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-balanced-string | Beats 99% in java and One of the most easiest and simplest logic🔥 | beats-99-in-java-and-one-of-the-most-eas-k1cf | Complexity
Time complexity:
O(N)
Space complexity:
O(1)
Code | aryanxrajj | NORMAL | 2025-03-31T16:01:16.203544+00:00 | 2025-03-31T16:01:16.203544+00:00 | 1 | false |
# Complexity
- Time complexity:
- O(N)
- Space complexity:
- O(1)
# Code
```java []
class Solution {
public boolean isBalanced(String num) {
int even = 0, odd = 0;
for(int i = 0;i < num.length();i++){
char ch = num.charAt(i);
if(i % 2 == 0){
even+=ch - '0'... | 0 | 0 | ['Java'] | 0 |
check-balanced-string | 0 ms Beats 100.00% | Easy Typescript Solution with Explanation | 0-ms-beats-10000-easy-typescript-solutio-hba5 | IntuitionApproachComplexity
Time complexity: O(n) where n is the length of the input string num. This is because the function iterates through each character of | chetannada | NORMAL | 2025-03-31T11:26:19.461824+00:00 | 2025-03-31T11:26:43.891317+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: $$O(n)$$ where n is the length of the input string `num`. This is because the function iterates through each character of the string exactl... | 0 | 0 | ['String', 'TypeScript'] | 0 |
check-balanced-string | 0 ms Beats 100.00% | Easy Javascript Solution with Explanation | 0-ms-beats-10000-easy-javascript-solutio-yic4 | IntuitionApproachComplexity
Time complexity: O(n) where n is the length of the input string num. This is because the function iterates through each character of | chetannada | NORMAL | 2025-03-31T11:23:18.805756+00:00 | 2025-03-31T11:23:43.620964+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: $$O(n)$$ where n is the length of the input string `num`. This is because the function iterates through each character of the string exactl... | 0 | 0 | ['String', 'JavaScript'] | 0 |
check-balanced-string | Luke - checking whether string is balanced | luke-checking-whether-string-is-balanced-3oif | IntuitionApproachI used two variables (odd and even). I used a for loop and if the indice is even, I add the value of the index to the even, otherwise, I do tha | thunderlukey | NORMAL | 2025-03-30T13:39:08.706074+00:00 | 2025-03-30T13:39:08.706074+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
I used two variables (odd and even). I used a for loop and if the indice is even, I add the value of the index to the even, otherwise, I do that for the odd. Finally, I che... | 0 | 0 | ['Java'] | 0 |
check-balanced-string | O(N) Python3 solution easy peasy | on-python3-solution-easy-peasy-by-winsto-d0ep | IntuitionSimply sum the odd/even digitsApproach
Initialize two variables and a bool used as flag
Cycle chars in string
if flag is true add digit to first vari | WinstonWolf | NORMAL | 2025-03-29T17:23:20.359038+00:00 | 2025-03-29T17:23:20.359038+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Simply sum the odd/even digits
# Approach
<!-- Describe your approach to solving the problem. -->
- Initialize two variables and a bool used as flag
- Cycle chars in string
- - if flag is true add digit to first variable else add it in the... | 0 | 0 | ['Python3'] | 0 |
check-balanced-string | Scala solution with implicit and map | scala-solution-with-implicit-and-map-by-sqmnm | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | iyIeO99AmH | NORMAL | 2025-03-29T13:48:10.730969+00:00 | 2025-03-29T13:48:10.730969+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 | ['Scala'] | 0 |
check-balanced-string | Balanced or Busted? ⚖️🔢 Let’s Find Out! 😂 Python !! | balanced-or-busted-lets-find-out-python-b01zv | Intuition
The problem requires checking if the sum of digits at even indices
is equal to the sum of digits at odd indices.
Since we only need to traverse the st | Shahin1212 | NORMAL | 2025-03-29T05:16:07.090832+00:00 | 2025-03-29T05:16:07.090832+00:00 | 1 | false | # Intuition
- The problem requires checking if the sum of digits at even indices
- is equal to the sum of digits at odd indices.
- Since we only need to traverse the string once, a simple iteration
- with two counters (one for even indices and one for odd) will work efficiently.
# Approach
1. Initialize two var... | 0 | 0 | ['Python3'] | 0 |
check-balanced-string | Balanced or Busted? ⚖️🔢 Let’s Find Out! 😂 Python !! | balanced-or-busted-lets-find-out-python-rrflu | Intuition
The problem requires checking if the sum of digits at even indices
is equal to the sum of digits at odd indices.
Since we only need to traverse the st | Shahin1212 | NORMAL | 2025-03-29T05:16:04.338184+00:00 | 2025-03-29T05:16:04.338184+00:00 | 1 | false | # Intuition
- The problem requires checking if the sum of digits at even indices
- is equal to the sum of digits at odd indices.
- Since we only need to traverse the string once, a simple iteration
- with two counters (one for even indices and one for odd) will work efficiently.
# Approach
1. Initialize two var... | 0 | 0 | ['Python3'] | 1 |
check-balanced-string | Quick code loops | quick-code-loops-by-kunal_1310-weka | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kunal_1310 | NORMAL | 2025-03-28T07:05:09.978981+00:00 | 2025-03-28T07:05:09.978981+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 | ['C++'] | 0 |
check-balanced-string | Simple Swift Solution | simple-swift-solution-by-felisviridis-2yh9 | Code | Felisviridis | NORMAL | 2025-03-27T11:23:00.923845+00:00 | 2025-03-27T11:23:00.923845+00:00 | 4 | false | 
# Code
```swift []
class Solution {
func isBalanced(_ num: String) -> Bool {
var digits = Array(num), evenSum = 0, oddSum = 0
for i in 0..<digits.count {... | 0 | 0 | ['Swift'] | 0 |
check-balanced-string | O(n) easy to understand solution with JavaScript :) | on-easy-to-understand-solution-with-java-d9rn | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(1)Code | bek-shoyatbek | NORMAL | 2025-03-27T00:05:34.041577+00:00 | 2025-03-27T00:05:34.041577+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. $$O(n)$$ -->
$$O(n)$$
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 0 | 0 | ['JavaScript'] | 0 |
check-balanced-string | my solution | my-solution-by-leman_cap13-if96 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | leman_cap13 | NORMAL | 2025-03-24T14:44:22.305451+00:00 | 2025-03-24T14:44:22.305451+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 | ['Python3'] | 0 |
check-balanced-string | Something Interesting | something-interesting-by-shakhob-yr6x | Code | Shakhob | NORMAL | 2025-03-24T07:04:30.971248+00:00 | 2025-03-24T07:04:30.971248+00:00 | 1 | false | # Code
```python3 []
class Solution:
def isBalanced(self, num: str) -> bool:
number = int(num)
even = 0
odd = 0
i = 0
while number > 0:
if i % 2 == 0:
even += number % 10
else:
odd += number % 10
number //= 1... | 0 | 0 | ['Python3'] | 0 |
check-balanced-string | 0 ms Beats 100.00% | 0-ms-beats-10000-by-bvc01654-xi31 | IntuitionApproachComplexity
Time complexity:O(n)
Space complexity:O(1)
Code | bvc01654 | NORMAL | 2025-03-23T09:56:21.631912+00:00 | 2025-03-23T09:56:21.631912+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:O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 0 | 0 | ['String', 'C++'] | 0 |
check-balanced-string | PYTHON WITH MK!! | python-with-mk-by-mohan_kumar_hd-k7td | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | MOHAN_KUMAR_HD | NORMAL | 2025-03-22T18:03:04.201367+00:00 | 2025-03-22T18:03:04.201367+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 | ['Python3'] | 0 |
check-balanced-string | Check Balanced String | check-balanced-string-by-jyenduri-2qgb | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | jyenduri | NORMAL | 2025-03-22T07:41:44.522930+00:00 | 2025-03-22T07:41:44.522930+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 | ['JavaScript'] | 0 |
check-balanced-string | Very Easy Solution | very-easy-solution-by-devanand8590-p2iw | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Devanand8590 | NORMAL | 2025-03-21T04:47:31.321491+00:00 | 2025-03-21T04:47:31.321491+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 | ['Dart'] | 0 |
check-balanced-string | Solution | solution-by-omkarpatil_1304-3jej | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | omkarpatil_1304 | NORMAL | 2025-03-20T18:14:24.808750+00:00 | 2025-03-20T18:14:24.808750+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 | ['Java'] | 0 |
check-balanced-string | Easy Solution | easy-solution-by-adhi_m_s-9rox | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Adhi_M_S | NORMAL | 2025-03-20T10:23:20.011189+00:00 | 2025-03-20T10:23:20.011189+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 | ['Python'] | 0 |
check-balanced-string | Simple java Solution | simple-java-solution-by-pritdesa-9it8 | Code | pritdesa | NORMAL | 2025-03-19T19:52:54.281237+00:00 | 2025-03-19T19:52:54.281237+00:00 | 1 | false |
# Code
```java []
class Solution {
public boolean isBalanced(String num) {
int even = 0, odd = 0;
for(int i=0;i<num.length();i++){
if(i%2==0){
even += Character.getNumericValue(num.charAt(i));
}else{
odd += Character.getNumericValue(... | 0 | 0 | ['Java'] | 0 |
check-balanced-string | EASY SOLUTION 0ms /100 beats | easy-solution-0ms-100-beats-by-cookie_co-zise | IntuitionApproachMain approach is to first convert string to integer by subtracting the current string value to '0' ASCII value ...Complexity
Time complexity: | Cookie_coder | NORMAL | 2025-03-19T18:10:18.023861+00:00 | 2025-03-19T18:10:18.023861+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
Main approach is to first convert string to integer by subtracting the current string value to '0' ASCII value ...
# Complexity
- Time complexity:
O(n)
- Space complexity:
O(1)
# Code
```cpp []
class Solution {
public:
bo... | 0 | 0 | ['C++'] | 0 |
check-balanced-string | clear logic | clear-logic-by-sophie84-b0c0 | Code | sophie84 | NORMAL | 2025-03-19T17:02:10.816687+00:00 | 2025-03-19T17:02:10.816687+00:00 | 1 | false | # Code
```python3 []
class Solution:
def isBalanced(self, num: str) -> bool:
sum_even = 0
sum_odd = 0
for i in range(len(num)):
if i%2==0:
sum_even += int(num[i])
else:
sum_odd += int(num[i])
return (sum_even==sum_odd)
``` | 0 | 0 | ['Python3'] | 0 |
check-balanced-string | easy solution | easy-solution-by-mukund_th04-odet | IntuitionWe need to check if a string of digits is balanced, meaning the sum of digits at even indices is equal to the sum of digits at odd indices. The simples | Mukund_TH04 | NORMAL | 2025-03-19T06:32:40.394077+00:00 | 2025-03-19T06:32:40.394077+00:00 | 1 | false | # Intuition
We need to check if a string of digits is balanced, meaning the sum of digits at even indices is equal to the sum of digits at odd indices. The simplest idea is to traverse the string once, summing digits at even and odd positions separately, and then compare the two sums.
---
# Approach
1. **Traverse the... | 0 | 0 | ['C++'] | 0 |
check-balanced-string | one liner... check it out | one-liner-check-it-out-by-senth-q1ye | Complexity
Time complexity:
O(n)
Space complexity:
O(n)
Code | senth | NORMAL | 2025-03-19T05:59:22.396325+00:00 | 2025-03-19T05:59:22.396325+00:00 | 1 | false | # Complexity
- Time complexity:
O(n)
- Space complexity:
O(n)
# Code
```python3 []
class Solution:
def isBalanced(self, num: str) -> bool:
return sum(int(n) for n in num[::2]) == sum(int(n) for n in num[1::2])
``` | 0 | 0 | ['Python3'] | 0 |
check-balanced-string | simple... | simple-by-anakha123-ifbz | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | anakha123 | NORMAL | 2025-03-18T04:24:00.127779+00:00 | 2025-03-18T04:24:00.127779+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-balanced-string | C++ Easiest approach | c-easiest-approach-by-aasthagupta_17-w1yn | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | aasthagupta_17 | NORMAL | 2025-03-16T07:57:13.648498+00:00 | 2025-03-16T07:57:13.648498+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 | ['C++'] | 0 |
check-balanced-string | Java Solution beats 99.04% | java-solution-beats-9904-by-moreshivam20-9t9b | IntuitionFind even index and odd index then add sumApproachSo we will run two for loops first start from 0th index as even and will increment counter each time | moreshivam20 | NORMAL | 2025-03-16T06:18:20.092720+00:00 | 2025-03-16T06:18:20.092720+00:00 | 1 | false | # Intuition
Find even index and odd index then add sum
# Approach
So we will run two for loops first start from 0th index as even and will increment counter each time by i+2 for skip one odd index and for odd index we will start from 1st index and will keep addiing respeciveli in evenSum and OddSum and in last will
... | 0 | 0 | ['Java'] | 0 |
check-balanced-string | i think this is the easy way to understand for the beginner but not the optimize solution | i-think-this-is-the-easy-way-to-understa-catf | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | mdarizaman | NORMAL | 2025-03-15T08:47:43.842033+00:00 | 2025-03-15T08:47:43.842033+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 | ['Python3'] | 0 |
check-balanced-string | 0ms Runtime with best Case Time Complexity using C Language and Suitable String Operations : ) | 0ms-runtime-with-best-case-time-complexi-8qxd | IntuitionThe problem requires checking whether the sum of digits at even indices equals the sum at odd indices. By iterating through the string of digits, you c | Vivek_Bartwal | NORMAL | 2025-03-15T06:12:24.812462+00:00 | 2025-03-15T06:12:24.812462+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem requires checking whether the sum of digits at even indices equals the sum at odd indices. By iterating through the string of digits, you can categorize the digits based on their indices and compute two separate sums. If these s... | 0 | 0 | ['C'] | 0 |
check-balanced-string | Beats 100% || C++ | beats-100-c-by-prem2907-gm2s | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | prem2907 | NORMAL | 2025-03-12T14:32:44.421229+00:00 | 2025-03-12T14:32:44.421229+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 | ['C++'] | 0 |
check-balanced-string | Java | java-by-soumya_699-0fc5 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Soumya_699 | NORMAL | 2025-03-12T06:08:06.032818+00:00 | 2025-03-12T06:08:06.032818+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 | ['Java'] | 0 |
check-balanced-string | easy and best solution in c++ | easy-and-best-solution-in-c-by-ashish754-sfhm | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Ashish754937 | NORMAL | 2025-03-12T05:39:05.879263+00:00 | 2025-03-12T05:39:05.879263+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 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | O(1) || Math | o1-math-by-fahad06-sdn7 | Intuition\n- Simplify the problem by understanding the cyclic nature of the ball passing. By calculating the number of complete back-and-forth trips and the rem | fahad_Mubeen | NORMAL | 2024-06-09T04:03:32.733818+00:00 | 2024-08-26T16:02:27.187913+00:00 | 8,139 | false | # Intuition\n- Simplify the problem by understanding the cyclic nature of the ball passing. By calculating the number of complete back-and-forth trips and the remaining steps, we can determine the final position of the ball. \n- If the total number of complete trips is even, the ball continues in the forward direction.... | 71 | 0 | ['Math', 'C++', 'Java', 'Python3'] | 11 |
find-the-child-who-has-the-ball-after-k-seconds | Easy Video Solution 🔥 || How to 🤔 in Interview || O(N)->O(1) 🔥 | easy-video-solution-how-to-in-interview-pq6kq | If you like the solution Please Upvote and subscribe to my youtube channel\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \nDry Run | ayushnemmaniwar12 | NORMAL | 2024-06-09T04:04:18.588221+00:00 | 2024-06-09T06:33:24.740726+00:00 | 677 | false | ***If you like the solution Please Upvote and subscribe to my youtube channel***\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nDry Run few examples and observe how you can derive a formula\n\n\n\n# ***Easy Video Explanation***\n\nhttps://youtu.be/lTD4AqpPxUA\n\n \n\n# Code (Bru... | 12 | 5 | ['Math', 'Iterator', 'Python', 'C++', 'Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | ✅✅💯EASY SOLUTION || C++ 🔥🔥💯✅✅ | easy-solution-c-by-nikhil73995-0g7y | Intuition\n Describe your first thoughts on how to solve this problem. \nImagine children playing a game of "pass the ball" in a line.\nThe ball starts with the | nikhil73995 | NORMAL | 2024-06-09T04:22:28.127008+00:00 | 2024-06-09T05:35:26.186774+00:00 | 933 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nImagine children playing a game of "pass the ball" in a line.\nThe ball starts with the first child and moves one step right each second.\nWhen the ball reaches the last child, they throw it back (reverse direction).\nWhen the ball reache... | 9 | 0 | ['C++'] | 4 |
find-the-child-who-has-the-ball-after-k-seconds | Python 3 || 2 lines, abs || T/S: 94% / 28% | python-3-2-lines-abs-ts-94-28-by-spauldi-nov4 | Here\'s the intuition:\n\n- After 2*n - 2 seconds, the ball is back to child 0, so we only need to be concerned with k%= 2*n -2.\n\n- The graph of f(k) = n-1 - | Spaulding_ | NORMAL | 2024-06-09T05:05:39.382770+00:00 | 2024-06-11T15:55:37.398543+00:00 | 237 | false | Here\'s the intuition:\n\n- After `2*n - 2` seconds, the ball is back to *child 0*, so we only need to be concerned with `k%= 2*n -2`.\n\n- The graph of ` f(k) = n-1 - abs(k-n+1)` is in the figure below, for the case in which *n* = 3. If *k* = 5, then *k* % 4 = 1 and *f*(1) = 1.\n\n sol||0ms beats 100% | 2-line-o1-sol0ms-beats-100-by-anwendeng-x70y | Intuition\n Describe your first thoughts on how to solve this problem. \nModular arithmetic with period N=2*(n-1)\n# Approach\n Describe your approach to solvin | anwendeng | NORMAL | 2024-06-09T04:33:13.325239+00:00 | 2024-06-09T04:33:13.325269+00:00 | 779 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nModular arithmetic with period `N=2*(n-1)`\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Modular arithmetic `x=k%N;`\n2. The answer is `(x<n)?x:N-x`\n# Complexity\n- Time complexity:\n<!-- Add your time complexi... | 7 | 0 | ['Math', 'C++'] | 3 |
find-the-child-who-has-the-ball-after-k-seconds | 5Lines simple code O(K) | 5lines-simple-code-ok-by-sumanth2328-3202 | \n\n# Code\n\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n posneg,j=1,0\n for i in range(k):\n j+=posneg\n | sumanth2328 | NORMAL | 2024-06-09T04:10:23.743711+00:00 | 2024-06-09T04:10:23.743760+00:00 | 980 | false | \n\n# Code\n```\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n posneg,j=1,0\n for i in range(k):\n j+=posneg\n if j==0 or j==n-1:\n posneg*=-1\n return j\n``` | 7 | 0 | ['Python3'] | 5 |
find-the-child-who-has-the-ball-after-k-seconds | 🔥Very Easy || 4 lines || Without Formula || Changing direction at the ends. | very-easy-4-lines-without-formula-changi-ae8m | Complexity\n- Time complexity:\nO(k) \n- Space complexity:\nO(1) \n\n# Code\n\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int dir | __esh_WAR_Amit | NORMAL | 2024-06-09T04:53:15.571067+00:00 | 2024-06-09T12:45:42.448970+00:00 | 356 | false | # Complexity\n- Time complexity:\n$$O(k)$$ \n- Space complexity:\n$$O(1)$$ \n\n# Code\n```\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int direction = 1; // 1 means right, -1 means left\n int position = 0; // Starting position\n\n for (int steps = 0; steps < k; ++steps) {\n... | 6 | 0 | ['C++'] | 1 |
find-the-child-who-has-the-ball-after-k-seconds | 💢☠💫Easiest👾Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅💥🔥💫Explained☠💥🔥 Beats 💯 | easiestfaster-lesser-cpython3javacpython-pisr | Intuition\n\n\n\n\n\n Describe your first thoughts on how to solve this problem. \nPython3 []\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> i | Edwards310 | NORMAL | 2024-06-09T04:16:37.537885+00:00 | 2024-06-09T05:23:12.889569+00:00 | 209 | false | # Intuition\n\n\n {\n let ballPosition = 0; // Start with child 0 holding the ball\n let direction = 1; // Initially, | deleted_user | NORMAL | 2024-06-09T04:09:49.657188+00:00 | 2024-06-09T04:09:49.657214+00:00 | 1,468 | false | # Code\n\n```javascript []\nvar numberOfChild = function (n, k) {\n let ballPosition = 0; // Start with child 0 holding the ball\n let direction = 1; // Initially, the ball moves towards the right\n\n for (let i = 0; i < k; i++) {\n if (direction === 1) {\n if (ballPosition === n - 1) direction = -1; \n ... | 6 | 0 | ['C++', 'Java', 'Python3', 'JavaScript'] | 2 |
find-the-child-who-has-the-ball-after-k-seconds | DIRECT APPROACH || BASIC SIMULATION | direct-approach-basic-simulation-by-abhi-tllg | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nJUST KEEP ON ITERATIONG | Abhishekkant135 | NORMAL | 2024-06-09T04:22:51.831611+00:00 | 2024-06-09T04:22:51.831645+00:00 | 418 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nJUST KEEP ON ITERATIONG AS IT SAYS AND YOU ARE DONE.\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\no(n)\n- Space complexity... | 4 | 0 | ['Simulation', 'Java'] | 3 |
find-the-child-who-has-the-ball-after-k-seconds | Java 1 Line simple code 🔥 O(1) Time and space | java-1-line-simple-code-o1-time-and-spac-kmii | Intuition\nsince time starts from 0 so K will always be related to n-1.\n\n# Approach\nSo firstly classify watch when ball reverses direction, If the final dire | hashed-meta | NORMAL | 2024-06-16T21:52:57.564119+00:00 | 2024-06-16T21:52:57.564143+00:00 | 99 | false | # Intuition\nsince time starts from ```0``` so `K` will always be related to n-1.\n\n# Approach\nSo firstly classify watch when ball reverses direction, If the final direction is same as the starting direction then it will be like 0 for 1st person, 1 for 2nd person and so on.\nAnd if the direction is reversed then chan... | 3 | 0 | ['Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple O(1) || Solution fast | simple-o1-solution-fast-by-cs_balotiya-7xzw | Complexity\n- Time complexity:\nO(1)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int tem | cs_balotiya | NORMAL | 2024-06-09T12:22:23.169060+00:00 | 2024-06-09T12:22:23.169084+00:00 | 42 | false | # Complexity\n- Time complexity:\n$$O(1)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int temp1 = k / (n - 1);\n int temp2 = k % (n - 1);\n if (temp1 % 2 == 0) {\n // forward\n return temp2;\n }\n ... | 3 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | O(1) Solution | Beats 100💯 | o1-solution-beats-100-by-_rishabh_96-6zg2 | Number of Children in the Roundabout Game\n\n## Problem Statement\nGiven n children standing in a circle, you start counting them from 1 to k. Once you reach k, | _Rishabh_96 | NORMAL | 2024-06-09T09:42:41.373129+00:00 | 2024-06-09T09:42:41.373163+00:00 | 118 | false | # Number of Children in the Roundabout Game\n\n## Problem Statement\nGiven `n` children standing in a circle, you start counting them from 1 to `k`. Once you reach `k`, the count restarts from 1 again, but starting with the next child. After counting to `k` several times, determine which child will be the last one coun... | 3 | 0 | ['Math', 'C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Clean Simple Solution O(1) | clean-simple-solution-o1-by-lokeshsk1-j9un | Approach\n\nFor n = 5, k = 6\n\n0\t[0, 1, 2, 3, 4]\n1\t[0, 1, 2, 3, 4]\n2\t[0, 1, 2, 3, 4]\n3\t[0, 1, 2, 3, 4]\n4\t[0, 1, 2, 3, 4]\n5\t[0, 1, 2, 3, 4]\n6\t[0, 1 | lokeshsk1 | NORMAL | 2024-06-09T08:50:55.538841+00:00 | 2024-06-09T09:04:07.764326+00:00 | 99 | false | # Approach\n\nFor n = 5, k = 6\n\n0\t[```0```, 1, 2, 3, 4]\n1\t[0, ```1```, 2, 3, 4]\n2\t[0, 1, ```2```, 3, 4]\n3\t[0, 1, 2, ```3```, 4]\n4\t[0, 1, 2, 3, ```4```]\n5\t[0, 1, 2, ```3```, 4]\n6\t[0, 1, ```2```, 3, 4]\n7 [0, ```1```, 2, 3, 4]\n\nFor 8 it again comes to "0"\nand for 9 it comes to "1"\n\nWe should make k ... | 3 | 0 | ['Python3'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Iterative Solution | Python | iterative-solution-python-by-pragya_2305-vczv | Complexity\n- Time complexity: O(k)\n\n- Space complexity: o(1)\n\n# Code\n\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n child | pragya_2305 | NORMAL | 2024-06-09T04:19:07.108114+00:00 | 2024-06-09T04:19:07.108148+00:00 | 238 | false | # Complexity\n- Time complexity: O(k)\n\n- Space complexity: o(1)\n\n# Code\n```\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n child = 0\n reverse = 0\n while k:\n for i in range(n-1):\n if k==0:\n return child\n ... | 3 | 0 | ['Array', 'Math', 'Python', 'Python3'] | 3 |
find-the-child-who-has-the-ball-after-k-seconds | TC: O(1) | SC: O(1) | BEATS 100% | SIMPLEST EXPLANATION | tc-o1-sc-o1-beats-100-simplest-explanati-qxf7 | Intuition\n Describe your first thoughts on how to solve this problem. \nIf k is less than n, then simply return the kth child (because ball is passing towards | vishwa-vijetha-g | NORMAL | 2024-06-09T04:03:35.034524+00:00 | 2024-06-09T04:06:03.097045+00:00 | 239 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf k is less than n, then simply return the kth child (because ball is passing towards right and it is the first time),\nif k is greater than n, then find the direction of ball pass by dividing k with the total children (n - 1).\nIf the q... | 3 | 2 | ['Math', 'Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | One line Solution 🚀|| 💡 Simple Math-Based Logic || 🔥 Beginner Friendly || 🧑💻 Clean Code | one-line-solution-simple-math-based-logi-fks6 | IntuitionThe problem can be solved using modular arithmetic to determine the position of a child after a given number of steps. The idea is to simulate the circ | Arunkarthick_k | NORMAL | 2025-01-28T22:34:21.827537+00:00 | 2025-01-28T22:34:21.827537+00:00 | 131 | false | # Intuition
The problem can be solved using modular arithmetic to determine the position of a child after a given number of steps. The idea is to simulate the circular behavior of counting, where positions wrap around after reaching the end.
# Approach
1. Calculate the effective position using the modulo operator to h... | 2 | 0 | ['Math', 'Simulation', 'Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple Easy to Understand Java Code || Beats 100% | simple-easy-to-understand-java-code-beat-0vqh | Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\njava []\nclass Solution {\n public int numberOfChild(int n, int k) {\n return | Saurabh_Mishra06 | NORMAL | 2024-09-21T06:09:56.941524+00:00 | 2024-09-21T06:09:56.941560+00:00 | 174 | false | # Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\n```java []\nclass Solution {\n public int numberOfChild(int n, int k) {\n return (n-1) - Math.abs((n-1) - k%((n-1)*2));\n }\n}\n``` | 2 | 0 | ['Math', 'Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Two Methods Math & pointer | two-methods-math-pointer-by-sangam_verma-pncg | FIRST METHOD - Using Maths\n# Code\n\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int d=k/(n-1);\n int r=k%(n-1);\n | sangam_verma | NORMAL | 2024-07-06T02:07:40.656156+00:00 | 2024-07-06T02:07:40.656179+00:00 | 278 | false | # FIRST METHOD - Using Maths\n# Code\n```\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int d=k/(n-1);\n int r=k%(n-1);\n if(d&1){\n return n-r-1; // odd then cournt from end side\n }\n else return r; // even then count from front\n }\n};\n```\n# SE... | 2 | 0 | ['Math', 'C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Assembly with explanation and comparison with C | assembly-with-explanation-and-comparison-b7vg | \n# Rationale\n\nI\'ve solved this LeetCode problem using inline assembly in C. My goal isn\'t to showcase complexity, but rather to challenge myself and deepen | pcardenasb | NORMAL | 2024-06-14T01:24:00.009696+00:00 | 2024-06-14T01:24:00.009713+00:00 | 116 | false | \n# Rationale\n\nI\'ve solved this LeetCode problem using inline assembly in C. My goal isn\'t to showcase complexity, but rather to challenge myself and deepen my understanding of assembly language. \n\nAs a beginner in assembly, I welcome any improvements or comments on my code.\n\nAdditionally, I aim to share insigh... | 2 | 0 | ['C'] | 1 |
find-the-child-who-has-the-ball-after-k-seconds | O(1) | Efficiently Finding the Child Who Holds the Ball After K Seconds | o1-efficiently-finding-the-child-who-hol-r42v | Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve the problem of finding the child who has the ball after k seconds, we can leve | pushpitjain2006 | NORMAL | 2024-06-10T12:48:41.237805+00:00 | 2024-06-10T12:48:41.237840+00:00 | 6 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve the problem of finding the child who has the ball after k seconds, we can leverage the observation that the ball\'s movement is periodic. The ball moves back and forth between the two ends of the line. Therefore, after a certain ... | 2 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | O(1) approach explained detailly | o1-approach-explained-detailly-by-hrushi-17q0 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem can be visualized as children sitting in a circle where the distribution of | Hrushi_7 | NORMAL | 2024-06-10T09:50:00.234706+00:00 | 2024-06-10T09:50:00.234742+00:00 | 35 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem can be visualized as children sitting in a circle where the distribution of items happens in a specific pattern, reversing direction after completing a round. By carefully calculating the direction and the remainder, we can de... | 2 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple solution using Math || O(1) | simple-solution-using-math-o1-by-user745-xgz5 | Intuition\nAfter 2 * (n-1) seconds, ball ends up with the first child. After n-1 seconds, the ball is with the last child.\n\n# Approach\nSo take reminder of k | user7454af | NORMAL | 2024-06-09T22:49:58.446617+00:00 | 2024-06-09T22:49:58.446635+00:00 | 38 | false | # Intuition\nAfter $$2 * (n-1)$$ seconds, ball ends up with the first child. After $$n-1$$ seconds, the ball is with the last child.\n\n# Approach\nSo take reminder of k with $$2 * (n-1)$$. If k is less than $$n-1$$ return k, else return $$(n-1) - (k - (n-1))$$ as the ball reaches the end $$n-1$$ seconds and runs backw... | 2 | 0 | ['Rust'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | PYTHON🔥|| BEAT 100%🔥|| 100% EFFICIENT🔥|| BEGINNER'S APPROACH || EASY TO UNDERSTAND || | python-beat-100-100-efficient-beginners-q13c7 | if it\'s help, please up \u2B06vote!\u2764\uFE0F and comment\uD83D\uDD25\n# Intuition\n Describe your first thoughts on how to solve this problem. \n- My first | Adit_gaur | NORMAL | 2024-06-09T12:04:34.125402+00:00 | 2024-06-09T12:04:34.125433+00:00 | 34 | false | # if it\'s help, please up \u2B06vote!\u2764\uFE0F and comment\uD83D\uDD25\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- My first thought was to simulate the movement of the ball among the children in the queue. Given that the ball changes direction when it reaches either end of t... | 2 | 0 | ['Python'] | 1 |
find-the-child-who-has-the-ball-after-k-seconds | O(1) Java solution using simple Math logic | o1-java-solution-using-simple-math-logic-oo9t | Intuition\nThe ball starts with child 0 and moves to the right. When it reaches the end (either child 0 or child n\u22121), the direction reverses. The problem | Sanjeev1903 | NORMAL | 2024-06-09T06:25:54.930669+00:00 | 2024-06-09T06:25:54.930700+00:00 | 130 | false | # Intuition\nThe ball starts with child 0 and moves to the right. When it reaches the end (either child 0 or child n\u22121), the direction reverses. The problem is to determine which child has the ball after k seconds. The key insight is to notice the periodic nature of the ball\'s movement.\n\n# Complexity\n- Time Co... | 2 | 0 | ['Java'] | 1 |
find-the-child-who-has-the-ball-after-k-seconds | 100% BEATS || O(1) Complexity || Mathematical Approach | 100-beats-o1-complexity-mathematical-app-m2ir | If k is less than n:#\n - If k is less than the total number of children n, it means the ball won\'t reach either end of the line within k seconds. So, the ch | shubham6762 | NORMAL | 2024-06-09T04:40:18.140811+00:00 | 2024-06-09T04:40:18.140844+00:00 | 47 | false | 1. **If `k` is less than `n`:**#\n - If `k` is less than the total number of children `n`, it means the ball won\'t reach either end of the line within `k` seconds. So, the child who receives the ball after `k` seconds will simply be the `k`th child in the queue.\n\n2. **If `k` is greater than or equal to `n`:**\n ... | 2 | 0 | ['Math', 'C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Very Simple C Solution 100% | very-simple-c-solution-100-by-rajnarayan-17ev | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. when current reaches | rajnarayansharma110 | NORMAL | 2024-06-09T04:32:58.689312+00:00 | 2024-06-09T04:32:58.689342+00:00 | 19 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. when current reaches end then traverse in reverse direction else traverse forward\n# Complexity\n- Time complexity:O(k)\n<!-- Add your time complexity here, e.g. $$... | 2 | 0 | ['C'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Best Solution Beats ✅100% --pkg | best-solution-beats-100-pkg-by-pradeepkr-mpgs | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nApproach is to traverse through the loop and if it reaches to the end of | pradeepkrgupta_39 | NORMAL | 2024-06-09T04:24:37.175010+00:00 | 2024-06-09T04:24:37.175041+00:00 | 192 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nApproach is to traverse through the loop and if it reaches to the end of the k then reverse it by using the codition (if pos==n-1 || pos==0)\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity : 0(k)\n- ... | 2 | 0 | ['Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Contest Solution! | contest-solution-by-qatyayani-d24j | \n\n# Code\n\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n f=0\n i=0\n while True:\n if k==0:\n | Qatyayani | NORMAL | 2024-06-09T04:07:37.803023+00:00 | 2024-06-09T04:07:37.803046+00:00 | 30 | false | \n\n# Code\n```\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n f=0\n i=0\n while True:\n if k==0:\n return i\n if f==0:\n i+=1\n if i==n-1:\n f=1\n elif f==1:\n i... | 2 | 0 | ['Python3'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | deque 🔥| C++ | deque-c-by-varuntyagig-a7xu | Code | varuntyagig | NORMAL | 2025-03-12T16:29:09.829468+00:00 | 2025-03-12T16:29:09.829468+00:00 | 16 | false | # Code
```cpp []
class Solution {
public:
int numberOfChild(int n, int k) {
deque<int> dq(1, 0);
bool flag1 = false;
for (int i = 1; i <= k; ++i) {
if (!flag1) {
dq.push_back(i);
if (dq.size() == n) {
flag1 = true;
... | 1 | 0 | ['Queue', 'C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | ☑️ Finding the Child Who Has the Ball After K Seconds. ☑️ | finding-the-child-who-has-the-ball-after-f1rh | Code | Abdusalom_16 | NORMAL | 2025-02-24T04:50:38.965059+00:00 | 2025-02-24T04:50:38.965059+00:00 | 39 | false | # Code
```dart []
class Solution {
int numberOfChild(int n, int k) {
int max = n-1;
int number = 0;
bool isReverse = false;
while(k > 0){
k--;
if(isReverse){
number--;
if(number == 0){
isReverse = false;
}
}else{
... | 1 | 0 | ['Math', 'Simulation', 'Dart'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | C modulo O(1) | c-modulo-o1-by-michelusa-6v3y | null | michelusa | NORMAL | 2024-12-12T23:46:13.027458+00:00 | 2024-12-12T23:46:13.027458+00:00 | 20 | false | C modulo O(1)\nCount number of whole trips.\n\n# Code\n```c []\nint numberOfChild(int n, int k) {\n --n;\n const int trips = k / n ;\n const int left = k % n ;\n\n \n\n if ((trips & 1) == 1) { // at end\n return n - left ;\n } else {\n return left;\n }\n}\n``` | 1 | 0 | ['C'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Twister JS | twister-js-by-joelll-qr73 | 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 | Joelll | NORMAL | 2024-08-10T04:01:00.264690+00:00 | 2024-08-10T04:01:00.264720+00:00 | 78 | 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)$$ --... | 1 | 0 | ['JavaScript'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | easy c++ solution || beats 100% | easy-c-solution-beats-100-by-yogeshwarib-y7yx | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | yogeshwaribisht21 | NORMAL | 2024-06-15T15:20:22.439520+00:00 | 2024-06-15T15:20:22.439550+00:00 | 198 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 1 |
find-the-child-who-has-the-ball-after-k-seconds | Simple Java Code ☠️ | simple-java-code-by-abhinandannaik1717-akgp | Code\n\nclass Solution {\n public int numberOfChild(int n, int k) {\n if(k>=(2*(n-1))){\n k = k%(2*(n-1));\n }\n if(k>=(n-1)) | abhinandannaik1717 | NORMAL | 2024-06-15T14:11:13.971539+00:00 | 2024-06-19T13:07:21.114983+00:00 | 378 | false | # Code\n```\nclass Solution {\n public int numberOfChild(int n, int k) {\n if(k>=(2*(n-1))){\n k = k%(2*(n-1));\n }\n if(k>=(n-1)){\n k = k%(n-1);\n return (n-k-1);\n }\n else{\n return k;\n }\n }\n}\n```\n\n# Intuition\n<!-- De... | 1 | 0 | ['Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | C++ Easy Code💯 || Beats 100% || Beginner friendly || Easy to understand || With explanation 💯💀 | c-easy-code-beats-100-beginner-friendly-pm385 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe numberOfChild function is designed to determine the position of a "child" in a sequ | Shodhan_ak | NORMAL | 2024-06-14T13:21:21.082716+00:00 | 2024-06-14T13:21:21.082757+00:00 | 5 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe numberOfChild function is designed to determine the position of a "child" in a sequence after k steps, given n total positions. Here\'s a step-by-step breakdown of the intuition behind the code:\n\nInitialization:\n\na represents the ... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | C++ | Circular Linked List | Easy and intuitive approach | c-circular-linked-list-easy-and-intuitiv-yaph | Intuition\nThe first thought to the question were recognising that this question involves a cyclic event where the ball goes from 0 to n-1 and then again comes | Anugrah_Gupta | NORMAL | 2024-06-13T09:04:53.885726+00:00 | 2024-06-13T09:04:53.885755+00:00 | 3 | false | # Intuition\nThe first thought to the question were recognising that this question involves a cyclic event where the ball goes from 0 to n-1 and then again comes back to 0. This led me to the idea of using Circular Linked List.\n\n# Approach\nSince the question says that there is a reversal of the passing order. So if ... | 1 | 0 | ['Linked List', 'C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | simple solution with basic mathod | simple-solution-with-basic-mathod-by-vin-ezc7 | 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 | vinay_kumar_swami | NORMAL | 2024-06-10T19:23:07.583768+00:00 | 2024-06-10T19:23:07.583794+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:\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++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | ✅ O(1) Solution | o1-solution-by-eleev-0x0r | Solution\nswift\nstruct Solution {\n @_optimize(speed)\n func numberOfChild(_ n: Int, _ k: Int) -> Int {\n let n = n - 1\n // > Ifthe number | eleev | NORMAL | 2024-06-09T17:39:46.210726+00:00 | 2024-06-09T17:39:46.210750+00:00 | 9 | false | # Solution\n```swift\nstruct Solution {\n @_optimize(speed)\n func numberOfChild(_ n: Int, _ k: Int) -> Int {\n let n = n - 1\n // > Ifthe number of complete back-and-forth cycles is even:\n // -> calculate the remaining passings\n // > Otherwise (if odd):\n // -> the ... | 1 | 0 | ['Math', 'Swift'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Day 253 Problem 2 Solved | day-253-problem-2-solved-by-lakshya311-slxe | 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# Code\n\nclass Solut | lakshya311 | NORMAL | 2024-06-09T13:15:44.500470+00:00 | 2024-06-09T13:15:44.500524+00:00 | 6 | 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# Code\n```\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int ans = k % (2 * n - 2);\n return (ans < n) ? ans : 2 * n - 2 - ans;\... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | ✅💯Very Easy Solution | Python 🔥🔥 | very-easy-solution-python-by-kg-profile-u313 | Code\n\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n eff = k % (2 * (n - 1))\n p = 0\n d = 1 \n \n | KG-Profile | NORMAL | 2024-06-09T13:04:22.727711+00:00 | 2024-06-09T13:04:22.727751+00:00 | 57 | false | # Code\n```\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n eff = k % (2 * (n - 1))\n p = 0\n d = 1 \n \n for _ in range(eff):\n if p == 0:\n d = 1 \n elif p == n - 1:\n d = -1 \n \n p +=... | 1 | 0 | ['Python3'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | [Python] simple brute force | python-simple-brute-force-by-pbelskiy-01do | \nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n a = list(range(n))\n\n while len(a) < k*2:\n a.extend(list(ran | pbelskiy | NORMAL | 2024-06-09T09:56:12.804698+00:00 | 2024-06-09T09:56:12.804741+00:00 | 13 | false | ```\nclass Solution:\n def numberOfChild(self, n: int, k: int) -> int:\n a = list(range(n))\n\n while len(a) < k*2:\n a.extend(list(range(n - 2, -1, -1)))\n a.extend(list(range(1, n)))\n\n return a[k]\n``` | 1 | 0 | ['Python'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Easy One Liner Solution along with Intuition. | easy-one-liner-solution-along-with-intui-6be3 | Let\'s understand the intuition behind the solution with an example:\nGiven - \n\nn = 4\nk = 5\n\n\nNow let us look at all the solution from k=1 to k =12\n\n\n+ | OmGujarathi | NORMAL | 2024-06-09T09:45:35.859460+00:00 | 2024-06-09T09:45:56.967097+00:00 | 10 | false | **Let\'s understand the intuition behind the solution with an example:**\nGiven - \n```\nn = 4\nk = 5\n```\n\nNow let us look at all the solution from k=1 to k =12\n\n```\n+-----+-------+\n| k | Answer|\n+-----+-------+\n| 1 | 1 |\n| 2 | 2 |\n| 3 | 3 |\n| 4 | 2 |\n| 5 | 1 |\n| 6 | 0 ... | 1 | 0 | ['Math'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | easy solution | beats 100% | easy-solution-beats-100-by-leet1101-jlcc | Intuition\nThe ball is passed back and forth between the children, reversing direction when it hits either end (child 0 or child n-1). This back-and-forth movem | leet1101 | NORMAL | 2024-06-09T08:52:32.119134+00:00 | 2024-06-09T08:52:32.119167+00:00 | 300 | false | ## Intuition\nThe ball is passed back and forth between the children, reversing direction when it hits either end (child 0 or child n-1). This back-and-forth movement can be tracked using a simple iterative approach.\n\n## Approach\n1. Start with the ball at position 0 and the direction set to 1 (moving to the right).\... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Easy Brute Force Approach | Easy to Understand | Beats 100% of users | easy-brute-force-approach-easy-to-unders-ak6a | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nStep 1: \ntake an array | chaturvedialok44 | NORMAL | 2024-06-09T08:49:27.896989+00:00 | 2024-06-09T08:49:27.897042+00:00 | 17 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nStep 1: \ntake an array arr[] of size n to hold the child.\nwill take a pointer p initially at arr[0] and variable count to couter the variable k and control the numbe... | 1 | 0 | ['Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | 🔥Very Easy to understand and intuitive O(1)🔥approach.Beginner friendly | very-easy-to-understand-and-intuitive-o1-eh7b | \n\n# Complexity\n- Time complexity:\n O(1) \n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n | Saksham_Gulati | NORMAL | 2024-06-09T08:08:43.647910+00:00 | 2024-06-09T08:08:43.647948+00:00 | 142 | false | \n\n# Complexity\n- Time complexity:\n $$O(1)$$ \n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n if((k/(n-1))%2==0)\n return k%(n-1);\n return n-1-k%(n-1);\n \n }\n};\n``` | 1 | 0 | ['Math', 'C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | 🔥Simulation - Beginner Friendly | Clean Code | C++ | | simulation-beginner-friendly-clean-code-yt9k3 | Code\n\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n if (k == 1)\n return 1;\n \n int curr = 1;\n b | Antim_Sankalp | NORMAL | 2024-06-09T07:07:21.201885+00:00 | 2024-06-09T07:07:21.201920+00:00 | 78 | false | # Code\n```\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n if (k == 1)\n return 1;\n \n int curr = 1;\n bool dir = false;\n k--;\n while (k--)\n {\n if (curr == n - 1)\n dir = true;\n if (curr == 0)\n ... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | ✅ Easy C++ Solution | easy-c-solution-by-moheat-5ur0 | Code\n\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int curr = 0;\n bool direction = true;\n while(k!=0)\n {\ | moheat | NORMAL | 2024-06-09T06:22:47.575151+00:00 | 2024-06-09T06:22:47.575175+00:00 | 91 | false | # Code\n```\nclass Solution {\npublic:\n int numberOfChild(int n, int k) {\n int curr = 0;\n bool direction = true;\n while(k!=0)\n {\n if(direction)\n {\n if(curr == n-2)\n {\n direction = false;\n }\n ... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | C++ || EASY | c-easy-by-abhishek6487209-dwen | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Abhishek6487209 | NORMAL | 2024-06-09T05:00:45.773161+00:00 | 2024-06-09T05:00:45.773185+00:00 | 75 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple Math || O(1) 100% Faster Solution | simple-math-o1-100-faster-solution-by-ya-3yv2 | Intuition\nFor a complete cycle there will be total of 2(n - 1) children and time will be k % 2(n - 1) because rest of the time complete cycles occur.\nNow if(t | yashvardhannn152004 | NORMAL | 2024-06-09T04:41:16.367546+00:00 | 2024-06-09T04:41:16.367580+00:00 | 5 | false | # Intuition\nFor a complete cycle there will be total of 2*(n - 1) children and time will be k % 2*(n - 1) because rest of the time complete cycles occur.\nNow if(time < n) means finally the ball was in left to right direction and answer would be k%l. Else it will be in right to left direction and answer would be 2*(n ... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | O(1) solution | o1-solution-by-ashis_kar-yazo | Approach\nBased on the quotient we can find the direction and the using the reminder we can find the result.\n\n# Complexity\n- Time complexity:O(1)\n\n- Space | ashis_kar | NORMAL | 2024-06-09T04:24:12.437097+00:00 | 2024-06-09T04:24:12.437183+00:00 | 69 | false | # Approach\nBased on the quotient we can find the direction and the using the reminder we can find the result.\n\n# Complexity\n- Time complexity:$$O(1)$$\n\n- Space complexity:$$O(1)$$\n\n# Code\n```\npublic class Solution {\n public int NumberOfChild(int n, int k) {\n n--;\n int q=k/n;\n int r... | 1 | 0 | ['C#'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Python O(n) Tc and o(1) Sc | python-on-tc-and-o1-sc-by-siva_manoj-qye0 | Intuition\nAfter every n-1 seconds the ball will change it\'s direction\nAfter every 2(n-1) seconds the ball will repeat it\'s previous cycle\n\n# Approach\n1 . | Siva_Manoj | NORMAL | 2024-06-09T04:11:49.060414+00:00 | 2024-06-09T04:11:49.060436+00:00 | 30 | false | # Intuition\nAfter every n-1 seconds the ball will change it\'s direction\nAfter every 2*(n-1) seconds the ball will repeat it\'s previous cycle\n\n# Approach\n1 . So we can update k value as k = k%(2*(n-1)) (bcz same cycle will repeat)\n\n2 . We can keep track of direction of ball by using direction variable which wil... | 1 | 0 | ['Math', 'Python3'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Math solution | math-solution-by-prancinglynx-7223 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | prancingLynx | NORMAL | 2024-06-09T04:10:21.286476+00:00 | 2024-06-09T04:10:21.286525+00:00 | 292 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $... | 1 | 0 | ['Python3'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Easy C++ Beginner Solution || O(1) || Beats 100% | easy-c-beginner-solution-o1-beats-100-by-tgu1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Ravi_Prakash_Maurya | NORMAL | 2024-06-09T04:06:57.578541+00:00 | 2024-06-09T04:06:57.578565+00:00 | 94 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 2 |
find-the-child-who-has-the-ball-after-k-seconds | C++ | O(1) | Intuition Explained in Detail | c-o1-intuition-explained-in-detail-by-me-2g6o | After n-1 seconds we reach right end \nAfter next n-1 seconds we reach left end\nAfter next n-1 seconds we right end again \n\n.... and so on \n\nSo,\nif n-1 / | mercer80 | NORMAL | 2024-06-09T04:04:35.217534+00:00 | 2024-06-09T05:01:06.172932+00:00 | 91 | false | After n-1 seconds we reach right end \nAfter next n-1 seconds we reach left end\nAfter next n-1 seconds we right end again \n\n.... and so on \n\nSo,\nif `n-1 / k` is an **odd** number we know for sure we are at **right end**.\n\nif `n-1/ k` is an **even** number then we know for sure we are at **left end.**\n\n\nOnce... | 1 | 0 | ['C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Java Solution | java-solution-by-solved-nzih | \nclass Solution {\n public int numberOfChild(int n, int k) {\n int index = 0;\n char direction = \'r\';\n while (k != 0) {\n | solved | NORMAL | 2024-06-09T04:03:46.504134+00:00 | 2024-06-09T04:03:46.504166+00:00 | 37 | false | ```\nclass Solution {\n public int numberOfChild(int n, int k) {\n int index = 0;\n char direction = \'r\';\n while (k != 0) {\n if (index == n - 1) {\n direction = \'l\';\n }\n if (index == 0) {\n direction = \'r\';\n }\n... | 1 | 0 | ['Simulation', 'Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple O(1) solution | simple-o1-solution-by-0x4c0de-e28q | Intuition\n Describe your first thoughts on how to solve this problem. \nThe loop size is 2 * n - 2.\n\n### Approach\n Describe your approach to solving the pro | 0x4c0de | NORMAL | 2024-06-09T04:01:58.244466+00:00 | 2024-06-09T05:11:50.672408+00:00 | 128 | false | ### Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe loop size is `2 * n - 2`.\n\n### Approach\n<!-- Describe your approach to solving the problem. -->\nWe can query the remainder of the loop, forward or backward.\n\n### Complexity\n- Time complexity: $$O((1)$$\n<!-- Add your time com... | 1 | 0 | ['Math', 'Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Swift😎 | swift-by-upvotethispls-y4o3 | Math (accepted answer) | UpvoteThisPls | NORMAL | 2025-04-08T06:26:57.024427+00:00 | 2025-04-08T06:26:57.024427+00:00 | 2 | false | **Math (accepted answer)**
```
class Solution {
func numberOfChild(_ n: Int, _ k: Int) -> Int {
let (cycle, index) = k.quotientAndRemainder(dividingBy: n-1)
return cycle.isMultiple(of:2) ? index : n-1-index
}
}
``` | 0 | 0 | ['Swift'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | easy approach | easy-approach-by-sidgogia20-kz7a | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | sidgogia20 | NORMAL | 2025-04-05T15:22:55.116220+00:00 | 2025-04-05T15:22:55.116220+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 0 | 0 | ['Math', 'Simulation', 'C++'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple Swift Solution | simple-swift-solution-by-felisviridis-51px | CodeCode | Felisviridis | NORMAL | 2025-04-02T10:42:49.116923+00:00 | 2025-04-02T11:48:07.132296+00:00 | 2 | false | 
# Code
```swift []
class Solution {
func numberOfChild(_ n: Int, _ k: Int) -> Int {
var pointer = 0, reverse = true
for i in 1...k {
if point... | 0 | 0 | ['Math', 'Swift', 'Simulation'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple and Easy Solution ( Runtime beats 100% and Memory beats 89%) | simple-and-easy-solution-runtime-beats-1-a04h | IntuitionThe problem involves passing a ball among n children arranged in a line. The ball starts at one end and moves sequentially until it reaches the other e | baytree1238 | NORMAL | 2025-03-11T09:23:30.975698+00:00 | 2025-03-11T09:23:30.975698+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem involves passing a ball among n children arranged in a line. The ball starts at one end and moves sequentially until it reaches the other end, at which point it reverses direction. By thinking of the ball’s path as a “back-and-f... | 0 | 0 | ['Python'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Simple solution in Java. Beats 100 % | simple-solution-in-java-beats-100-by-kha-co2e | Complexity
Time complexity:
O(1)
Space complexity:
O(1)
Code | Khamdam | NORMAL | 2025-03-08T04:54:07.632249+00:00 | 2025-03-08T04:54:07.632249+00:00 | 3 | false | # Complexity
- Time complexity:
O(1)
- Space complexity:
O(1)
# Code
```java []
class Solution {
public int numberOfChild(int n, int k) {
n = n - 1;
int rounds = k / n;
int remainingSteps = k % n;
if (rounds % 2 == 0) {
return remainingSteps;
} else {
... | 0 | 0 | ['Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Repeating sequence can be solved with mathematical formulae for O(1) complexity [Kotlin] | repeating-sequence-can-be-solved-with-ma-ic9c | IntuitionThe problem describes a simple sequence of numbers that start from zero, increase to a maximum value and then decline again to zero.
The maximum value | user5661 | NORMAL | 2025-03-03T17:08:00.381915+00:00 | 2025-03-03T17:08:00.381915+00:00 | 1 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem describes a simple sequence of numbers that start from zero, increase to a maximum value and then decline again to zero.
The maximum value is the number of transitions in any direction before you reach the last child in the chai... | 0 | 0 | ['Kotlin'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | O(n) Time Solution - Simple and Fast! | on-time-solution-simple-and-fast-by-ches-d3xe | ApproachChild 0 holds the ball, and passes it on to Child 1, who passes it on to Child 2, and so on. But whenever it reaches the n-1th child, the direction reve | ChessTalk890 | NORMAL | 2025-02-28T17:51:57.977636+00:00 | 2025-02-28T17:51:57.977636+00:00 | 3 | false | # Approach
<!-- Describe your approach to solving the problem. -->
Child 0 holds the ball, and passes it on to Child 1, who passes it on to Child 2, and so on. But whenever it reaches the n-1th child, the direction reverses. So, if n = 5, then the sequence will look something like this:
`0 -> 1 -> 2 -> 3 -> 4 -> 3 -> ... | 0 | 0 | ['Python3'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | O(1) Solution fast as f*ck | o1-solution-fast-as-fck-by-yarikkotsur-kjyc | IntuitionApproachCyclic solution
Ecach children has bal: 0,1,2,3...n-1,n-2,...,3,2,1
CycleLen = 2n - 2
You need to take mod from cycle len
If k < n then k is so | yarikkotsur | NORMAL | 2025-02-25T16:45:19.454616+00:00 | 2025-02-25T16:45:19.454616+00:00 | 3 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
Cyclic solution
Ecach children has bal: 0,1,2,3...n-1,n-2,...,3,2,1
CycleLen = 2n - 2
You need to take mod from cycle len
If k < n then k is solution otherwise you have to substract k from cycle len.
# Complexity
- Time comple... | 0 | 0 | ['Go'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Find the child who has the ball after k seconds📍 || Simple logic🎯|| Easy solution | find-the-child-who-has-the-ball-after-k-il5og | IntuitionThe problem simulates the movement of a child along a straight path of n positions. The child starts at position 0 and moves one step at a time. When r | palle_sravya | NORMAL | 2025-02-20T05:01:48.708438+00:00 | 2025-02-20T05:01:48.708438+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The problem simulates the movement of a child along a straight path of n positions. The child starts at position 0 and moves one step at a time. When reaching either end (0 or n-1), the direction reverses. The goal is to determine the child... | 0 | 0 | ['Math', 'Simulation', 'Java'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Easy Beginner friendly | Python code| Similar solution for Passing Pillow | easy-beginner-friendly-python-code-simil-ps5x | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | S-urjith_1001 | NORMAL | 2025-02-11T06:46:21.170691+00:00 | 2025-02-11T06:46:21.170691+00:00 | 7 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->Solve it like we solve passing pillow problem
# Approach
<!-- Describe your approach to solving the problem. -->See this similar solution
https://youtu.be/SR78dGOaxUs?si=WmiauViYAdmLrYAW
# Complexity
- Time complexity:
<!-- Add your time c... | 0 | 0 | ['Python3'] | 0 |
find-the-child-who-has-the-ball-after-k-seconds | Java | O(1) | easy to understand | java-o1-easy-to-understand-by-lost_in_sk-635v | IntuitionApproachComplexity
Time complexity : O(1)
Space complexity : O(1)
Code | lost_in_sky | NORMAL | 2025-02-10T18:57:27.789092+00:00 | 2025-02-10T18:57:27.789092+00:00 | 6 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity : O(1)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity : O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -... | 0 | 0 | ['Java'] | 0 |
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