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‎README.md

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This repo is a collection of my LeetCode solutions, primarily written in Python, Java, and C. On any page, `click the main title` to be redirected to the official `LeetCode` page for the question, topic, list, etc. See the `Additional Categories` section for pages that group questions by different criteria -- e.g. by their *related topics*.
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## Category Notes
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1. **Daily** - Daily challenge questions that were done on the day of
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2. **Weekly Premium** - Weekly premium questions that were done on week of

‎markdowns/Questions_By_Code_Length.md

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‎markdowns/_1002. Find Common Characters.md

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> *First completed : June 04, 2024*
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>
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> *Last updated : July 01, 2024*
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> *Last updated : July 03, 2024*
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# Daily
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# Doing this with bad wifi so I can barely submit so idc about efficiency lol
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class Solution:
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def commonChars(self, words: List[str]) -> List[str]:
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cnter = []

‎markdowns/_1184. Distance Between Bus Stops.md

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> *First completed : May 31, 2024*
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>
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> *Last updated : July 01, 2024*
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> *Last updated : July 03, 2024*
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*To see the question prompt, click the title.*
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> ```
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> Notes
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> A bit bulky but did it for the theoretical runtime savings. Still O(n) but
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> can save on the additional calclations if one direction is significantly
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> shorter than the other by not having to have the additional addition operations.
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> ```
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>
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------
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## Solutions
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- [e1184.py](<../my-submissions/e1184.py>)
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### Python
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#### [e1184.py](<../my-submissions/e1184.py>)
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```Python
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''' Notes
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A bit bulky but did it for the theoretical runtime savings. Still O(n) but
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can save on the additional calclations if one direction is significantly
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shorter than the other by not having to have the additional addition operations.
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'''
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class Solution:
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def distanceBetweenBusStops(self, distance: List[int], start: int, destination: int) -> int:
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counter, clockwise = start, start

‎markdowns/_1404. Number of Steps to Reduce a Number in Binary Representation to One.md

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> *First completed : May 29, 2024*
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>
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> *Last updated : July 01, 2024*
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> *Last updated : July 03, 2024*
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*To see the question prompt, click the title.*
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> ```
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> Ideas
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> We can just go right to left.
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> - If rightmost bit is 0, +1 step and shift position left 1
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> - If 1, +1 step and have a "carry" for the next
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> ```
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>
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>
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------
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## Solutions
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- [m1404 Daily.py](<../my-submissions/m1404 Daily.py>)
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### Python
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#### [m1404 Daily.py](<../my-submissions/m1404 Daily.py>)
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```Python
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''' Ideas
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We can just go right to left.
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- If rightmost bit is 0, +1 step and shift position left 1
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- If 1, +1 step and have a "carry" for the next
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'''
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class Solution:
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def numSteps(self, s: str) -> int:
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counter = 0

‎markdowns/_1438. Longest Continuous Subarray With Absolute Diff Less Than or Equal to Limit.md

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> *First completed : June 22, 2024*
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>
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> *Last updated : June 22, 2024*
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> *Last updated : July 03, 2024*
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*To see the question prompt, click the title.*
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> ```
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> Notes
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> We can store the max and min value going in each direction
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>
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> Option 1: Brute force all options
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> Option 2: Sliding window
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> If we're above limit, we shift whichever pointer is leftmost += 1
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> ```
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------
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## Solutions
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- [m1438 Daily.java](<../my-submissions/m1438 Daily.java>)
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### Java
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#### [m1438 Daily.java](<../my-submissions/m1438 Daily.java>)
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```Java
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/** Notes
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We can store the max and min value going in each direction
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Option 1: Brute force all options
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Option 2: Sliding window
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If we're above limit, we shift whichever pointer is leftmost += 1
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*/
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class Solution {
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public int longestSubarray(int[] nums, int limit) {
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int output = 0;

‎markdowns/_1442. Count Triplets That Can Form Two Arrays of Equal XOR.md

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> *First completed : May 29, 2024*
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>
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> *Last updated : July 01, 2024*
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> *Last updated : July 03, 2024*
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*To see the question prompt, click the title.*
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> ```
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> Notes
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> Negation of XOR is Biconditional
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> i.e. p <-> q
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> (p -> q) and (q -> p)
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> (!p or q) and (!q or p)
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> (!p and !q) or (p and q)
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> p == q
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>
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>
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> 1^0 --> 1 --> --> 1^0 --> 1
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> 1^1 --> 0 --> --> 0^1 --> 1
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> 0^0 --> 0 --> --> 0^0 --> 0
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> 0^1 --> 1 --> --> 1^1 --> 0
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>
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> XORing by the same value again reverses the result!
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>
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> Since 0 <= i < j <= k < arr.len --> k-i+1 >= 2
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>
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> If two arrays XOR together to get 0, then the two individual sections should have
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> an XOR that equal each other.
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>
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> That is, XOR(arr1) == XOR(arr2) so that XOR(arr1, arr2) == 0
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>
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------
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## Solutions
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- [m1442 Daily.py](<../my-submissions/m1442 Daily.py>)
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### Python
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#### [m1442 Daily.py](<../my-submissions/m1442 Daily.py>)
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```Python
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''' Notes
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Negation of XOR is Biconditional
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i.e. p <-> q
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(p -> q) and (q -> p)
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(!p or q) and (!q or p)
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(!p and !q) or (p and q)
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p == q
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1^0 --> 1 --> --> 1^0 --> 1
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1^1 --> 0 --> --> 0^1 --> 1
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0^0 --> 0 --> --> 0^0 --> 0
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0^1 --> 1 --> --> 1^1 --> 0
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XORing by the same value again reverses the result!
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Since 0 <= i < j <= k < arr.len --> k-i+1 >= 2
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If two arrays XOR together to get 0, then the two individual sections should have
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an XOR that equal each other.
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That is, XOR(arr1) == XOR(arr2) so that XOR(arr1, arr2) == 0
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'''
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class Solution:
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def countTriplets(self, arr: List[int]) -> int:
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counter = 0

‎markdowns/_1482. Minimum Number of Days to Make m Bouquets.md

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> *First completed : June 18, 2024*
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>
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> *Last updated : June 18, 2024*
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> *Last updated : July 03, 2024*
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*To see the question prompt, click the title.*
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> ```
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> Notes
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> This seems a lot like merge intervals where we need to find how many days till
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> sum(interval - remainder for interval in intervals) >= m * k
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> ```
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>
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## Solutions
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- [m1482 Dailly.py](<../my-submissions/m1482 Dailly.py>)
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- [m1482 v2 Removed Helper.py](<../my-submissions/m1482 v2 Removed Helper.py>)
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### Python
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#### [m1482 Dailly.py](<../my-submissions/m1482 Dailly.py>)
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```Python
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''' Notes
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This seems a lot like merge intervals where we need to find how many days till
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sum(interval - remainder for interval in intervals) >= m * k
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'''
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class Solution:
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def minDays(self, bloomDay: List[int], m: int, k: int) -> int:
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#### [m1482 v2 Removed Helper.py](<../my-submissions/m1482 v2 Removed Helper.py>)
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```Python
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''' Notes
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This seems a lot like merge intervals where we need to find how many days till
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sum(interval - remainder for interval in intervals) >= m * k
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'''
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class Solution:
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def minDays(self, bloomDay: List[int], m: int, k: int) -> int:
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‎markdowns/_1551. Minimum Operations to Make Array Equal.md

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> *First completed : June 12, 2024*
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>
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> *Last updated : July 01, 2024*
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> *Last updated : July 03, 2024*
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*To see the question prompt, click the title.*
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>
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> ```
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> sum = (2 * n + 2) * n // 2
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> mean = (2 * n + 2) / 2
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> margineToMove = sum(2n+1 - ((2 * n + 2) / 2))
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> = sum(2n+1 - (n+1))
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> = sum(n)
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> = (1 + n) * n // 2
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> number of steps = margineToMove / 2
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> = (1 + n) * n // 2 // 2
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> = (n^2 + 1) // 4
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> = n^2 // 4
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> = n ** 2 // 4
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>
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> ```
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------
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## Solutions
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- [m1551 v2.c](<../my-submissions/m1551 v2.c>)
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```Python
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class Solution:
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def minOperations(self, n: int) -> int:
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# sum = (2 * n + 2) * n // 2
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# mean = (2 * n + 2) / 2
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# margineToMove = sum(2n+1 - ((2 * n + 2) / 2))
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# = sum(2n+1 - (n+1))
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# = sum(n)
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# = (1 + n) * n // 2
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# number of steps = margineToMove / 2
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# = (1 + n) * n // 2 // 2
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# = (n^2 + 1) // 4
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# = n^2 // 4
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# = n ** 2 // 4
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return n ** 2 // 4
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```
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‎markdowns/_1552. Magnetic Force Between Two Balls.md

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> *First completed : June 20, 2024*
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>
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> *Last updated : June 20, 2024*
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> *Last updated : July 03, 2024*
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*To see the question prompt, click the title.*
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> ```
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> Notes
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> In essence, we have x positions and m balls, and we need to find AN ideal
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> way where we can place the m balls into m positions so that the min distance
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> between the balls is minimized
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>
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> The ideal case will be (max(position) - min(position)) / (m - 1)
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> ```
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------
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## Solutions
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- [m1552 Daily.py](<../my-submissions/m1552 Daily.py>)
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### Python
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#### [m1552 Daily.py](<../my-submissions/m1552 Daily.py>)
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```Python
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''' Notes
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In essence, we have x positions and m balls, and we need to find AN ideal
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way where we can place the m balls into m positions so that the min distance
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between the balls is minimized
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The ideal case will be (max(position) - min(position)) / (m - 1)
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'''
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class Solution:
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def maxDistance(self, position: List[int], m: int) -> int:
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position.sort()

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