Hanan

“The lust for comfort murders the passion of the soul.” ― Khalil Gibran

you promised yourself you’re gonna go harder this year. don’t forget.

My mom recently asked me how to take a screenshot At first I laughed, but then I remembered she taught me how to tie my shoes.

“Nothing helps; I must help myself, or I am finished.” — Friedrich Nietzsche

📌

This generation’s daily routine: - Wake up. - Stare at a 6.7 inch screen. - Work on a 16 inch screen. - Relax with a 55inch screen. - Stare at a 6.7 Inch Screen - Sleep Repeat.

Why this works? In a circular path: Faster pointer always catches the slower one If no cycle → fast hits null Elegant. Efficient. #Day29 #DSA #LeetCode #LinkedList

2️⃣ Optimal approach: Floyd’s Cycle Detection (Tortoise & Hare) Use two pointers: slow → moves 1 step fast → moves 2 steps If there’s a cycle, they must meet inside the loop. Time: O(n) Space: O(1)

Day 29 of DSA Problem: Linked List Cycle Goal: Detect if a linked list contains a cycle Solve it with O(1) extra memory A classic interview favorite 1️⃣ Brute-force idea (using extra memory) 2️⃣ Optimal approach: Floyd’s Cycle Detection (Tortoise & Hare)

Reversing the second half avoids extra memory — clean & efficient.

// Compare let left = head, right = prev; while (right) { if (left.val !== right.val) return false; left = left.next; right = right.next; } return true; };

Day 28 – DSA Progress Palindrome Linked List O(n) time O(1) space Approach: • Find middle using slow & fast pointers • Reverse the second half in-place • Compare both halves

#LeetCode #DSA #CodingJourney #100DaysOfCode

current.next = list1; list1 = list1.next; } else { current.next = list2; list2 = list2.next; } current = current.next; } current.next = list1 ?? list2; return dummy.next; }

Day 27 | LeetCode Solved Merge Two Sorted Lists using an iterative approach. ✔️ Uses a dummy node ✔️ Runs in O(n + m) time ✔️ Clean and beginner-friendly