🔥 The Challenge

Problem: Merge two sorted linked lists into one sorted list.

Input: list1 = [1->2->4], list2 = [1->3->4]  
Output: [1->1->2->3->4->4]

🧠 My Battle Strategy

I deployed the Two-Pointer Technique with a Dummy Node vanguard:

  1. Sentinel Node: Created a dummy head to eliminate edge cases.
  2. Pointer March: Advanced through both lists like shadow soldiers
  3. Optimal Selection: Always chose the smaller value to maintain order 
function mergeTwoLists(list1: ListNode | null, list2: ListNode | null): ListNode | null {
  const dummy = new ListNode(0); // 🛡️ My shield against null cases
  let current = dummy;

  while (list1 && list2) {
    if (list1.val <= list2.val) {
      current.next = list1;
      list1 = list1.next!;
    } else {
      current.next = list2;
      list2 = list2.next!;
    }
    current = current.next;
  }

  current.next = list1 || list2; // ⚡ Finishing move
  return dummy.next;
}

💡 Key Insights
Space Efficiency: O(1) space complexity by reusing existing nodes

Time Optimization: O(n+m) time by single-pass merging

Elegant Edge Handling: Dummy node prevents null reference headaches

🚀 Level Up Challenge

Can you solve this recursively? Here's a sneak peek:

function mergeRecursive(l1: ListNode | null, l2: ListNode | null): ListNode | null {
  if (!l1) return l2;
  if (!l2) return l1;

  if (l1.val < l2.val) {
    l1.next = mergeRecursive(l1.next, l2);
    return l1;
  } else {
    l2.next = mergeRecursive(l1, l2.next);
    return l2;
  }
}

Daily Progress: ▰▰▰▰▰ 4.44% (4/90)
Tags: algorithms, linkedlists, typescript, dsa, programming