Find All Duplicates in an Array | LeetCode 442 | C++, Java, Python

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August LeetCoding Challenge | Problem 6 | Find All Duplicates in an Array | 6 August,
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[mandeep]

ur background is really intelligently selected as it stops the reflection..& explaination was great..

[himanshupatel]

why we took abs of n .
n is index and we are making the no. at n negative.

[Ahmedmeshref]

Thanks for your effort. I do understand the technique used, but I don't understand why it works??

[risavkumarjha]

Nice Approach, sir
Thanks for sharing.

[tulikachhaukhhadiya]

sir but this approach at 5:27 won't work when we have nos. repeating thrice

[GR_EDITZZZ]

Just xoring all the elemnts in order to extract the unique elements & store them & them later on xoring this resultant unique elements with the whole array so that we left with only duplicate elements & store & then return them into an array...

[mainakbanerjee]

Very Good technique Sir, Nicely Explained

[hrithikrudra]

Isn't it should be nums[nums[n-1]]*=-1.
Please someone help me.Thanks

[hereweare]

What if the array elements are {12, 13, ...} Like this.. how can we go to the 12th 13th index. Will this solution work for this?

[andreykalmatskiy]

When you flip sign you use one bit in the input data, so it's still O(N) space solution.

[subhambanerjee]

what if we are not allowed to modify the original array?

[PiBiNi]

They have specified that we cannot use exrta space. But result array is extra space right? Please correct me if I'm wrong here

[kajalpareek]

Can we do sorting first and then check if arr[i]!= arr[i-1] ?

[akshitkushwaha]

Here's how Im gonna do it:-

ar = [4, 3, 2, 7, 8, 2, 3, 1]

count = []

for i in range(len(ar)):
if ar[i] not in count:
count.append(ar[i])
else:
print(ar[i], 'is occuring more than once')

[akashgupta]

But u modified the array
And in Ques there is given that you cannot modified the array

[SangelRally]

What tools are you using ?? Black board and Software to write using pen tablet.

[rajeevranjan-tuyv]

Not satisfied with the answer. Came here in search of a simple solution that works for any array but found a complex solution.
Can we do it without using a vector or list, with the simple array?
What if we have {102, 203, -405, 203, 43, 78, 102, 202}. How we will find the duplicate only using simple array concept?

[DhananjaySharma-kbyf]

This is not working with the following input ->
int[] numRay = {17, 27, 11, 23, 14, 29, 17, 24, 3, 6, 18, 8, 18, 16, 29, 11, 24, 5, 0, 1, 28, 3, 28, 4, 13, 7, 7, 27, 10, 21};

Expected solution-> [3, 7, 11, 17, 18, 24, 27, 28, 29]
Actual solution -> [3, 7, 11, 17, 24, 27, 28, 29]


My solution (instead of multiply, using addition)
int[] numRay = {17, 27, 11, 23, 14, 29, 17, 24, 3, 6, 18, 8, 18, 16, 29, 11, 24, 5, 0, 1, 28, 3, 28, 4, 13, 7, 7, 27, 10, 21};
int constant = numRay.length;
Set<Integer> solution = new TreeSet<>();
for (int n : numRay) {
if (n >= constant) {
n = n - constant;
}
if (numRay[n] < constant) {
numRay[n] = numRay[n] + constant;
} else {
solution.add(n);
}
}

[TheMsnitish]

Bhai, naya question ka solution ? Vertical order B Tree. Would like to learn from your example

[jay-rathod-]

Sir this solution was so cool and simplified. I thought of using rabbit and tortoise so went in a wrong direction.

One request: please do explanation and code to check Bipartite.