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- Jun 03 2015 02:30
adeydas on master

Contains duplicates 2 3 sum and 4 sum (compare)

- May 28 2015 20:35
adeydas on master

Isomorphic Strings (compare)

- May 27 2015 22:02
adeydas on master

Calculate 7n/8 without using mu… (compare)

- May 27 2015 21:47
adeydas on master

Find largest pair sum (compare)

- May 27 2015 20:47
adeydas on master

Trie (compare)

- May 27 2015 20:24
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Trie (compare)

- May 27 2015 19:57
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Trie insert and search (compare)

- May 26 2015 20:31
adeydas on master

Divide Two numbers (compare)

- May 26 2015 20:07
adeydas on master

Search for a range (compare)

- May 26 2015 19:52
adeydas on master

Max subarray - Kadane algorithm (compare)

- May 26 2015 17:50
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Sort colors (compare)

- May 26 2015 17:38
adeydas on master

Excel sheet col title (compare)

- May 26 2015 16:29
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Largest number (compare)

- May 26 2015 06:58
adeydas on master

Updated README (compare)

- May 26 2015 06:57
adeydas on master

Min size subarr sum - O(n) solu… (compare)

- May 26 2015 03:05
adeydas on master

Kth largest element in an array (compare)

- May 26 2015 03:00
adeydas on master

Contains duplicates (compare)

- May 26 2015 02:56
adeydas on master

ATOI (compare)

- May 25 2015 21:09
adeydas on master

DT code (compare)

- May 21 2015 21:43
adeydas on gh-pages

Update styles.css (compare)

Hey guys can someone help me out ? For the below algorithm can you tell me why the Big O runtime is O($n^2$) ? I'm think that the last block where there is three nested for loops would make it O($n^3$)

```
n = 1000
for c from 1 to n
for d from 1 to n
result $$c^3$$ + $$d^3$$
append (c, d) to list at value map[result]
for each result, list in map
for each pair1 in list
for each pair2 in list
print pair1, pair2
```

hey guys, I made a similar practice repository that might help you out :) https://github.com/AlJohri/practice#practice

Can someone explain the time complexity of this function ?

int fun(int n)

{

int count = 0;

for (int i = 0; i < n; i++)

for (int j = i; j > 0; j--)

count = count + 1;

return count;

}

{

int count = 0;

for (int i = 0; i < n; i++)

for (int j = i; j > 0; j--)

count = count + 1;

return count;

}