title: Remove Duplicates From Sorted Array
tags:
- remove-duplicates-from-sorted-array
- No.26
- simple
- in-place
- loop-invariant
- array

Problem

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

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

Example 1:

Given nums = [1,1,2],

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

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length.

Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Corner Cases

• single element

input:  [0]
output: [0], 1

Solutions

Naive

Many sorting algorithms like Quick Sort and Merge Sort take the strategy of in-place. This kind of method mantains a set in the array as the loop invariant.

Here we maintain index i so that the set nums[0 : i] does not have duplicates. Since the array is sorted, we do not even have to visit all elements in nums[0 : i] to see if nums[j] is a duplicate or not. We just need to compare nums[j] with the tail nums[i].

Running time is O(n).

class Solution {
    public int removeDuplicates(int[] nums) {
        int     i  = 0;
        boolean re = true;

        for (int j=0; j < nums.length; j++) {
            re        = (nums[i] < nums[j]);
            nums[i+1] = (re) ? nums[j] : nums[i+1];
            i         = (re) ? i + 1   : i;
        }

        return i + 1;
    }
}

When the array is not sorted, at most O(n^2):

class Solution {
    public int removeDuplicates(int[] nums) {
        int     i  = 0;
        boolean re = false;

        for (int j=0; j < nums.length; j++) {
            re = false;
            for (int k=0; k<=i; k++) {
                re = (nums[k] == nums[j]) ? true : re;
            }
            nums[i+1] = (re) ? nums[i+1] : nums[j];
            i         = (re) ? i : i+1;
        }

        return i + 1;
    }
}