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 |
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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<>();
/**
* co... | 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.token... | 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 s... | 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 _... | 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 timeT... | 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... | 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 comple... | 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
`... | 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
`... | 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
`... | 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
`... | 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 {
unord... | 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;
... | 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. $$O(n)$$ --... | 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, tim... | 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... | 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... | 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... | 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... | 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`, `_wi... | 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. $$O(n)$$ --... | 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. $$O(n)$$ --... | 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 ... | 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] ... | 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` ... | 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 ... | 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... | 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 = ... | 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 ... | 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... | 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 \... | 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 po... | 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 Solutio... | 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 pos... | 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... | 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\... | 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 ... | 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 ... | 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 ... | 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 ... | 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` i... | 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 high... | 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 b... | 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 ... | 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 ... | 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> getMaxi... | 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 ... | 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 ... | 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 ... | 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. $$O(n)$$ --... | 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# Appro... | 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 ... | 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 ... | 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 ... | 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 ... | 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) -... | 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 ... | 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://leet... | 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)... | 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\... | 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... | 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 s... | 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 Maxi... | 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 ... | 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> an... | 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... | 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 ... | 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... | 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, ... | 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.... | 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.... | 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[... | 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 ... | 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 ... | 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 ... | 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 o... | 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 c... | 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 ... | 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 ... | 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)$$ --... | 2 | 0 | ['C++'] | 0 |
maximum-xor-for-each-query | Easiest solution | FEW LINES | JavaScript | C++ | Python | easiest-solution-few-lines-javascript-c-g3exn | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Nurliaidin | NORMAL | 2024-11-08T12:42:19.839701+00:00 | 2024-11-08T12:42:19.839740+00:00 | 49 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(n)$$\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O... | 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 ma... | 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 (u... | 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 r... | 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... | 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... | 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 resul... | 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.Int... | 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 ... | 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(... | 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 r... | 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... | 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 t... | 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: 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 ... | 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. $$O(n)$$ --... | 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 pro... | 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)... | 2 | 0 | ['Python3'] | 0 |
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