Question

Formatted question description: https://leetcode.ca/all/80.html

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. 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.

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 = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

1 <= nums.length <= 3 * 10⁴
-10⁴ <= nums[i] <= 10⁴
nums is sorted in non-decreasing order.

Algorithm

The maximum number of repetitions allowed is two, then a variable cnt can be used to record and several repetitions are allowed, cnt is initialized to 1, if there is one repetition, cnt is decremented by 1, then the next time there is a repetition, the fast pointer is directly moving forward, if it is not repeated at this time, cnt is restored to 1.

Since the entire array is ordered, once a non-repeating number appears, it must be greater than this number, and there will be no more duplicates after this number.

Code

public class Remove_Duplicates_from_Sorted_Array_II {

    class Solution {
        public int removeDuplicates(int[] nums) {
            int i = 0;
            for (int num : nums) {
                if (i < 2 || num > nums[i - 2]) {
                    nums[i++] = num;
                }
            }
            return i;
        }
    }

    public int removeDuplicates(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }

        int pre = 0, cur = 1, dupAllowed = 1, n = nums.length;
        while (cur < n) {
            if (nums[pre] == nums[cur] && dupAllowed == 0) {
                ++cur;
            } else {
                if (nums[pre] == nums[cur]) {
                    --dupAllowed;
                } else {
                    dupAllowed = 1;
                }

                nums[++pre] = nums[cur++];
            }
        }

        return pre + 1;
    }
}


############

class Solution {
    public int removeDuplicates(int[] nums) {
        int i = 0;
        for (int num : nums) {
            if (i < 2 || num != nums[i - 2]) {
                nums[i++] = num;
            }
        }
        return i;
    }
}

// OJ: https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int removeDuplicates(vector<int>& A) {
        int j = 0;
        for (int i = 0; i < A.size(); ++i) {
            if (j - 2 < 0 || A[j - 2] != A[i]) A[j++] = A[i];
        }
        return j;
    }
};

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        i = 0
        for num in nums:
            if i < 2 or num != nums[i - 2]:
                nums[i] = num
                i += 1
        return i

############

class Solution(object):
  def removeDuplicates(self, nums):
    """
    :type nums: List[int]
    :rtype: int
    """
    if len(nums) <= 2:
      return len(nums)
    cnt = 0
    j = 1
    for i in range(1, len(nums)):
      if nums[i] == nums[i - 1]:
        cnt += 1
        if cnt < 2:
          nums[j] = nums[i]
          j += 1
      else:
        nums[j] = nums[i]
        j += 1
        cnt = 0
    return j

func removeDuplicates(nums []int) int {
	i := 0
	for _, num := range nums {
		if i < 2 || num != nums[i-2] {
			nums[i] = num
			i++
		}
	}
	return i
}

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function (nums) {
    let i = 0;
    for (const num of nums) {
        if (i < 2 || num != nums[i - 2]) {
            nums[i++] = num;
        }
    }
    return i;
};

public class Solution {
    public int RemoveDuplicates(int[] nums) {
        int i = 0;
        foreach(int num in nums)
        {
            if (i < 2 || num != nums[i - 2])
            {
                nums[i++] = num;
            }
        }
        return i;
    }
}

impl Solution {
    pub fn remove_duplicates(nums: &mut Vec<i32>) -> i32 {
        let mut len = 0;
        for i in 0..nums.len() {
            if i < 2 || nums[i] != nums[len - 2] {
                nums[len] = nums[i];
                len += 1;
            }
        }
        len as i32
    }
}

function removeDuplicates(nums: number[]): number {
    let k = 0;
    for (const x of nums) {
        if (k < 2 || x !== nums[k - 2]) {
            nums[k++] = x;
        }
    }
    return k;
}

80 - Remove Duplicates from Sorted Array II

Question

Algorithm

Code

All Problems

All Solutions

Welcome to Subscribe On Youtube

Question

Algorithm

Code

All Problems

All Solutions