leetcode-150

4. Remove Duplicates from Sorted Array II

Problem Statement

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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.

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Solution

Algorithm Explanation

1. Base Case Handling:
   • If the array has 2 or fewer elements, return its length immediately since no modifications are needed.
2. Use of Two Pointers:
   • index is used to track the position where the next valid element should be placed.
   • The loop iterates through nums starting from index 2 because the first two elements can always remain.
3. Checking Condition:
   • If the current element nums[i] is different from nums[index - 2], it means it can be included in the final array (as the frequency is at most 2).
   • We update nums[index] with nums[i] and increment index.
4. Returning the New Length:
   • After processing all elements, index holds the length of the modified array.

Complexity Analysis

• Time Complexity: (O(n)) since we iterate through the array once.
• Space Complexity: (O(1)) as we modify the array in place without using additional space.

Implementation

function removeDuplicates(nums) {
    if (nums.length <= 2) return nums.length; // If array has 2 or fewer elements, return its length
    
    let index = 2; // Start from the third position since the first two can always remain
    
    for (let i = 2; i < nums.length; i++) {
        if (nums[i] !== nums[index - 2]) { // Check if the current number is different from the one two places back
            nums[index] = nums[i]; // Update the array in place
            index++; // Move the index forward
        }
    }
    
    return index; // Return the new length of the modified array
}

// Example usage:
let nums = [1,1,1,2,2,3];
let k = removeDuplicates(nums);
console.log(k, nums.slice(0, k)); // Output: 5, [1,1,2,2,3]

nums = [0,0,1,1,1,1,2,3,3];
k = removeDuplicates(nums);
console.log(k, nums.slice(0, k)); // Output: 7, [0,0,1,1,2,3,3]

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