LeetCode – Remove Duplicates from Sorted Array II (Java)

Follow up for "Remove Duplicates": What if duplicates are allowed at most twice?

For example, given sorted array A = [1,1,1,2,2,3], your function should return length = 5, and A is now [1,1,2,2,3].

So this problem also requires in-place array manipulation.

Java Solution 1

We can not change the given array's size, so we only change the first k elements of the array which has duplicates removed.

public int removeDuplicates(int[] nums) {
    if(nums==null){
        return 0;
    }
    if(nums.length<3){
        return nums.length;
    }
 
    int i=0;
    int j=1;
/*
 
         i, j    1 1 1 2 2 3
step1    0  1      i j
step2    1  2      i   j
step3    1  3        i   j
step4    2  4          i   j
 
*/
    while(j<nums.length){
        if(nums[j]==nums[i]){
            if(i==0){
                i++;
                j++;
            }else if(nums[i]==nums[i-1]){
                j++;
            }else{    
                i++;
                nums[i]=nums[j];
                j++;
            }
        }else{
            i++;
            nums[i]=nums[j];
            j++;
        }
    }
 
    return i+1;
}

The problem with this solution is that there are 4 cases to handle. If we shift our two points to right by 1 element, the solution can be simplified as the Solution 2.

Java Solution 2

public int removeDuplicates(int[] nums) {
    if(nums==null){
        return 0;
    }
 
    if (nums.length <= 2){
        return nums.length;
    }
/*
1,1,1,2,2,3
  i j
*/
    int i = 1; // point to previous
    int j = 2; // point to current
 
    while (j < nums.length) {
        if (nums[j] == nums[i] && nums[j] == nums[i - 1]) {
            j++;
        } else {
            i++;
            nums[i] = nums[j];
            j++;
        }
    }
 
    return i + 1;
}

Category >> Algorithms

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Ankit Shah

similar to the duplicates I solution

private static int[] removeDuplicateII(int[] a) { if (a.length < 3) return a; int i = 0; int j = 1; int k = 2; while (k < a.length) { if (a[i] == a[j] && a[j] == a[k]) { j++; k++; } else { i++; a[i] = a[j]; j++; a[j] = a[k]; k++; } }
int[] b = Arrays.copyOf(a, i + 2); return b; }
Larry Okeke

public static int remove_duplicates(int[] nums){ int duplicates =0; for(int i = 0; i < nums.length; i++){ if(i < nums.length-1 && nums[i] == nums[i+1]){ int step = i++; while((i < nums.length-1 )&&nums[step] == nums[i++]){ duplicates++; } } } return nums.length - duplicates;
}
Swathi

How about this? A slight modification on the remove duplicates I

int[] removeDuplicatesN(int[] arr, int N){ int i=0,j=1, l=0; int n= arr.length; while(j<n){ if(arr[i]==arr[j]){ j++; l++; if(l<N) i++; } else{ l=0; i++; arr[i] = arr[j]; j++; } } return Arrays.copyOf(arr, i+1); }
cp

the better solution has a little bug. If an array [1,1] then it does not remove the element.
vincent

Good!
Liang Tao

I think what you said is the answer for “Remove Duplicates” problem, not this II version.
TK

class Solution {
public:
int removeDuplicates(int A[], int n) {
int currentPosition = 2;
for (int i = 2; i < n; ++ i)
if (A[i] != A[i-2])
A[currentPosition++] = A[i];
return min(currentPosition, n);
}
};
claire

great solution
Walter

The better solution is actually confusing. We can make the rule like this:
Assign previous pointer to 1st element(a[0]) and current pointer to 2nd element(a[1]).
1) if a[prev] != a[curr], {prev++; a[prev]=a[cur]; cur++;}
2) else, {cur++;}

you can remove the A[curr] == A[prev – 1] in this case.
W Han

In the first solution, the loop starts from the second element. If input array has only one element it will cause index out of bounds error.

LeetCode – Remove Duplicates from Sorted Array II (Java)

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