - 1\n \n // Process each query in reverse\n for i := 0; i < n; i++ {\n // Calculate the k that maximizes XOR\n answer[i] = XORed ^ max_k\n \n // Update XORed by removing the effect of the last element\n XORed ^= nums[n - 1 - i]\n }\n \n return answer\n}\n\n```\n---\n\n# Step-by-Step Detailed Explanation\n\n1. **Initialize Variables**: Calculate the cumulative XOR (`XORed`) of all elements in `nums`. Initialize `max_k` to be `2^maximumBit - 1`, which represents the highest possible number within the allowed bit range.\n\n2. **Process Queries**: Iterate over each query in reverse order:\n - For each query, calculate `k` by XORing `XORed` with `max_k`. This gives the maximum XOR result for the current `nums` subset.\n - Store `k` in the `answer` array.\n - Update `XORed` by XORing it with the last element in the current `nums` subset to simulate the removal of that element.\n\n3. **Return Result**: Once all queries are processed, the `answer` array contains the result for each query in the correct order. This final array is returned as the solution. \n\nBy following this method, we efficiently compute the required maximum XOR values for each query in `O(n)` time complexity.\n\n---\n\n\n | 2 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript'] | 1 |
maximum-xor-for-each-query | scala twoliner | scala-twoliner-by-vititov-njkg | scala []\nobject Solution {\n def getMaximumXor(nums: Array[Int], maximumBit: Int): Array[Int] = {\n lazy val mask = (1<<maximumBit)-1\n nums.scanLeft(0) | vititov | NORMAL | 2024-11-08T01:13:32.292534+00:00 | 2024-11-08T01:18:15.397508+00:00 | 13 | false | ```scala []\nobject Solution {\n def getMaximumXor(nums: Array[Int], maximumBit: Int): Array[Int] = {\n lazy val mask = (1<<maximumBit)-1\n nums.scanLeft(0){case (b,a) => b^a}.drop(1).map(_ ^ mask).reverse\n }\n}\n``` | 2 | 0 | ['Array', 'Bit Manipulation', 'Prefix Sum', 'Scala'] | 0 |
maximum-xor-for-each-query | Java easy solution | java-easy-solution-by-aji_hsu-wj63 | Approach\nnotice that:\n1. 0 ^ a = a\n2. if a ^ b = c then a ^ c = b\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\njava []\n | AJI_Hsu | NORMAL | 2024-11-08T00:19:13.650210+00:00 | 2024-11-08T00:19:32.436940+00:00 | 14 | false | # Approach\nnotice that:\n1. `0 ^ a = a`\n2. `if a ^ b = c then a ^ c = b`\n\n# Complexity\n- Time complexity:\n$$O(n)$$\n\n- Space complexity:\n$$O(n)$$\n\n# Code\n```java []\nclass Solution {\n private int pow(int x, int y) {\n if (y == 0) return 1;\n else if (y % 2 == 1) return pow(x, y - 1) * x;\n else {\n int t = pow(x, y / 2);\n return t * t;\n }\n }\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int suf = 0;\n int[] ans = new int[nums.length];\n maximumBit = pow(2, maximumBit);\n for (int i = 0; i < nums.length; i++) {\n suf ^= nums[i];\n ans[nums.length - i - 1] = suf ^ (maximumBit - 1);\n }\n return ans;\n }\n}\n``` | 2 | 0 | ['Java'] | 0 |
maximum-xor-for-each-query | Easy to understand || JavaScript | easy-to-understand-javascript-by-harshad-bxm0 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Harshad_Patel | NORMAL | 2024-07-04T03:43:59.443147+00:00 | 2024-07-04T03:43:59.443171+00:00 | 33 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n/**\n * @param {number[]} nums\n * @param {number} maximumBit\n * @return {number[]}\n */\nvar getMaximumXor = function(nums, maximumBit) {\n let xor = (2**maximumBit) - 1;\n for(let i =0; i<nums.length; i++){\n xor ^= nums[i];\n nums[i] = xor;\n } \n return nums.reverse();\n};\n``` | 2 | 0 | ['JavaScript'] | 0 |
maximum-xor-for-each-query | Java Solution | Easy and Simple Approach | Beginner Friendly | O(N) Time complexity | java-solution-easy-and-simple-approach-b-99cm | Intuition\n Describe your first thoughts on how to solve this problem. \nIntuition behind this is that we are calculating the XOR of same elements everytime alo | sidver-18 | NORMAL | 2024-02-19T20:48:46.535914+00:00 | 2024-02-19T20:48:46.535935+00:00 | 124 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIntuition behind this is that we are calculating the XOR of same elements everytime along with the current element at ith index so we can save the xor of previous i-1 elements.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUsed prefix sum approach for saving the xor of previous i-1 elements in a single variable and updating that for every ith element and using a single variable so that we don\'t use any extra space here. Then to maximise the XOR we know that taking XOR with 2^maxbit - 1 will give max value.\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int[] getMaximumXor(int[] nums, int maximumBit) {\n int[] ans = new int[nums.length];\n int j = 0, val = 0;\n for(int i = nums.length - 1; i >= 0; i--){\n val = val ^ nums[j++];\n ans[i] = val ^ ((1 << maximumBit) - 1);\n }\n return ans;\n }\n}\n``` | 2 | 0 | ['Bit Manipulation', 'Java'] | 1 |
maximum-xor-for-each-query | Simple fast solution TIME COMPLEXITY O(n) | simple-fast-solution-time-complexity-on-gc8ou | Complexity\n- Time complexity: O(n)\n- Space complexity: O(n)\n\n# Code\npython []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: in | fuglaeff | NORMAL | 2023-05-25T00:12:58.799907+00:00 | 2023-05-25T00:12:58.799936+00:00 | 204 | false | # Complexity\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n\n# Code\n```python []\nclass Solution:\n def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:\n ans = [(1 << maximumBit) - 1]\n for n in nums:\n ans.append(ans[-1] ^ n)\n\n return ans[len(ans)-1:0:-1]\n``` | 2 | 0 | ['Python3'] | 0 |
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