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.
Examples:
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).Solution in Python:
To solve the problem of removing duplicates in-place in a sorted array such that each unique element appears at most twice, we can use a two-pointer technique. This method ensures that we modify the array in-place without using extra space, maintaining O(1) additional memory usage.
Explanation:
- Initialize Pointers:
- Use a pointer
ito track the position where the next valid element should be placed. - Iterate through the array with another pointer
j.
- Use a pointer
- Process Each Element:
- Keep a count of occurrences of each element.
- For each element
nums[j], if it appears at most twice, place it at the positioniand incrementi.
- Return the Length:
- The pointer
iwill eventually point to the length of the array where the elements meet the criteria of appearing at most twice.
- The pointer
from typing import List
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
# Initialize the index `i` where the next valid element will be placed
i = 0
# Iterate through the array
for num in nums:
# If the current number can be placed in the result array
# Either we have less than 2 elements in the result array
# or the current number is greater than the element at position `i-2`
# which means it can appear at most twice
if i < 2 or num > nums[i - 2]:
# Place the number at the current position `i`
nums[i] = num
# Increment the index `i`
i += 1
# Return the length of the modified array
return iSolution in Javascript:
/**
* @param {number[]} nums
* @return {number}
*/
var removeDuplicates = function(nums) {
// Initialize the index `i` where the next valid element will be placed
let i = 0;
// Iterate through the array
for (let num of nums) {
// If the current number can be placed in the result array
// Either we have less than 2 elements in the result array
// or the current number is greater than the element at position `i-2`
// which means it can appear at most twice
if (i < 2 || num > nums[i - 2]) {
// Place the number at the current position `i`
nums[i] = num;
// Increment the index `i`
i += 1;
}
}
// Return the length of the modified array
return i;
};Solution in Java:
class Solution {
public int removeDuplicates(int[] nums) {
// Initialize the index `i` where the next valid element will be placed
int i = 0;
// Iterate through the array
for (int num : nums) {
// If the current number can be placed in the result array
// Either we have less than 2 elements in the result array
// or the current number is greater than the element at position `i-2`
// which means it can appear at most twice
if (i < 2 || num > nums[i - 2]) {
// Place the number at the current position `i`
nums[i] = num;
// Increment the index `i`
i++;
}
}
// Return the length of the modified array
return i;
}
}Solution in C#:
public class Solution {
public int RemoveDuplicates(int[] nums) {
// Initialize the index `i` where the next valid element will be placed
int i = 0;
// Iterate through the array
foreach (int num in nums) {
// If the current number can be placed in the result array
// Either we have less than 2 elements in the result array
// or the current number is greater than the element at position `i-2`
// which means it can appear at most twice
if (i < 2 || num > nums[i - 2]) {
// Place the number at the current position `i`
nums[i] = num;
// Increment the index `i`
i++;
}
}
// Return the length of the modified array
return i;
}
}