leetcode_gym/_problems/findAllDupsInArray.md

layout, title, last_solved, category, leetcode_url, status

layout	title	last_solved	category	leetcode_url	status
page	Find all Duplicates in Array	2020-08-06	array	https://leetcode.com/problems/find-all-duplicates-in-an-array	Attempted

Problem

Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.

Find all the elements that appear twice in this array.

Could you do it without extra space and in O(n) runtime?

Example:
Input:
[4,3,2,7,8,2,3,1]

Output:
[2,3]

Solution

If you can use extra space for a hashmap. Then you can use this one-liner

{% highlight python %}

def findDuplicates(self, nums: List[int]) -> List[int]:

c = Counter(nums)

print(c)

# output = []
# for k, v in c.items():
#     if v==2:
#         output.append(k)

return [k for k,v in c.items() if v==2]

{% endhighlight %}

---

To not use any additional space, we can only modify the existing array. Use the constraint 1<a[i]<n

• Map the current value to it's position in the array. (ie: [4,3,2,7,...], '7' is in the 4's place, so make 7->-7)
• If there is another 4 in the array, it will check the 7 and see that it's negative. If it's negative, add it to the output array.

{% highlight python %}

def findDuplicates(self, nums: List[int]) -> List[int]:

# [4,3,2,7,8,2,3,1]
# [2,2]

output = []

for i,v in enumerate(nums):           
    
    pos_v = abs(v)
    
    if nums[pos_v-1]<0:
        output.append(pos_v)
    
    # mark as visited
    nums[pos_v-1] = abs(nums[pos_v-1])*-1

{% endhighlight %}

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