Approach: Two Pointers, Time complextiy : O(n)1
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10class Solution {
public int removeDuplicates(int[] nums) {
if(nums.length<2) return 1;
int j=0;
for(int i=1;i<nums.length;i++){
if(nums[i]!=nums[i-1]) nums[++j]=nums[i];
}
return ++j;
}
}