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34,071 Views • Oct 30, 2024 • Click to toggle off description
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Uploaded At Oct 30, 2024 ^^
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RYD date created : 2024-11-23T21:26:02.40555Z
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25 Comments
Top Comments of this video!! :3
Remember, with enough practice you can fit a hashmap anywhere kids
39 |
Answer should be 4 but by your way it will be 3, actually we need to calculate n choose 2 or n(n-1)/2, where n is the number of 1's or let say m's in the given array where you are just subtracting 1 from each n.
So the edit would be, inside of the for loop remove everything and add the following
k = counter[num]
good_pairs += (k*(k-1))//2
This would give the correct result.
Love your content by the way ❤️
67 |
it is spread to every number to C(n,k) and sum it.
3 |
you should not subtract 1, you should check number of compositions😊
1 |
In javascript create an object and do nums.foreach to add stuff to it o[num] = (o[num] || 0) + 1
and then return Object.values(o).reduce acc + num--*num/2 with initial 0
|
For toy examples there's not much need to optimise, but big arrays will be able to use the vectorisation built into pandas and numpy.
In fact, it's basically a one-liner in numpy:
def goodpairs(arr: list) -> int:
return np.sum(np.vectorize(npairs)(np.bincount(np.array(arr))))
with npairs as:
def npairs(n: int) -> int:
return (n * n - n) // 2
1 |
Explanation of the code and theory behind it
If a number occurs 4 times then good pairs are - 6 (3+2+1)
If a number occurrences 6 times its (5 + 4.... +1)
So we can make a counter to get the frequencies and to calculate 5+4+3+2+1 when occurrences = 6 (we use the formula sum of an arithmetic progression
as the required n + (n-d) + (n-2d) is an arithmetic progression)
(n * (2a + (n-1)*d))/2
Where n = total elements in the series (n = 5 when we have 5+4+3+2+1)
a = starting element (in our case 5)
d = difference between 2 elements (in our case -1 because 4-5 = -1 and also 3-4 = -1)
Then we use this formula and apply it to all occurrences in the counter hashmap and return the sum
9 |
Nice how he ends early to skip the ending of his explanation, where he realizes that he's WRONG, and rather than NOT POST A WRONG VIDEO, he just chops off the ending. If you follow out his logic, subtracting 1 from each entry and adding them up, then you get an answer of 3, which is clearly wrong.
|
Smart
1 |
Get all the counts, do n choose 2 on each, sum them up. n choose 2 = n(n-1)/2, so we can factor out the 2 and do that at the end.
```python
def numIdenticalPairs(self, nums: List[int]) -> int:
return sum(
count * (count - 1)
for count in Counter(nums).values()
) // 2
```
3 |
It doesn't make sense to make these videos with errors when you have everything on leetcode. Waste of time
1 |
let me know what do u think about my solution:
def numIdenticalPairs(self, nums: List[int]) -> int:
D = {}
D_ = {}
count =0
for i in range(len(nums)):
if nums[i] not in D:
D[nums[i]] = i
D_[nums[i]] = 0
else:
if D[nums[i]] <i:
count += 1+D_[nums[i]]
D_[nums[i]] +=1
return count
|
@GregHogg
3 weeks ago
Master data structures and algorithms for FREE at algomap.io/ :)
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