Remove Duplicates from Sorted Array¶

Problem Link

Description¶

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.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Approach 1: Two Pointers¶

• Time Complexity: O(n)
• Space Complexity: O(1)

This approach uses a "two-pointer" technique. One pointer (i) iterates through the array, and another pointer (writeIndex) keeps track of the position where the next unique element should be placed.

When nums[i] is greater than nums[writeIndex], it means we've found a new unique element. We then increment writeIndex and copy the value of nums[i] to this new position.

function removeDuplicates(nums: number[]): number {
  if (nums.length === 0) {
    return 0;
  }
  let writeIndex = 0;
  for (let i = 1; i < nums.length; i++) {
    if (nums[i] > nums[writeIndex]) {
      nums[++writeIndex] = nums[i];
    }
  }
  return writeIndex + 1;
}

// Example 1
let nums1 = [1, 1, 2];
let k1 = removeDuplicates(nums1);
console.log(k1); // 2
console.log(nums1.slice(0, k1)); // [1, 2]

// Example 2
let nums2 = [0, 0, 1, 1, 1, 2, 2, 3, 3, 4];
let k2 = removeDuplicates(nums2);
console.log(k2); // 5
console.log(nums2.slice(0, k2)); // [0, 1, 2, 3, 4]