Remove Duplicates from Sorted Array II

LeetCode 80 | Difficulty: Medium

Medium

Problem Description

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 * 10^4`

- `-10^4 <= nums[i] <= 10^4`

- `nums` is sorted in **non-decreasing** order.

Topics: Array, Two Pointers

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Approach

Two Pointers

Use two pointers to traverse the array, reducing the search space at each step. This avoids the need for nested loops, bringing complexity from O(n²) to O(n) or O(n log n) if sorting is involved.

When to use

Array is sorted or can be sorted, and you need to find pairs/triplets that satisfy a condition.

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Solutions

Solution 1: C# (Best: 280 ms)

Metric	Value
Runtime	280 ms
Memory	N/A
Date	2018-07-02

Solution

public class Solution {
    public int RemoveDuplicates(int[] nums) {
        if(nums.Length<=2) return nums.Length;
    
    int index=0;
    int i=0;
    int totalCount = 0;
    while(i<nums.Length)
    {
        var current = nums[i];
        int currentCount = 1;
        i++;
        while(i<nums.Length && current==nums[i]) {i++; currentCount++; }
        for (int j = 0; j < currentCount && j<2; j++)
        {
            nums[index] = current;
            index++;
        }   
        totalCount += currentCount>=2 ? 2 : currentCount;

    }
    return totalCount;
    }
}

📜 1 more C# submission(s)

Submission (2018-07-02) — 288 ms, N/A

public class Solution {
    public int RemoveDuplicates(int[] nums) {
        if (nums.Length <= 2) return nums.Length;

    int index = 0;
    int i = 0;
    int totalCount = 0;
    while (i < nums.Length)
    {
        var current = nums[i];
        int currentCount = 1;
        i++;
        while (i < nums.Length && current == nums[i]) { i++; currentCount++; }
        currentCount = currentCount >= 2 ? 2 : currentCount;
        for (int j = 0; j < currentCount; j++)
        {
            nums[index] = current;
            index++;
        }
        totalCount += currentCount;
    }
    return totalCount;
    }
}

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Complexity Analysis

Approach	Time	Space
Two Pointers	$O(n)$	$O(1)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Ask: "Can I sort the array?" — Sorting often enables two-pointer techniques.