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] <= 100numsis sorted in non-decreasing order.
class Solution:
def removeDuplicates(self, nums: list[int]) -> int:
i = 0
for num in nums:
if i == 0 or num != nums[i-1]:
nums[i] = num
i += 1
return i
Explanation (Simple Version)
i keeps track of the position of the next unique number.
Loop through each number:
If it’s the first number or different from the last unique number, place it at nums[i] and move i.
At the end, i → length of unique numbers, and nums[:i] → all unique numbers.
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