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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check-if-digits-are-equal-in-string-after-operations-i | Python 3 lines, quardratic solution | python-3-lines-quardratic-solution-by-sa-jbxc | IntuitionWe can just simulate step by step where at each step we reduce the number of digits by one.Complexity
Time complexity: O(n2). As each steps takes O(n) | salvadordali | NORMAL | 2025-04-11T07:45:22.829708+00:00 | 2025-04-11T07:45:22.829708+00:00 | 1 | false | # Intuition
We can just simulate step by step where at each step we reduce the number of digits by one.
# Complexity
- Time complexity: $O(n^2)$. As each steps takes $O(n)$
- Space complexity: $O(n)$
# Code
```python3 []
class Solution:
def hasSameDigits(self, s: str) -> bool:
digits = [int(c) for c in s]
... | 0 | 0 | ['Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Simplest C++ Solution | simplest-c-solution-by-0j60pwopfd-yp5p | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | 0j60PWOPfD | NORMAL | 2025-04-10T22:05:48.244802+00:00 | 2025-04-10T22:05:48.244802+00:00 | 1 | false | # Intuition
# Approach
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
class Solution {
public:
bool hasSameDigits(string s) {
while (s.size()>= 2) {
... | 0 | 0 | ['C++'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Optimized simple solution - beats 100%🔥 | optimized-simple-solution-beats-100-by-c-kky1 | Complexity
Time complexity: O(N)
Space complexity: O(N)
Code | cyrusjetson | NORMAL | 2025-04-10T10:45:17.682827+00:00 | 2025-04-10T10:45:36.760893+00:00 | 2 | false | # Complexity
- Time complexity: O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(N)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public boolean hasSameDigits(String s) {
int[] digits = new int[s.length()];
int distance = ... | 0 | 0 | ['Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | easy stringbuilder solution (java) | easy-stringbuilder-solution-java-by-dpas-2z8v | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | dpasala | NORMAL | 2025-04-09T03:10:16.629839+00:00 | 2025-04-09T03:10:16.629839+00:00 | 1 | 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 | ['Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Python Solution -0(n) | python-solution-0n-by-vini__7-h81j | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Vini__7 | NORMAL | 2025-04-08T07:05:57.773571+00:00 | 2025-04-08T07:05:57.773571+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
`... | 0 | 0 | ['Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Java | Recursion approach | Beats 90% | java-recursion-approach-beats-90-by-igor-r928 | Code | IgorChurakov | NORMAL | 2025-04-07T07:49:37.973381+00:00 | 2025-04-07T07:49:37.973381+00:00 | 1 | false | # Code
```java []
class Solution {
public boolean hasSameDigits(String s) {
var chars = new ArrayList<Integer>();
for (int i = 0; i < s.length(); i++) {
chars.add(s.charAt(i) - '0');
}
return hasSameDigits(chars);
}
boolean hasSameDigits(ArrayList<Integer> chars)... | 0 | 0 | ['Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Java Intermediate Solution | java-intermediate-solution-by-trevorkmci-9xrv | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | trevorkmcintyre | NORMAL | 2025-04-07T01:16:17.948693+00:00 | 2025-04-07T01:16:17.948693+00:00 | 1 | 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 | ['Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Fastest way - infinite loop | fastest-way-infinite-loop-by-zemamba-6ix5 | null | zemamba | NORMAL | 2025-04-06T22:25:22.399898+00:00 | 2025-04-06T22:25:22.399898+00:00 | 1 | false | ```javascript []
/**
* @param {string} s
* @return {boolean}
*/
var hasSameDigits = function(s) {
arr = [...s]
while (arr.length > 2) {
arr = newDigit(arr)
}
return arr[0] == arr[1]
};
function newDigit(ar) {
temp = []
for (let i = 0; i < ar.length - 1; i++) {
temp.push((~... | 0 | 0 | ['JavaScript'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | s.compactMap{$0.wholeNumberValue} | scompactmap0wholenumbervalue-by-victor-s-kkaj | null | Victor-SMK | NORMAL | 2025-04-06T17:56:00.311048+00:00 | 2025-04-06T17:56:00.311048+00:00 | 1 | false |
```swift []
class Solution {
func hasSameDigits(_ s: String) -> Bool {
var arr = s.compactMap{$0.wholeNumberValue}
while arr.count > 2 {
for i in 1..<arr.count {
arr[i-1] = (arr[i-1] + arr[i]) % 10
}
arr.removeLast()
}
re... | 0 | 0 | ['Math', 'String', 'Swift', 'Python', 'Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Kotlin Solution | kotlin-solution-by-osagiog-9j3t | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | osagiog | NORMAL | 2025-04-06T12:51:51.169889+00:00 | 2025-04-06T12:51:51.169889+00:00 | 1 | 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 | ['Kotlin'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Simple C simluation 100% runtime 94.97% memory | simple-c-simluation-100-runtime-9497-mem-0s1j | IntuitionOkay so.... we need to repeatedly modify the string by replacing adjacent digits with their sum modulo 10 until the string length reduces to exactly 2 | PerryThePpatypus | NORMAL | 2025-04-06T01:01:45.856795+00:00 | 2025-04-06T01:02:57.253994+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Okay so.... we need to repeatedly modify the string by replacing adjacent digits with their sum modulo 10 until the string length reduces to exactly 2 characters. At that point, we check if those two characters are the same.
Instead of all... | 0 | 0 | ['C'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | C# | Simulation | c-simulation-by-we6mmvl7tl-n90w | Complexity
Time complexity: O(n2)
Space complexity: O(1)
Code | we6MMvL7tl | NORMAL | 2025-04-04T13:54:23.080282+00:00 | 2025-04-04T13:54:23.080282+00:00 | 3 | false | # Complexity
- Time complexity: $$O(n^2)$$
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: $$O(1)$$
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```csharp []
public class Solution {
public bool HasSameDigits(string s) {
while (s.Length != 2)
{
... | 0 | 0 | ['Simulation', 'C#'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Check If Digits Are Equal in String After Operations I in Java | check-if-digits-are-equal-in-string-afte-f3tj | Code | sowmy3010 | NORMAL | 2025-04-02T11:26:03.410988+00:00 | 2025-04-02T11:26:03.410988+00:00 | 1 | false |
# Code
```java []
class Solution {
public String Modulo(String s){
String st="";
for(int i=1;i<s.length();i++){
char ch1 = s.charAt(i-1);
char ch2 = s.charAt(i);
int p = (ch1-'0'+ch2-'0')%10;
st+=p;
}
return st;
}
public boole... | 0 | 0 | ['Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Checking if a Number Reduces to Two Identical Digits | checking-if-a-number-reduces-to-two-iden-wlaq | IntuitionThe problem requires checking if a given string of digits, when repeatedly transformed by replacing adjacent pairs with their sum modulo 10, eventually | Nylrem | NORMAL | 2025-03-31T20:36:10.834163+00:00 | 2025-03-31T20:36:10.834163+00:00 | 1 | false |
## Intuition
The problem requires checking if a given string of digits, when repeatedly transformed by replacing adjacent pairs with their sum modulo 10, eventually reduces to two identical digits.
## Approach
1. **Iterative Transformation**:
- Start with the given string `s`.
- Repeatedly form a new... | 0 | 0 | ['Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | my solution | my-solution-by-leman_cap13-574e | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | leman_cap13 | NORMAL | 2025-03-30T19:02:15.010852+00:00 | 2025-03-30T19:02:15.010852+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
`... | 0 | 0 | ['Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Scala solution with implicit, recursion and pattern matching | scala-solution-with-implicit-recursion-a-k30z | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | iyIeO99AmH | NORMAL | 2025-03-29T23:47:52.669218+00:00 | 2025-03-29T23:47:52.669218+00:00 | 1 | 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 | ['Scala'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Solution using extra space with JS | solution-using-extra-space-with-js-by-be-e6ei | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | bek-shoyatbek | NORMAL | 2025-03-29T19:33:01.811139+00:00 | 2025-03-29T19:33:01.811139+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
`... | 0 | 0 | ['JavaScript'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Simple C++ Solution using queue. | simple-c-solution-using-queue-by-optimiz-yddk | Complexity
Time complexity:
O(n)
Space complexity:
O(n)
Code | OptimizingCompiler | NORMAL | 2025-03-29T10:44:19.908531+00:00 | 2025-03-29T10:44:19.908531+00:00 | 3 | false | # Complexity
- Time complexity:
$$O(n)$$
- Space complexity:
$$O(n)$$
# Code
```cpp []
class Solution {
public:
bool hasSameDigits(string s) {
deque<int> q;
for (char c : s) {
q.push_back(c - '0');
}
int n = 0;
int c = q.size();
while (q.size() > 2) {
... | 0 | 0 | ['C++'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Simple Swift Solution | simple-swift-solution-by-felisviridis-qjyc | Code | Felisviridis | NORMAL | 2025-03-25T09:51:01.184421+00:00 | 2025-03-25T09:51:01.184421+00:00 | 2 | false | 
# Code
```swift []
class Solution {
func hasSameDigits(_ s: String) -> Bool {
var digits = s.compactMap(\.wholeNumberValue)
while digits.count > 2 {
... | 0 | 0 | ['Swift'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Go | go-by-stuu-7m0i | Code | stuu | NORMAL | 2025-03-23T01:48:21.827964+00:00 | 2025-03-23T01:48:21.827964+00:00 | 3 | false | 
# Code
```golang []
func hasSameDigits(s string) bool {
operationResult := s
for {
operationResult = reduceDigit(operationResult)
if len(operationResult) == 2 {
break
}
}
if operationResu... | 0 | 0 | ['Go'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | C++ Solution | c-solution-by-chandreyee_12-uaqy | Code | chandreyee_12 | NORMAL | 2025-03-22T15:50:14.666810+00:00 | 2025-03-22T15:50:14.666810+00:00 | 5 | false |
# Code
```cpp []
class Solution {
public:
bool hasSameDigits(string s) {
while(s.size()>2)
{
string str="";
for(int i=0;i<s.size()-1;i++)
{
int x=s[i]-'0';
int y=s[i+1]-'0';
int sum=(x+y)%10;
char c... | 0 | 0 | ['C++'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Update the list in place, 93% speed | update-the-list-in-place-93-speed-by-evg-zx83 | Code | evgenysh | NORMAL | 2025-03-22T12:59:59.911040+00:00 | 2025-03-22T12:59:59.911040+00:00 | 2 | false | # Code
```python3 []
class Solution:
def hasSameDigits(self, s: str) -> bool:
lst = list(map(int, s))
for end in range(len(lst) - 1, 1, -1):
for i in range(end):
lst[i] = (lst[i] + lst[i + 1]) % 10
return lst[0] == lst[1]
``` | 0 | 0 | ['Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Easy Java Solution for Beginners | easy-java-solution-for-beginners-by-ther-u0ad | Upvote PleaseApproachComplexity
Time complexity:
Space complexity:
Code | therajnishraj | NORMAL | 2025-03-21T05:42:39.452279+00:00 | 2025-03-21T05:42:39.452279+00:00 | 3 | false | # Upvote Please
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add yo... | 0 | 0 | ['Hash Table', 'Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Easy In-Place Solution | easy-in-place-solution-by-j_cox-9qpy | IntuitionApproach
Loop through S until it has 2 characters
On each loop, take pairs of chars, convert them to int
Perform the operation of adding them and modul | J_Cox | NORMAL | 2025-03-20T16:03:05.019220+00:00 | 2025-03-20T16:03:05.019220+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
1. Loop through S until it has 2 characters
2. On each loop, take pairs of chars, convert them to int
3. Perform the operation of adding them and modulo by 10
4. Replace the first of the two digits with this new value
5. Repeat ... | 0 | 0 | ['C++'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | True subquadratic complexity using a bit of number theory :) | true-subquadratic-complexity-using-a-bit-zug9 | ApproachThere is a closed form for the two final digits - they are given by summing digits of s multiplied by binomial coefficients, similar to Pascal's triangl | user5094uY | NORMAL | 2025-03-19T17:58:22.537458+00:00 | 2025-03-19T17:58:22.537458+00:00 | 2 | false | # Approach
There is a closed form for the two final digits - they are given by summing digits of `s` multiplied by binomial coefficients, similar to Pascal's triangle.
This can be leveraged to get log-linear time complexity and logarithmic space complexity. This is a situation where we can't just assume all arithmetic... | 0 | 0 | ['Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Using Character Array and HashMap | using-character-array-and-hashmap-by-hem-mlet | IntuitionUsing Character Array and HashMapApproachIterative Transformation:The while loop runs until the length of s is reduced to 2.
In each iteration, calcula | hemu1116 | NORMAL | 2025-03-18T17:14:18.241757+00:00 | 2025-03-18T17:14:18.241757+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Using Character Array and HashMap
# Approach
<!-- Describe your approach to solving the problem. -->
Iterative Transformation:
The while loop runs until the length of s is reduced to 2.
In each iteration, calculateNewS(s, n) is called, wh... | 0 | 0 | ['Java'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | 0ms Runtime using C Programming Language with Easy String Operations | 0ms-runtime-using-c-programming-language-1hh9 | IntuitionThe problem revolves around transforming a string of digits into a sequence of pairs, repeatedly reducing the length of the string until only two digit | Vivek_Bartwal | NORMAL | 2025-03-15T05:32:23.636762+00:00 | 2025-03-15T05:32:23.636762+00:00 | 9 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem revolves around transforming a string of digits into a sequence of pairs, repeatedly reducing the length of the string until only two digits remain. The main focus is to compute the new digits as the modulo 10 of the sum of cons... | 0 | 0 | ['C'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Easy logic and code | easy-logic-and-code-by-aniketkumarsingh9-mqed | Intuitionchanging the string based on its size that should not get to 2.ApproachIterative Digit ReductionComplexity
Time complexity:
O(n).
Space complexity: | AniketKumarSingh99 | NORMAL | 2025-03-14T13:42:09.234943+00:00 | 2025-03-14T13:42:09.234943+00:00 | 3 | false | # Intuition
changing the string based on its size that should not get to 2.
# Approach
Iterative Digit Reduction
# Complexity
- Time complexity:
O(n).
- Space complexity:
O(1).
# Code
```cpp []
class Solution {
public:
bool hasSameDigits(string s) {
int size=s.size();
while(size!=2){
... | 0 | 0 | ['C++'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | SIMPLE C PROGRAM | simple-c-program-by-mr_jaikumar-ot5u | IntuitionApproachComplexity
Time complexity:
0 MS
Space complexity:
Code | Mr_JAIKUMAR | NORMAL | 2025-03-14T05:34:28.155999+00:00 | 2025-03-14T05:34:28.155999+00:00 | 11 | 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. $$O(n)$$ -->
0 MS
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# C... | 0 | 0 | ['C'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Kotlin Imperative Solution | kotlin-imperative-solution-by-curenosm-joib | Code | curenosm | NORMAL | 2025-03-13T14:35:19.032310+00:00 | 2025-03-13T14:35:19.032310+00:00 | 2 | false | # Code
```kotlin []
class Solution {
fun hasSameDigits(s: String): Boolean {
var s = s; var i = 1
while (s.length > 2) {
val str = StringBuilder()
for (i in 1 until s.length)
str.append(
(s[i - 1].digitToInt() + s[i].digitToInt()) % 10
... | 0 | 0 | ['Kotlin'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | Easy to understand... | easy-to-understand-by-udisha_1234-rc5v | Code | Udisha_1234 | NORMAL | 2025-03-13T12:52:43.788627+00:00 | 2025-03-13T12:52:43.788627+00:00 | 5 | false |
# Code
```cpp []
class Solution {
public:
bool hasSameDigits(string s) {
string ans = s;
while (ans.size() != 2) {
s = ans;
ans = "";
for (int i = 0; i + 1 < s.size(); i++) {
int a = s[i] - 48;
int b = s[i + 1] - 48;
... | 0 | 0 | ['C++'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | simple and easy answer | simple-and-easy-answer-by-shiva135-vk92 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Shiva135 | NORMAL | 2025-03-12T15:07:17.592900+00:00 | 2025-03-12T15:07:17.592900+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
`... | 0 | 0 | ['Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | simple | simple-by-diorsalimov2006-15e2 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | diorsalimov2006 | NORMAL | 2025-03-12T07:54:18.446961+00:00 | 2025-03-12T07:54:18.446961+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
`... | 0 | 0 | ['Python3'] | 0 |
check-if-digits-are-equal-in-string-after-operations-i | C++ | c-by-tinachien-nkko | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | TinaChien | NORMAL | 2025-03-12T00:40:31.996732+00:00 | 2025-03-12T00:40:31.996732+00:00 | 4 | 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 |
check-if-digits-are-equal-in-string-after-operations-i | Java | java-by-soumya_699-5nd6 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Soumya_699 | NORMAL | 2025-03-11T08:47:58.021618+00:00 | 2025-03-11T08:47:58.021618+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 | ['Java'] | 0 |
longest-uploaded-prefix | Python Elegant & Short | Amortized O(1) | Commented | python-elegant-short-amortized-o1-commen-wz9c | \nclass LUPrefix:\n """\n Memory: O(n)\n Time: O(1) per upload call, because adding to the set takes O(1) time, and the prefix\n\t\t\t\t can be incre | Kyrylo-Ktl | NORMAL | 2022-10-01T16:07:49.332700+00:00 | 2022-10-01T18:19:06.617886+00:00 | 3,394 | false | ```\nclass LUPrefix:\n """\n Memory: O(n)\n Time: O(1) per upload call, because adding to the set takes O(1) time, and the prefix\n\t\t\t\t can be increased no more than n times for all n calls to the upload function\n """\n\n def __init__(self, n: int):\n self._longest = 0\n self._nums =... | 78 | 2 | ['Python', 'Python3'] | 12 |
longest-uploaded-prefix | Most Easy and Short solution + Meme | most-easy-and-short-solution-meme-by-har-2iuy | Self explanatory solution\n\n\nclass LUPrefix {\n Set<Integer> set;\n int max=0;\n public LUPrefix(int n) {\n set=new HashSet<>();\n }\n p | HarshitMaurya | NORMAL | 2022-10-01T16:00:53.534938+00:00 | 2022-11-01T06:23:08.799523+00:00 | 2,741 | false | **Self explanatory solution**\n\n```\nclass LUPrefix {\n Set<Integer> set;\n int max=0;\n public LUPrefix(int n) {\n set=new HashSet<>();\n }\n public void upload(int video) {\n set.add(video);\n while(set.contains(max+1)) max++;\n }\n public int longest() {\n return max... | 37 | 1 | ['Ordered Set', 'Java'] | 7 |
longest-uploaded-prefix | C++ | Set | Easy Understanding | c-set-easy-understanding-by-kiranpalsing-wh7n | Approach \n- We will insert number in set s.\n- When the longest() function will be called, the set will try to increase the answer by checking the next values. | kiranpalsingh1806 | NORMAL | 2022-10-01T16:04:20.078027+00:00 | 2022-10-01T16:16:32.096928+00:00 | 2,449 | false | **Approach** \n- We will insert number in set `s`.\n- When the `longest()` function will be called, the set will try to increase the answer by checking the next values.\n\n**C++ Code**\n\n```cpp\nclass LUPrefix {\n public:\n set<int> s;\n int t = 0;\n LUPrefix(int n) {\n }\n\n void upload(int video) {\... | 26 | 4 | ['C'] | 7 |
longest-uploaded-prefix | Java TreeSet (super simple) | java-treeset-super-simple-by-ricola-ba5r | Intuition\nKeep a TreeSet of all the numbers (videos) NOT uploaded yet.\nYou can then look at the lowest not uploaded value, and the prefix is the value just be | ricola | NORMAL | 2022-10-01T16:01:10.469006+00:00 | 2022-10-03T09:46:49.356863+00:00 | 1,025 | false | # Intuition\nKeep a TreeSet of all the numbers (videos) NOT uploaded yet.\nYou can then look at the lowest not uploaded value, and the prefix is the value just before this one.\nSince it\'s a tree, it allows to get the lowest value (leftmost node in the tree) in O(log n)\n\n# Complexity\n- Time complexity:\n - initi... | 19 | 1 | ['Java'] | 1 |
longest-uploaded-prefix | ✅C++ | ✅Use Array | ✅Easy approach | c-use-array-easy-approach-by-yash2arma-e07s | \nclass LUPrefix {\npublic:\n \n vector<int> pre;\n int val=0;\n \n LUPrefix(int n) \n {\n pre.resize(n+2, 0);\n }\n \n void u | Yash2arma | NORMAL | 2022-10-01T17:11:30.530961+00:00 | 2022-10-01T17:11:30.531012+00:00 | 1,080 | false | ```\nclass LUPrefix {\npublic:\n \n vector<int> pre;\n int val=0;\n \n LUPrefix(int n) \n {\n pre.resize(n+2, 0);\n }\n \n void upload(int video) \n {\n pre[video] = 1;\n \n }\n \n int longest() \n {\n while(pre[val+1]==1)\n val++;\n ... | 12 | 0 | ['Array', 'C', 'C++'] | 1 |
longest-uploaded-prefix | Python 3 || iteration || T/S: 97% / 59% | python-3-iteration-ts-97-59-by-spaulding-lg18 | \nclass LUPrefix:\n\n def __init__(self, n: int):\n self.stream = [False]*n\n self.maxLength = n\n self.prefLength = 0\n \n de | Spaulding_ | NORMAL | 2022-10-01T19:06:27.019656+00:00 | 2024-06-15T14:41:14.594046+00:00 | 579 | false | ```\nclass LUPrefix:\n\n def __init__(self, n: int):\n self.stream = [False]*n\n self.maxLength = n\n self.prefLength = 0\n \n def upload(self, video: int) -> None:\n self.stream[video-1] = True\n\n def longest(self) -> int:\n for i in range(self.prefLength,self.maxLen... | 9 | 0 | ['Python', 'Python3'] | 0 |
longest-uploaded-prefix | ✅✅✅ C++ with Explanation || Very Simple & Easy to Understand Solution | c-with-explanation-very-simple-easy-to-u-s23c | Up Vote if you like the solution \n\n\n/* take an array that keeps a mark of each of the video that uploaded.\n keep another varible - latest, that keeps trac | kreakEmp | NORMAL | 2022-10-01T16:02:18.394646+00:00 | 2022-10-01T16:27:24.477152+00:00 | 898 | false | <b>Up Vote if you like the solution \n```\n\n/* take an array that keeps a mark of each of the video that uploaded.\n keep another varible - latest, that keeps track of all the video that\n has been already uploaded till that point of time.\n Update latest, when the new video uploaded just next to it, also keep \... | 9 | 0 | [] | 3 |
longest-uploaded-prefix | 3 Solutions: Fenwick Tree or Binary Indexed Tree | Segment Tree | Disjoint Set ADT | 3-solutions-fenwick-tree-or-binary-index-1jvx | This discussion thread includes 3 separate solutions using different data structures namely,\n\n1. Fenwick Tree or Binary Indexed Tree\n2. Segment Tree\n3. Disj | mrtwinklesharma | NORMAL | 2022-10-09T21:18:00.444046+00:00 | 2022-10-09T21:18:48.414730+00:00 | 522 | false | This discussion thread includes 3 separate solutions using different data structures namely,\n\n**1. Fenwick Tree or Binary Indexed Tree\n2. Segment Tree\n3. Disjoint Set**\n\nThe intuition behind fenwick tree and segment tree solution is,\n=> We want to find the Longest Uploaded Prefix,\n=> Let\'s create an array of s... | 8 | 0 | ['Tree', 'Binary Tree'] | 3 |
longest-uploaded-prefix | Java | O(1) time | 1D Array | java-o1-time-1d-array-by-akrchaubey-ndon | ```\nclass LUPrefix {\n \n boolean[] uploaded;\n int max;\n int size;\n public LUPrefix(int n) {\n uploaded = new boolean[n + 1];\n | akrchaubey | NORMAL | 2022-10-01T16:34:13.658878+00:00 | 2022-10-01T16:42:57.339528+00:00 | 775 | false | ```\nclass LUPrefix {\n \n boolean[] uploaded;\n int max;\n int size;\n public LUPrefix(int n) {\n uploaded = new boolean[n + 1];\n max = 0;\n size = n;\n }\n \n public void upload(int video) {\n uploaded[video] = true;\n\t\t\n\t\t/** \n\t\tIf the longest uploaded pre... | 8 | 2 | ['Array', 'Java'] | 0 |
longest-uploaded-prefix | Extremely Easy Solution | Vector instead of Set | Linear Solution | extremely-easy-solution-vector-instead-o-5pcz | We can just use a vector and resize it to n to initialize it. We can maintain a global variable named last wihch will tell us the longest video which we can see | modernbeast02 | NORMAL | 2022-10-03T02:14:10.231590+00:00 | 2022-10-03T02:14:10.231624+00:00 | 581 | false | We can just use a vector and resize it to n to initialize it. We can maintain a global variable named last wihch will tell us the longest video which we can see. Total Time Complexity will be O(N) but Amortized Time Complexity will be O(1). If it helped u, don\'t forget to upvote.\uD83D\uDE03\n```\nclass LUPrefix {\n ... | 7 | 0 | [] | 0 |
longest-uploaded-prefix | very easy solution ||C++||O(Nlogn) | very-easy-solution-conlogn-by-baibhavkr1-ooh0 | Intuition\nWe will going to use set here because we know that set have element in sorted order\n\n# Approach\nnow i am going to describe my intution using a exa | baibhavkr143 | NORMAL | 2022-10-01T16:03:35.029265+00:00 | 2022-10-01T16:13:24.670707+00:00 | 612 | false | # Intuition\nWe will going to use set here because we know that set have element in sorted order\n\n# Approach\nnow i am going to describe my intution using a example\nlet n=6;\nnow put all element from 1 to 6 in set\ns={1,2,3,4,5,6}\n\nnow when ever the function ***upload*** is called we gonna delete that element from... | 7 | 0 | ['Ordered Set', 'C++'] | 2 |
longest-uploaded-prefix | TreeSet | treeset-by-java_programmer_ketan-o0c4 | \n/*\nWe will maintain a sorted set of un-uploaded videos.\nTo handle longest() query just return the first_element of the set -1\nFor uploading videos remove t | Java_Programmer_Ketan | NORMAL | 2022-10-01T16:01:16.135472+00:00 | 2022-10-01T16:01:16.135535+00:00 | 278 | false | ```\n/*\nWe will maintain a sorted set of un-uploaded videos.\nTo handle longest() query just return the first_element of the set -1\nFor uploading videos remove the video from the set\n*/\nclass LUPrefix {\n TreeSet<Integer> set;\n public LUPrefix(int n) {\n this.set = new TreeSet<>();\n for(int i=... | 6 | 2 | [] | 2 |
longest-uploaded-prefix | Disjoint Set || Java | disjoint-set-java-by-abdulazizms-jbau | Intuition\nFirst thing came to my mind was to use a disjoint set since different chains are merged at some point.\n\n\n# Approach\nI want to connect the current | abdulazizms | NORMAL | 2022-10-01T16:10:11.796035+00:00 | 2022-10-01T16:23:42.937746+00:00 | 532 | false | # Intuition\nFirst thing came to my mind was to use a disjoint set since different chains are merged at some point.\n\n\n# Approach\nI want to connect the current index to its left and right and maintain the total size of the chain. Disjoint set is a great candidate.\n\nThis approach can be used to solve this problem t... | 5 | 0 | ['Java'] | 1 |
longest-uploaded-prefix | 🧽 Java Clean & Simple | Union Find | java-clean-simple-union-find-by-palmas-ala9 | \nclass LUPrefix {\n int[] map;\n\n public LUPrefix(int n) {\n map = new int[n + 1];\n }\n\n public void upload(int video) {\n map[vid | palmas | NORMAL | 2022-10-01T16:41:58.310568+00:00 | 2022-10-01T16:41:58.310606+00:00 | 279 | false | ```\nclass LUPrefix {\n int[] map;\n\n public LUPrefix(int n) {\n map = new int[n + 1];\n }\n\n public void upload(int video) {\n map[video - 1] = find(video);\n }\n\n public int longest() {\n return find(0);\n }\n\n int find(int index) {\n if (map[index] == 0)\n ... | 4 | 0 | ['Java'] | 1 |
longest-uploaded-prefix | [Python3] Using List || O(1) amortized || 354ms || beats 100% | python3-using-list-o1-amortized-354ms-be-keh0 | Here is a small trick with adding additional False to the list end like indicator of boundary. So we don\'t need to check the boundry in the while loop and got | yourick | NORMAL | 2024-03-09T19:14:47.874518+00:00 | 2024-03-09T19:35:46.554497+00:00 | 104 | false | ###### Here is a small trick with adding additional False to the list end like indicator of boundary. So we don\'t need to check the boundry in the while loop and got additional speed profit.\n`while self.videos[self.prefix]` instead of \n`while self.prefix < len(self.videos) and self.videos[self.prefix]`\n```python3 [... | 3 | 0 | ['Python', 'Python3'] | 0 |
longest-uploaded-prefix | 98% FASTER || C++ || USE ARRAY || EASY APPROACH | 98-faster-c-use-array-easy-approach-by-y-yarh | \nclass LUPrefix {\npublic:\n int x = 1;\n vector<bool> v;\n LUPrefix(int n) {\n v.resize(100002,false);\n }\n \n void upload(int video | yash___sharma_ | NORMAL | 2023-03-06T05:02:05.969121+00:00 | 2023-03-06T05:02:05.969181+00:00 | 569 | false | ```\nclass LUPrefix {\npublic:\n int x = 1;\n vector<bool> v;\n LUPrefix(int n) {\n v.resize(100002,false);\n }\n \n void upload(int video) {\n v[video] = true;\n while(v[x]==true){\n x++;\n }\n }\n \n int longest() {\n return x-1;\n }\n};\n``` | 3 | 0 | ['Array', 'C', 'C++'] | 0 |
longest-uploaded-prefix | ✅C++ || Easy to understand CODE || SHORT | c-easy-to-understand-code-short-by-abhin-tl9e | \n\n\tclass LUPrefix {\n\t\tpublic:\n\t\t\tvector v;\n\t\t\tint it=0;\n\t\t\tLUPrefix(int n) {\n\t\t\t\tvector temp(n,-1);\n\t\t\t\tv=temp;\n\t\t\t}\n\n\t\t\tvo | abhinav_0107 | NORMAL | 2022-10-01T18:58:03.353364+00:00 | 2022-10-01T18:59:15.520582+00:00 | 906 | false | \n\n\tclass LUPrefix {\n\t\tpublic:\n\t\t\tvector<int> v;\n\t\t\tint it=0;\n\t\t\tLUPrefix(int n) {\n\t\t\t\tvector<int> temp(n,-1);\n\t\t\t\tv=temp;\n\t\t\t}\n\n\t\t\tvoid upload(int video) {\n\t\t\t\tv[video-... | 3 | 0 | ['C', 'C++'] | 0 |
longest-uploaded-prefix | JAVA solution | HashSet | java-solution-hashset-by-sourin_bruh-b73l | Please Upvote !!! (\u25E0\u203F\u25E0)\n\nclass LUPrefix {\n Set<Integer> set;\n int maxConsecutiveVideo = 0;\n \n public LUPrefix(int n) {\n | sourin_bruh | NORMAL | 2022-10-01T18:05:30.507548+00:00 | 2022-10-01T18:05:30.507574+00:00 | 368 | false | ### ***Please Upvote !!!*** **(\u25E0\u203F\u25E0)**\n```\nclass LUPrefix {\n Set<Integer> set;\n int maxConsecutiveVideo = 0;\n \n public LUPrefix(int n) {\n set = new HashSet<>();\n }\n \n public void upload(int video) {\n set.add(video);\n while (set.contains(maxConsecutiveV... | 3 | 0 | ['Java'] | 0 |
longest-uploaded-prefix | easy implementation O(n) in vector || comments|| beginner friendly | easy-implementation-on-in-vector-comment-lzkw | \tclass LUPrefix {\n private:\n vector v;\n int i;\n\tpublic:\n LUPrefix(int n) {\n v.resize(n+1);\n\t\t//take i from start\n i=0;\n | sjain8078 | NORMAL | 2022-10-01T16:03:35.221525+00:00 | 2022-10-01T17:13:21.977631+00:00 | 218 | false | \tclass LUPrefix {\n private:\n vector<int> v;\n int i;\n\tpublic:\n LUPrefix(int n) {\n v.resize(n+1);\n\t\t//take i from start\n i=0;\n // memset(v.begin(),v.end(),0);\n }\n \n void upload(int video) {\n\t//upload it to the previous index so we can track it\n v[video-1]... | 3 | 0 | ['Design', 'C'] | 0 |
longest-uploaded-prefix | Using Segment Tree Approach | using-segment-tree-approach-by-jayush10a-tgdb | 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 | jayush10ashu | NORMAL | 2024-01-18T21:44:45.770255+00:00 | 2024-01-18T21:44:45.770297+00:00 | 72 | 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 | ['Segment Tree', 'C++'] | 0 |
longest-uploaded-prefix | Easy Solution Using Max & Min Heap | easy-solution-using-max-min-heap-by-shri-0tq5 | \n\n# Code\n\nclass LUPrefix {\npublic:\n priority_queue<int>q;\n priority_queue<int,vector<int>,greater<int>>q2;\n int prev;\n LUPrefix(int n) {\n | Shristha | NORMAL | 2023-02-05T13:08:45.611241+00:00 | 2023-02-05T13:08:45.611273+00:00 | 443 | false | \n\n# Code\n```\nclass LUPrefix {\npublic:\n priority_queue<int>q;\n priority_queue<int,vector<int>,greater<int>>q2;\n int prev;\n LUPrefix(int n) {\n prev=0;\n }\n \n void upload(int video) {\n q.push(video);\n q2.push(video);\n }\n \n int longest() {\n if(!q.e... | 2 | 0 | ['Heap (Priority Queue)', 'C++'] | 0 |
longest-uploaded-prefix | [Python] Union Find | python-union-find-by-coolguazi-w2d8 | \n# Code\n\nclass LUPrefix:\n\n def __init__(self, n: int):\n self.videos = list(range(n + 1))\n self.uploaded = [1] + [0] * (n + 1)\n\n def | coolguazi | NORMAL | 2023-01-30T10:55:56.179039+00:00 | 2023-01-30T10:55:56.179070+00:00 | 73 | false | \n# Code\n```\nclass LUPrefix:\n\n def __init__(self, n: int):\n self.videos = list(range(n + 1))\n self.uploaded = [1] + [0] * (n + 1)\n\n def upload(self, video: int) -> None:\n self.uploaded[video] = 1\n if self.uploaded[video + 1]: self.union(video, video + 1)\n if self.uplo... | 2 | 0 | ['Union Find', 'Python3'] | 0 |
longest-uploaded-prefix | very simple solution using bitarray | very-simple-solution-using-bitarray-by-d-4nfc | Intuition\nUse an array to keep the indexes of the videos that has been uploaded, and a variable llp (Last Longest Prefix) to keep the last longer prefix\n\n# A | dwluciano | NORMAL | 2022-12-12T20:14:47.667377+00:00 | 2022-12-12T20:15:18.776228+00:00 | 37 | false | # Intuition\nUse an array to keep the indexes of the videos that has been uploaded, and a variable llp (Last Longest Prefix) to keep the last longer prefix\n\n# Approach\nUse bit array to keep the indexes of the uploaded videos, and precalculate the LUP on each upload, this means to loop through the bit array until a n... | 2 | 0 | ['Bit Manipulation', 'C#'] | 0 |
longest-uploaded-prefix | Easiest approach | C++ | Vector | easiest-approach-c-vector-by-yashrajyash-fshv | \nclass LUPrefix {\npublic:\n vector<int> arr;\n int ptr = 0;\n LUPrefix(int n) {\n arr.resize(n, 0);\n }\n \n void upload(int video) { | yashrajyash | NORMAL | 2022-11-04T12:21:27.907972+00:00 | 2022-11-04T12:21:27.908019+00:00 | 30 | false | ```\nclass LUPrefix {\npublic:\n vector<int> arr;\n int ptr = 0;\n LUPrefix(int n) {\n arr.resize(n, 0);\n }\n \n void upload(int video) {\n arr[video-1]++;\n }\n \n int longest() {\n while(ptr < arr.size() && arr[ptr] != 0)\n ptr++;\n return ptr;\n }... | 2 | 0 | ['Array', 'C', 'C++'] | 0 |
longest-uploaded-prefix | [JAVA] Three Approaches [ Segment Tree + Binary Search ] [ Union Find] [Array Based] | java-three-approaches-segment-tree-binar-78xe | \n\n# Segment Tree + Binary Search\n\n\nclass LUPrefix {\n int tree [] ;\n int update(int l , int r , int index,int value){\n if(l==r && l==value){ | aniket7419 | NORMAL | 2022-10-02T17:37:30.688751+00:00 | 2022-10-02T17:37:30.688793+00:00 | 179 | false | \n\n# Segment Tree + Binary Search\n```\n\nclass LUPrefix {\n int tree [] ;\n int update(int l , int r , int index,int value){\n if(l==r && l==value){\n tree[index] = 1;\n return 1;\n }\n int mid = (l+r)/2;\n if(r<value || value<l) return tree[index];\n els... | 2 | 0 | ['Java'] | 0 |
longest-uploaded-prefix | C++ | Using Vector | Time Complexity - O(N) | Space Complexity - O(N) | c-using-vector-time-complexity-on-space-qnt78 | \nclass LUPrefix {\nprivate:\n int i;\n int n;\n vector<int> nums;\npublic:\n LUPrefix(int N) {\n i = 0;\n n = N+2;\n nums.resi | CPP_Bot | NORMAL | 2022-10-02T01:53:34.025574+00:00 | 2022-10-02T01:53:34.025615+00:00 | 284 | false | ```\nclass LUPrefix {\nprivate:\n int i;\n int n;\n vector<int> nums;\npublic:\n LUPrefix(int N) {\n i = 0;\n n = N+2;\n nums.resize(n, -1);\n }\n \n void upload(int video) {\n nums[video] = 1;\n while(i<n && (nums[i+1]!=-1)){\n i++;\n } \n ... | 2 | 0 | ['C'] | 0 |
longest-uploaded-prefix | [Python] Union-Find | python-union-find-by-dazzbourgh-cfpt | When adding an element, it becomes a member of its own set. Then if a video behind it was uploaded, go ahead and union new set with the previous one. Similarly, | dazzbourgh | NORMAL | 2022-10-01T19:51:57.798551+00:00 | 2022-10-01T19:51:57.798590+00:00 | 146 | false | When adding an element, it becomes a member of its own set. Then if a video behind it was uploaded, go ahead and union new set with the previous one. Similarly, if there\'s another set right after it - merge these two, too.\n\n```\nclass LUPrefix:\n\n def __init__(self, n: int):\n self.uploaded = [0] * (n + 1... | 2 | 0 | ['Union Find', 'Python'] | 0 |
longest-uploaded-prefix | Few Lines of Code That you should See.O(1) | few-lines-of-code-that-you-should-seeo1-bwrtt | \n | abhinaychaturvedi88 | NORMAL | 2022-10-01T17:44:05.114756+00:00 | 2022-10-01T17:50:12.188768+00:00 | 145 | false | \n | 2 | 0 | [] | 0 |
longest-uploaded-prefix | C++ | Easy Understanding | c-easy-understanding-by-coderabhi2308-6bgs | Self Explanatory Solution\n\nclass LUPrefix {\npublic:\n int ans=0;\n vector< int >isUpdated;\n \n LUPrefix( int n ) {\n isUpdated.assign( n+ | CoderAbhi2308 | NORMAL | 2022-10-01T16:33:42.669069+00:00 | 2022-10-01T16:34:56.349959+00:00 | 90 | false | Self Explanatory Solution\n\n```class LUPrefix {\npublic:\n int ans=0;\n vector< int >isUpdated;\n \n LUPrefix( int n ) {\n isUpdated.assign( n+2 , 0 );\n }\n \n void upload( int video ) {\n isUpdated[ video ] = 1;\n while ( isUpdated[ ans + 1 ] == 1 )\n {\n a... | 2 | 0 | ['C', 'Binary Tree'] | 0 |
longest-uploaded-prefix | Python clean | python-clean-by-404akhan-mdny | \n# N2. Longest Uploaded Prefix\nclass LUPrefix:\n def __init__(self, n: int):\n self.arr = [0] * (n + 2)\n self.ans = 0\n\n def upload(self | 404akhan | NORMAL | 2022-10-01T16:17:55.888047+00:00 | 2022-10-01T16:17:55.888075+00:00 | 150 | false | ```\n# N2. Longest Uploaded Prefix\nclass LUPrefix:\n def __init__(self, n: int):\n self.arr = [0] * (n + 2)\n self.ans = 0\n\n def upload(self, video: int) -> None:\n self.arr[video] = 1\n\n def longest(self) -> int:\n # answer only grows up, by 1\n while self.arr[self.ans +... | 2 | 0 | [] | 0 |
longest-uploaded-prefix | C++ | Easy to Understand | Using Vector | c-easy-to-understand-using-vector-by-dhe-680l | \nclass LUPrefix {\npublic:\n vector<int> v;\n int maxi=1;\n LUPrefix(int n) \n {\n v.resize(n+2, 0); \n v[0]=1;\n maxi=1;\n | dheerajkarwasra | NORMAL | 2022-10-01T16:16:18.417072+00:00 | 2022-10-01T17:23:05.324226+00:00 | 29 | false | ```\nclass LUPrefix {\npublic:\n vector<int> v;\n int maxi=1;\n LUPrefix(int n) \n {\n v.resize(n+2, 0); \n v[0]=1;\n maxi=1;\n }\n \n void upload(int video) \n {\n v[video]=1; \n while(v[maxi])\n maxi++;\n }\n \n int longest() \n {\n ... | 2 | 0 | ['C'] | 0 |
longest-uploaded-prefix | One Liner functions using vectors O(n) solution | one-liner-functions-using-vectors-on-sol-6kre | Explanation :\n1) i(variable) stores the current index upto which video prefixes from 1 to i-1 are all uploades.\n2) If a videos is uploaded while loop wil | Priyanshu_Chaudhary_ | NORMAL | 2022-10-01T16:13:35.320865+00:00 | 2022-10-01T17:19:20.456942+00:00 | 255 | false | Explanation :\n1) i(variable) stores the current index upto which video prefixes from 1 to i-1 are all uploades.\n2) If a videos is uploaded while loop will loop till it gets an unuploaded video \n3) Since all values are being travel only once the time complextity will be O(n).\n\n```\nclass LUPrefix {\npublic:\... | 2 | 0 | ['Array', 'C'] | 1 |
longest-uploaded-prefix | ✅✅✅🤓😎O(n) Solution with Most Easy Solution and Question too | on-solution-with-most-easy-solution-and-j0yar | ****Very Simple and Straight Forward Approch\n\n\nclass LUPrefix {\n\n private: vector<bool>server;\n int ind=-1,maxi=0,_n;\n public:\n | abhigupta7049612180 | NORMAL | 2022-10-01T16:06:18.488814+00:00 | 2022-10-01T16:15:50.118838+00:00 | 363 | false | ****Very Simple and Straight Forward Approch\n\n```\nclass LUPrefix {\n\n private: vector<bool>server;\n int ind=-1,maxi=0,_n;\n public:\n LUPrefix(int n) {\n server.resize(n+5,0);\n server[0]=1;\n _n=n;\n }\n \n void upload(int video) {\n server[video]=1;\n... | 2 | 0 | ['Array', 'C++'] | 1 |
longest-uploaded-prefix | [Python3] One-Pass Solution O(n), Clean & Concise | python3-one-pass-solution-on-clean-conci-18ws | \nclass LUPrefix:\n\n def __init__(self, n: int):\n self.videos = [False] * (n + 1)\n self.ans = 0\n\n def upload(self, video: int) -> None: | xil899 | NORMAL | 2022-10-01T16:03:04.790824+00:00 | 2022-10-01T16:03:04.790864+00:00 | 480 | false | ```\nclass LUPrefix:\n\n def __init__(self, n: int):\n self.videos = [False] * (n + 1)\n self.ans = 0\n\n def upload(self, video: int) -> None:\n self.videos[video] = True\n if video == self.ans + 1:\n while video < len(self.videos):\n if not self.videos[video... | 2 | 0 | ['Python', 'Python3'] | 0 |
longest-uploaded-prefix | simple code | streak method | simple-code-streak-method-by-priyanshuhe-pxzu | \nunordered_map<int, int> mp;\n LUPrefix(int n) {\n for(int i=0;i<n;i++)\n mp[i+1] = 0;\n }\n \n void upload(int video) {\n | priyanshuHere27 | NORMAL | 2022-10-01T16:02:34.845581+00:00 | 2022-10-01T16:13:26.254144+00:00 | 33 | false | ```\nunordered_map<int, int> mp;\n LUPrefix(int n) {\n for(int i=0;i<n;i++)\n mp[i+1] = 0;\n }\n \n void upload(int video) {\n int right = video + 1;\n int left = video - 1;\n int rightStreak = 0, leftStreak = 0;\n \n if(mp[right]!=0)\n rightSt... | 2 | 0 | ['C', 'Java'] | 1 |
longest-uploaded-prefix | C++ | Caching | Beats 98% - TC | c-caching-beats-98-tc-by-ghozt777-6hap | Intuition
use array to keep track of the videos uploaded
use caching to store previous results of longest() function to avoid recalculation
Approach
our DS cont | ghozt777 | NORMAL | 2025-03-25T06:07:11.172885+00:00 | 2025-03-25T06:07:11.172885+00:00 | 26 | false | # Intuition
- use array to keep track of the videos uploaded
- use caching to store previous results of `longest()` function to avoid recalculation
# Approach
- our DS contains a boolean array to store wether the ith video is uploaded or not, variable prefix to keep track of the largest prefix, and another variable to ... | 1 | 0 | ['C++'] | 0 |
longest-uploaded-prefix | (beats 95% for time) Segment tree | beats-95-for-time-segment-tree-by-divyan-qgps | Intuition\nwe can check for the left child and update the parent node on the basis of the left child if the left child dont have the node True the irrespective | Divyansh55BE29 | NORMAL | 2024-10-02T11:51:17.495872+00:00 | 2024-10-02T11:51:17.495905+00:00 | 41 | false | # Intuition\nwe can check for the left child and update the parent node on the basis of the left child if the left child dont have the node True the irrespective of the length of the right child and if the left child has a prefix value we can simply add the left and right child for parent value.\n\n# Approach\nSegment ... | 1 | 0 | ['Segment Tree', 'C++'] | 0 |
longest-uploaded-prefix | Segment Tree + binary search longest | O(1) time build, upload, O(logn) time logest(), O(n) space. | segment-tree-binary-search-longest-o1-ti-tzhf | 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 | ziegfeld | NORMAL | 2024-06-21T19:22:50.694902+00:00 | 2024-06-21T19:22:50.694918+00:00 | 20 | 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. $$O(n)$$ --... | 1 | 0 | ['C++'] | 1 |
longest-uploaded-prefix | Python | Union-Find | python-union-find-by-aljipa-9xhf | Code\n\nclass LUPrefix:\n\n def __init__(self, n: int):\n self.prnt = [-1] * n\n self.size = [0] * n\n\n def _find(self, i):\n while | aljipa | NORMAL | 2024-03-10T02:12:54.637166+00:00 | 2024-03-10T02:12:54.637192+00:00 | 9 | false | # Code\n```\nclass LUPrefix:\n\n def __init__(self, n: int):\n self.prnt = [-1] * n\n self.size = [0] * n\n\n def _find(self, i):\n while i != self.prnt[i]:\n i = self._find(self.prnt[i])\n return self.prnt[i]\n \n def _union(self, x, y):\n if x < 0 or y >= len(... | 1 | 0 | ['Python3'] | 0 |
longest-uploaded-prefix | Fenwick Tree Solution || Binary Search || Explained | fenwick-tree-solution-binary-search-expl-3cfj | First Learn Fenwick Tree from here :\nSimple Explanation\nhttps://www.hackerearth.com/practice/notes/binary-indexed-tree-or-fenwick-tree/\nAdvance Versions\nhtt | coder_rastogi_21 | NORMAL | 2024-02-24T13:52:15.923926+00:00 | 2024-06-25T06:39:31.056085+00:00 | 73 | false | # First Learn Fenwick Tree from here :\nSimple Explanation\n[https://www.hackerearth.com/practice/notes/binary-indexed-tree-or-fenwick-tree/]()\nAdvance Versions\n[https://cp-algorithms.com/data_structures/fenwick.html]()\n\n\n# Complexity\n- Time complexity:\nupload function: $$O(log(n))$$ \nlongest function: $$O(log(... | 1 | 0 | ['Binary Search', 'Binary Indexed Tree', 'C++'] | 1 |
longest-uploaded-prefix | [Java] Easy 100% solution | java-easy-100-solution-by-ytchouar-1rkl | java\nclass LUPrefix {\n private boolean[] videos;\n private int maxIdx = 0;\n\n public LUPrefix(int n) {\n this.videos = new boolean[n + 1];\n | YTchouar | NORMAL | 2024-02-21T03:26:45.035628+00:00 | 2024-02-21T03:27:19.382606+00:00 | 191 | false | ```java\nclass LUPrefix {\n private boolean[] videos;\n private int maxIdx = 0;\n\n public LUPrefix(int n) {\n this.videos = new boolean[n + 1];\n }\n \n public void upload(int video) {\n this.videos[video] = true;\n\n while(maxIdx < this.videos.length - 1 && videos[maxIdx + 1])\n... | 1 | 0 | ['Java'] | 0 |
longest-uploaded-prefix | Java | Array + Max Value | java-array-max-value-by-tbekpro-k5k5 | Code\n\nclass LUPrefix {\n\n int res;\n int max;\n int[] arr;\n\n public LUPrefix(int n) {\n res = 0;\n max = 0;\n arr = new in | tbekpro | NORMAL | 2023-11-25T14:23:22.935939+00:00 | 2023-11-25T14:23:22.935956+00:00 | 117 | false | # Code\n```\nclass LUPrefix {\n\n int res;\n int max;\n int[] arr;\n\n public LUPrefix(int n) {\n res = 0;\n max = 0;\n arr = new int[n + 1];\n }\n\n public void upload(int video) {\n arr[video]++;\n max = Math.max(video, max);\n if (video == res + 1) {\n ... | 1 | 0 | ['Java'] | 0 |
longest-uploaded-prefix | "Python Concise and Efficient | Achieving Amortized O(1) Complexity | Annotated for Clarity" | python-concise-and-efficient-achieving-a-gxtm | Intuition & Approach :\nThe class LUPrefix is designed to efficiently track the longest prefix of consecutive integers. It uses a set _nums to store uploaded vi | aditya0542pandey | NORMAL | 2023-10-28T06:29:00.598004+00:00 | 2023-10-28T06:29:00.598026+00:00 | 7 | false | # Intuition & Approach :\nThe class LUPrefix is designed to efficiently track the longest prefix of consecutive integers. It uses a set _nums to store uploaded videos, and the variable _longest to maintain the length of the longest consecutive prefix.\n\nIn the constructor __init__, it initializes _longest to 0 and cre... | 1 | 0 | ['Python3'] | 0 |
longest-uploaded-prefix | Java Average Time complexity O(1) | java-average-time-complexity-o1-by-milan-f6y4 | 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 | milangupta95 | NORMAL | 2023-08-07T20:10:55.124385+00:00 | 2023-08-07T20:10:55.124407+00:00 | 4 | 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)$$ --... | 1 | 0 | ['Java'] | 1 |
longest-uploaded-prefix | Easy Javascript Solution with array - max 731 ms | easy-javascript-solution-with-array-max-kwe5m | ```\nclass LUPrefix{\n constructor(n){\n this.i = 0;\n this.arr = new Array(n);\n //console.log(this.arr);\n }\n / \n * @param {nu | okarademirci | NORMAL | 2022-10-20T17:48:07.868916+00:00 | 2022-10-20T17:48:07.868956+00:00 | 14 | false | ```\nclass LUPrefix{\n constructor(n){\n this.i = 0;\n this.arr = new Array(n);\n //console.log(this.arr);\n }\n /** \n * @param {number} video\n * @return {void}\n */\n upload = function(video) {\n this.arr[video-1] = video;\n }\n\n/**\n * @return {number}\n */\n longest =... | 1 | 0 | [] | 0 |
longest-uploaded-prefix | Java || TreeSet || union Find | java-treeset-union-find-by-lagahuahubro-8co4 | set contains all the videos which are not yet uploaded so smallest number in the TreeSet represent the smallest numbered video that needs to be uploaded.\n\n\nc | lagaHuaHuBro | NORMAL | 2022-10-07T09:42:10.110252+00:00 | 2022-10-07T10:04:03.503252+00:00 | 45 | false | ```set contains all the videos which are not yet uploaded so smallest number in the TreeSet represent the smallest numbered video that needs to be uploaded.```\n\n```\nclass LUPrefix {\n TreeSet<Integer> set = new TreeSet<>();\n public LUPrefix(int n) {\n for (int i = 1; i <= n + 1; i++) {\n set... | 1 | 0 | [] | 0 |
longest-uploaded-prefix | Rust, HashSet, O(n) | rust-hashset-on-by-goodgoodwish-pb6k | Only record the current end of the prefix [1..i]. Keep extending it while can be extended.\n\nuse std::collections::HashSet;\n\nstruct LUPrefix {\n points: | goodgoodwish | NORMAL | 2022-10-07T03:53:04.465331+00:00 | 2022-10-07T03:53:04.465366+00:00 | 136 | false | Only record the current end of the prefix [1..i]. Keep extending it while can be extended.\n```\nuse std::collections::HashSet;\n\nstruct LUPrefix {\n points: HashSet<i32>,\n far: i32,\n}\n\n\n/** \n * `&self` means the method takes an immutable reference.\n * If you need a mutable reference, change it to `&mut ... | 1 | 0 | ['Rust'] | 0 |
longest-uploaded-prefix | C++ || Array | c-array-by-sachin-vinod-h487 | Approch :-\n1.] Here I am going to use a three number which is -1,0,1 which is as follows ( -1 -> not uploded, 0 -> uploaded but not created prefix , 1 -> uploa | Sachin-Vinod | NORMAL | 2022-10-06T11:32:41.360238+00:00 | 2022-10-06T11:32:41.360285+00:00 | 267 | false | **Approch** :-\n1.] Here I am going to use a three number which is -1,0,1 which is as follows ( -1 -> not uploded, 0 -> uploaded but not created prefix , 1 -> uploaded and created bond with prefix ).\n2.] In upload function what i am going to do is if **curr-1th** is connectd with prefix so we can expand prefix from *... | 1 | 0 | ['Array', 'C', 'C++'] | 0 |
longest-uploaded-prefix | C++ easy and concise solution | c-easy-and-concise-solution-by-sanket070-flpj | Happy Leetcoding !\n\n\nclass LUPrefix {\npublic:\n set<int>st;\n LUPrefix(int n) {\n for(int i=1;i<=n+1;i++)\n {\n st.insert(i); | sanket0708 | NORMAL | 2022-10-06T04:54:49.166637+00:00 | 2022-10-06T04:54:49.166683+00:00 | 61 | false | Happy Leetcoding !\n\n```\nclass LUPrefix {\npublic:\n set<int>st;\n LUPrefix(int n) {\n for(int i=1;i<=n+1;i++)\n {\n st.insert(i);\n }\n }\n \n void upload(int video) {\n st.erase(video);\n }\n \n int longest() {\n return (*st.begin())-1;\n }\n};... | 1 | 0 | [] | 0 |
longest-uploaded-prefix | Easy JS Solution: Queue | easy-js-solution-queue-by-dollysingh-tuzp | Explanation:\nData structure : Queue. \n\n\nvar LUPrefix = function(n) {\n let q = new Array(n).fill(0);\n this.q = q;\n this.f = 0; //"f" is front of | dollysingh | NORMAL | 2022-10-04T18:00:09.101073+00:00 | 2022-10-04T18:00:09.101113+00:00 | 84 | false | **Explanation:**\nData structure : Queue. \n\n```\nvar LUPrefix = function(n) {\n let q = new Array(n).fill(0);\n this.q = q;\n this.f = 0; //"f" is front of queue\n};\n\n//O(1)\nLUPrefix.prototype.upload = function(video) {\n this.q[video - 1] = 1;\n};\n\n//O(n)\nLUPrefix.prototype.longest = function() {\n... | 1 | 0 | ['Queue', 'JavaScript'] | 0 |
longest-uploaded-prefix | ✅ [Rust] fastest (100%) solution using boolean array (with detailed comments) | rust-fastest-100-solution-using-boolean-is2la | This solution employs a simple boolean array to store the upload status that is used to update longest prefix index. It demonstrated 98 ms runtime (100.00%) and | stanislav-iablokov | NORMAL | 2022-10-03T12:16:22.056899+00:00 | 2022-10-23T12:46:40.488448+00:00 | 79 | false | This [**solution**](https://leetcode.com/submissions/detail/814188631/) employs a simple boolean array to store the upload status that is used to update longest prefix index. It demonstrated **98 ms runtime (100.00%)** and used **54.5 MB memory (100.00%)**. Time complexity is linear: **O(N)**. Space complexity is const... | 1 | 0 | ['Array', 'Rust'] | 0 |
longest-uploaded-prefix | C++ solution using Fenwick Tree + Binary Search | c-solution-using-fenwick-tree-binary-sea-u5lf | Solution:\n```\ntemplate\nclass BIT {\npublic:\n vector bit;\n int n;\n \n BIT() {}\n \n BIT(int _n) {\n n = _n;\n bit = vector( | dhavalkumar | NORMAL | 2022-10-03T05:10:37.023618+00:00 | 2022-10-03T05:10:37.023657+00:00 | 157 | false | **Solution:**\n```\ntemplate<typename T>\nclass BIT {\npublic:\n vector<T> bit;\n int n;\n \n BIT() {}\n \n BIT(int _n) {\n n = _n;\n bit = vector<T>(n, 0LL);\n }\n \n void inc(int idx, T val) {\n for(int i = idx + 1; i <= n; i += (i & -i)) \n bit[i - 1] += val... | 1 | 0 | ['Binary Indexed Tree', 'Binary Tree'] | 0 |
longest-uploaded-prefix | [Python 3] Only update left and right endpoints | Time O(1), Space O(N) | python-3-only-update-left-and-right-endp-q2lv | If we maintain a list for all the segments so far, the longest uploaded prefix essentially is the size of the first element a.k.a self.P[1] in my case for avoid | zhuzhengyuan824 | NORMAL | 2022-10-02T20:54:45.459935+00:00 | 2022-10-02T20:58:54.059587+00:00 | 165 | false | If we maintain a list for all the segments so far, the longest uploaded prefix essentially is the size of the first element a.k.a `self.P[1]` in my case for avoiding index overflow.\n\nThe only tricky part is the way to update left&right most endpoints: \n1. find left most endpoint(`x-self.P[x-1]`) and rightmost endpoi... | 1 | 0 | ['Python'] | 0 |
longest-uploaded-prefix | c++ simple solution | c-simple-solution-by-hanzhoutang-ue1x | The idea is that after one video has been uploaded, it will never be removed, which means the logested prefix keeps growing and we can simply track the by a si | hanzhoutang | NORMAL | 2022-10-02T20:23:27.212958+00:00 | 2022-10-02T20:25:36.980115+00:00 | 35 | false | The idea is that after one video has been uploaded, it will never be removed, which means the logested prefix keeps growing and we can simply track the by a single variable. (The intuition here is very samilar to sliding window)\nAnd since the simple variable will move from 0 to n. We know the time complixity is actua... | 1 | 0 | [] | 0 |
longest-uploaded-prefix | c#, Sorted Set | c-sorted-set-by-bytchenko-bcn3 | Intuition\n Describe your first thoughts on how to solve this problem. \nWe can store unloaded chunks, so to compupute the prefix we should find the min unloade | bytchenko | NORMAL | 2022-10-02T10:38:04.664043+00:00 | 2022-10-02T10:38:04.664089+00:00 | 29 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe can store **unloaded** chunks, so to compupute the prefix we should find the min unloaded chunk.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nWe can use `SortedSet<int>` to store **unloaded** chunks\n\n# Comp... | 1 | 0 | ['C#'] | 0 |
longest-uploaded-prefix | C++ | Vector | Pointer | Well Commented | O(n) solution | c-vector-pointer-well-commented-on-solut-ydhx | Please upvote if you like!!\n\nTC: O(n)\nSC: O(n)\n\nclass LUPrefix {\npublic:\n\t// we will use a vector to keep track of the uploaded videos\n vector<int> | _prit | NORMAL | 2022-10-02T05:32:50.845855+00:00 | 2022-10-02T05:33:13.710056+00:00 | 55 | false | **Please upvote if you like!!**\n\nTC: O(n)\nSC: O(n)\n``` \nclass LUPrefix {\npublic:\n\t// we will use a vector to keep track of the uploaded videos\n vector<int> v;\n int pointer;\n \n LUPrefix(int n) {\n\t\t// initialize the vector of size (n+1) as the video indexes start from 1 to n\n v.resize(n... | 1 | 0 | ['C'] | 0 |
longest-uploaded-prefix | C++ || Easy & Simple || Amortized O(1) || 100% faster | c-easy-simple-amortized-o1-100-faster-by-dtl7 | \nclass LUPrefix {\npublic:\n vector<bool>uploaded;\n int size;\n int longestPrefix;\n LUPrefix(int n) {\n size = n;\n uploaded.resize | AJAY_MAKVANA | NORMAL | 2022-10-02T05:09:00.167742+00:00 | 2022-10-02T05:10:33.439512+00:00 | 58 | false | ```\nclass LUPrefix {\npublic:\n vector<bool>uploaded;\n int size;\n int longestPrefix;\n LUPrefix(int n) {\n size = n;\n uploaded.resize(size + 1, false);\n longestPrefix = 0;\n }\n\n void upload(int video) {\n uploaded[video] = true;\n }\n\n int longest() {\n ... | 1 | 0 | ['C', 'C++'] | 0 |
longest-uploaded-prefix | C++|| SET|| Easy | c-set-easy-by-aakash_172-2guk | \nclass LUPrefix {\npublic:\n set<int>st;\n LUPrefix(int n) {\n for(int i=1;i<=n+1;i++){\n st.insert(i);\n }\n }\n \n vo | aakash_172 | NORMAL | 2022-10-01T21:36:37.768224+00:00 | 2022-10-01T21:36:37.768263+00:00 | 38 | false | ```\nclass LUPrefix {\npublic:\n set<int>st;\n LUPrefix(int n) {\n for(int i=1;i<=n+1;i++){\n st.insert(i);\n }\n }\n \n void upload(int video) {\n st.erase(video);\n }\n \n int longest() {\n return (*st.begin())-1;\n }\n};\n``` | 1 | 0 | ['C', 'Ordered Set'] | 0 |
longest-uploaded-prefix | Easy and optimized python3 solution | easy-and-optimized-python3-solution-by-m-4mi8 | class LUPrefix:\n\n def init(self, n: int):\n \n self.v = [0]*(n+1)\n self.cur = 1\n \n\n def upload(self, video: int) -> None | mayankmaheshwari | NORMAL | 2022-10-01T20:21:41.712865+00:00 | 2022-10-01T20:23:44.703539+00:00 | 123 | false | class LUPrefix:\n\n def __init__(self, n: int):\n \n self.v = [0]*(n+1)\n self.cur = 1\n \n\n def upload(self, video: int) -> None:\n self.v[video] = 1\n if video == self.cur:\n self.cur+=1\n \n while self.cur<len(self.v) and self.v[self.cur]!=0:\... | 1 | 0 | ['Python3'] | 0 |
longest-uploaded-prefix | Python All O(1) endpoint algorithm, no amortization | python-all-o1-endpoint-algorithm-no-amor-3hod | It\'s quite straightforward to get amortized O(logn) with SortedList or amortized O(1) by counting up till the missing video. However, if you are familiar with | chuan-chih | NORMAL | 2022-10-01T19:59:08.486019+00:00 | 2022-10-01T20:22:44.692065+00:00 | 167 | false | It\'s quite straightforward to get amortized `O(logn)` with `SortedList` or amortized `O(1)` by counting up till the missing `video`. However, if you are familiar with the "endpoint algorithm" that keeps track of all the contiguous segments, it\'s just as simple to get an all `O(1)` algorithm:\n\n1. Use `left` and `rig... | 1 | 0 | ['Python'] | 1 |
longest-uploaded-prefix | Python Easy Solution | python-easy-solution-by-titanolodon-xkq4 | \nclass LUPrefix:\n\n def __init__(self, n: int) -> None:\n \n self.uploaded = set()\n self.prefix = list(range(n, -1, -1))\n \n | Titanolodon | NORMAL | 2022-10-01T19:31:25.046012+00:00 | 2022-10-01T19:31:25.046051+00:00 | 78 | false | ```\nclass LUPrefix:\n\n def __init__(self, n: int) -> None:\n \n self.uploaded = set()\n self.prefix = list(range(n, -1, -1))\n \n def upload(self, video: int) -> None:\n \n self.uploaded.add(video - 1)\n\t\t\n while self.prefix[-1] in self.uploaded:\n ... | 1 | 0 | ['Python'] | 0 |
longest-uploaded-prefix | Java Array Solution | java-array-solution-by-solved-yrk6 | \nclass LUPrefix {\n boolean[] array;\n int maxSize;\n public LUPrefix(int n) {\n this.array = new boolean[n];\n this.maxSize = 0;\n } | solved | NORMAL | 2022-10-01T19:13:03.012999+00:00 | 2022-10-01T19:13:03.013043+00:00 | 61 | false | ```\nclass LUPrefix {\n boolean[] array;\n int maxSize;\n public LUPrefix(int n) {\n this.array = new boolean[n];\n this.maxSize = 0;\n }\n public void upload(int video) {\n array[video - 1] = true;\n int count = 0;\n for (int i = 0; i < array.length; i++) {\n ... | 1 | 0 | ['Array', 'Java'] | 0 |
longest-uploaded-prefix | O(nlogn) using merging lower upper bound intervals | onlogn-using-merging-lower-upper-bound-i-oh59 | Main Idea:\n+ Managing lower/upper bound intervals\n+ Uploading method:\n + vmin, vmax = video, video\n + Find upper bound interval: video - 1 -> found: vmin | dntai | NORMAL | 2022-10-01T18:13:23.528707+00:00 | 2022-10-01T18:13:23.528750+00:00 | 121 | false | **Main Idea**:\n+ Managing lower/upper bound intervals\n+ Uploading method:\n + vmin, vmax = video, video\n + Find upper bound interval: video - 1 -> found: vmin = min(interval), delete lower/upper bound intervals corresponding\n + Find lower bound interval: video + 1 -> found: vmax = max(interval), delete lower/upp... | 1 | 0 | ['Python3'] | 0 |
longest-uploaded-prefix | Simple swift solution | simple-swift-solution-by-zhedre1n-565v | \nclass LUPrefix {\n var last = 0\n var cache = Set<Int>()\n\n init(_ n: Int) {}\n \n func upload(_ video: Int) {\n appendIfNeeded(video)\ | ZheDre1N | NORMAL | 2022-10-01T18:09:17.364618+00:00 | 2022-10-01T18:25:32.436772+00:00 | 50 | false | ```\nclass LUPrefix {\n var last = 0\n var cache = Set<Int>()\n\n init(_ n: Int) {}\n \n func upload(_ video: Int) {\n appendIfNeeded(video)\n }\n \n func longest() -> Int {\n last\n }\n \n private func appendIfNeeded(_ video: Int) {\n if last + 1 == video {\n ... | 1 | 0 | ['Swift'] | 0 |
longest-uploaded-prefix | Two very efficient Solutions | Python | two-very-efficient-solutions-python-by-s-v321 | Sorted List solution:\n\nfrom sortedcontainers import SortedList\n\nclass LUPrefix:\n def __init__(self, n: int):\n self.notUploaded = SortedList(i fo | swissnerd | NORMAL | 2022-10-01T17:55:12.521228+00:00 | 2022-10-01T18:51:18.230826+00:00 | 71 | false | **Sorted List** solution:\n```\nfrom sortedcontainers import SortedList\n\nclass LUPrefix:\n def __init__(self, n: int):\n self.notUploaded = SortedList(i for i in range(n + 1))\n # Time: O(n)\n # Space: O(n)\n\n def upload(self, video: int) -> None:\n self.notUploaded.remove(video - 1)\n #... | 1 | 0 | ['Python'] | 0 |
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