Remove duplications in array and linked list-26,82,83 - 8/28/2024

26:Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

function removeDuplicates(nums: number[]): number {
  let l = 0;
  for (let r = 1; r < nums.length; r++) {
    if (nums[l] !== nums[r]) {
      l++;
      nums[l] = nums[r];
    }
  }

  return l + 1;
}

83:Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

function deleteDuplicates(head: ListNode | null): ListNode | null {
  if (head === null) return null;
  //s is initialized to point to the current node being processed.
  let s = head;
  // f is initialized to the next node, which will be used to check for duplicates.
  let f = head.next;
  while (f !== null) {
    if (f.val !== s.val) {
      s.next = f;
      s = s.next;
    }
    f = f.next;
  }
  // s.next = null ensures that any remaining part of the list after s is cut off, which might have been part of a duplicate sequence.
  s.next = null;
  return head;
}

82:Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

function deleteDuplicates(head: ListNode | null): ListNode | null {
  if (head === null) return null;

  // Create a dummy node to handle edge cases like removing the first node
  let dummy = new ListNode(0, head);

  let prev = dummy;
  let current = head;

  while (current !== null) {
    // Check if the current node has duplicates
    if (current.next !== null && current.val === current.next.val) {
      // Skip all nodes with the current value
      while (current.next !== null && current.val === current.next.val) {
        current = current.next;
      }
      // Link prev to the node after the last duplicate,effectively removing the duplicates from the list.
      prev.next = current.next;
    } else {
      // Move prev pointer forward only if no duplicates were found
      prev = prev.next;
    }
    current = current.next;
  }

  return dummy.next;
}