Remove Duplicates from Sorted Array - LeetCode Solution and Approach

  

 26. Remove Duplicates from Sorted Array - LeetCode - Easy

Solve this problem in LeetCode

Problem 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. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
• Return k.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array

int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;

for (int i = 0; i < k; i++) {

    assert nums[i] == expectedNums[i];

}

If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [1,1,2]

Output: 2, nums = [1,2,_]

Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]

Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]

Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

 

Constraints:

1 <= nums.length <= 3 * 10^4

-100 <= nums[i] <= 100

nums is sorted in non-decreasing order.

Approach:

In this problem we have to remove the duplicates in-place. The elements are in sorted order, so the duplicates will be in sequence. Now the problem is reduced to finding next unique element, which is quiet simpler.

• We will store the first unique element in a variable say prevUnique.
• We will iterate through given array and check if number is not same as previous unique element stored in prevUnique. Now we have found new unique element. We will store the element at index say p which is equal to number of unique elements.
• At last we will update the previous unique element with new unique and will continue iterating.
• After iterating, we will get the number of unique elements stored in index p.

JAVA:

Time Complexity : 0(n)

Space Complexity: 0(1)

class Solution {
    public int removeDuplicates(int[] nums) {
        int prevUnique = nums[0], p = 1;
        for (int n : nums) {
            if (n != prevUnique) {
                nums[p++] = n;
                prevUnique = n;
            }
        }
        return p;
    }
}

Keep Learning and Practicing with Programming Chaska !!!