Remove Duplicates from Sorted Array - Array - Easy - LeetCode
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Remove Duplicates from Sorted Array - Array - Easy - LeetCode

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  • 1The task is to remove duplicates from a sorted array in-place, ensuring each element appears only once.
  • 2The function must return the new length of the modified array without using extra space for another array.
  • 3Examples demonstrate that the first few elements of the array should be unique, while the rest can be ignored.

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"The task is to remove duplicates from a sorted array in-place, ensuring each element appears only once."

Remove Duplicates from Sorted Array - Array - Easy - LeetCode

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example 1:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.

It doesn't matter what you leave beyond the returned length. Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.

It doesn't matter what values are set beyond the returned length. Clarification:

Confused why the returned value is an integer but your answer is an array?

Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy) int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller. // using the length returned by your function, it prints the first len elements. for (int i = 0; i < len; i++) { print(nums[i]); }

public class Solution {
 public int RemoveDuplicates(int[] nums) {
 if(nums.Length == 0){
 return 0;
 }
 
 int prev = nums[0];
 int j = 1;
 for(int i = 1;i<nums.Length;i++){
 if(nums[i] != prev){
 prev = nums[i];
 nums[j]=prev;
 j++;
 }
 }
 
 return j;
 }
}

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sumitc91

Published on 24 September 2020 · 1 min read · 262 words

Part of AskGif Blog · coding

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