Find All Duplicates in an Array
To solve this problem without using any additional space and within O(n) time complexity, we can use the array itself as a marker for the presence of an integer. Since the integers in the array are in the range [1, n] and each integer appears once or twice, we can use the value of each integer as an index to mark its occurrence.
Here’s the stepbystep guide:
 Iterate Through Array: Traverse each element in the array.
 Find Corresponding Index: For the current integer
num
, find the corresponding index by taking the absolute valueidx = abs(num)  1
.  Mark the Occurrence: If the value at the corresponding index
nums[idx]
is positive, negate it to mark the occurrence. If it’s already negative, then the number has occurred twice, so we add it to the result.  Collect Duplicates: Collect all the numbers that appear twice.
Here’s the code:


Explanation:
 For each number in
nums
, we use its absolute value minus one as the index (idx
) to look up in the array.  If the value at
nums[idx]
is negative, it means the current number has appeared before, so we add it to the result.  Otherwise, we negate the value at
nums[idx]
to mark its occurrence.
This approach satisfies the constraint of using only constant extra space and runs in O(n) time, where n is the length of the array nums
.


Brute Force
T: O(n^2) S: O(1)
We can use extra space, by using the set to check for duplicates?
Can you do it in O(1) space and O(n) runtime?
Sort the list and remove the duplicates? T:O(n log n) S:O(1) or O(n)
0.1 2 3 4 5 6 7 [4,3,2,7,8,2,3,1] i,j 4,3 4,2 4,7 4,8 4,2 4,3 4,1
size = 8
[4,3,2,7,8,2,3,1]
0, 6 1, 7
0 1 2 3 4 5 6 7 [1,2,2,3,3,4,7,8]
Two Pointers Approach T: O()

