LeetCode 0026. Remove Duplicates from Sorted Array Solution Java | Python | C/C++ | JavaScripts | Go | Rust | Explaination + CODE

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.

Consider the number of unique elements in nums to be k​​​​​​​​​​​​​​. After removing duplicates, return the number of unique elements k.

The first k elements of nums should contain the unique numbers in sorted order. The remaining elements beyond index k - 1 can be ignored.

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 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

Solutions

Solution 1: Single Pass

We use a variable k to record the current length of the processed array. Initially, k=0 represents an empty array.

Then we traverse the array from left to right. For each element x we encounter, if k=0 or x ≠ nums[k-1], we place x in the position of nums[k], and then increment k by 1. Otherwise, x is the same as nums[k-1], so we skip this element. Continue to traverse until the entire array is traversed.

In this way, when the traversal ends, the first k elements in nums are the answer we are looking for, and k is the length of the answer.

The time complexity is O(n), and the space complexity is O(1). Here, n is the length of the array.

Supplement:

The original problem requires that the same number appear at most once. We can extend it to keep at most k identical numbers.

• Since the same number can be kept at most k times, we can directly keep the first k elements of the original array;
• For the following numbers, the premise of being able to keep them is: the current number x is compared with the last kth element of the previously retained numbers. If they are different, keep them, otherwise skip them.

Similar problems:

• 80. Remove Duplicates from Sorted Array II

PythonJavaC++GoTypeScriptRustJavaScriptC#PHP

class Solution: def removeDuplicates(self, nums: List[int]) -> int: k = 0 for x in nums: if k == 0 or x != nums[k - 1]: nums[k] = x k += 1 return k(code-box)

class Solution { public int removeDuplicates(int[] nums) { int k = 0; for (int x : nums) { if (k == 0 || x != nums[k - 1]) { nums[k++] = x; } } return k; } }(code-box)

class Solution { public: int removeDuplicates(vector<int>& nums) { int k = 0; for (int x : nums) { if (k == 0 || x != nums[k - 1]) { nums[k++] = x; } } return k; } };(code-box)

func removeDuplicates(nums []int) int { k := 0 for _, x := range nums { if k == 0 || x != nums[k-1] { nums[k] = x k++ } } return k }(code-box)

function removeDuplicates(nums: number[]): number { let k: number = 0; for (const x of nums) { if (k === 0 || x !== nums[k - 1]) { nums[k++] = x; } } return k; }(code-box)

impl Solution { pub fn remove_duplicates(nums: &mut Vec<i32>) -> i32 { let mut k = 0; for i in 0..nums.len() { if k == 0 || nums[i] != nums[k - 1] { nums[k] = nums[i]; k += 1; } } k as i32 } }(code-box)

/** * @param {number[]} nums * @return {number} */ var removeDuplicates = function (nums) { let k = 0; for (const x of nums) { if (k === 0 || x !== nums[k - 1]) { nums[k++] = x; } } return k; };(code-box)

public class Solution { public int RemoveDuplicates(int[] nums) { int k = 0; foreach (int x in nums) { if (k == 0 || x != nums[k - 1]) { nums[k++] = x; } } return k; } }(code-box)

class Solution { /** * @param Integer[] $nums * @return Integer */ function removeDuplicates(&$nums) { $k = 0; foreach ($nums as $x) { if ($k == 0 || $x != $nums[$k - 1]) { $nums[$k++] = $x; } } return $k; } }(code-box)