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I don't particularly know what test case 8 is exactly, but I would assume that this test case has set the input parameters of n and d to be very large. If you look at the constraints on n, then you will see that n can be as large as 10,000 and d also can be as large as n. With an approach that manually shifts the array one by one a certain amount of times can lead to a lot of work. Possibly shifting all 10,000 elements to the left 10,000 times racks up to a lot of work. You would have to shift something up to 10,000,000,000 times in this scenario. That is a lot of work and would most likely be the reason the test case times out before finishing.
Its easy man i cant figure out the last test case though !!
1.Input an array of size n
2.Input number of rotations as r
3. r=r%n -> it reduces the number of repeatative rotations thus decreasing the complexity of our code . suppose the size of the array is 3 and number of rotations is also 3 then our loop wont run it will simply print the same elements .
take a temperory variable and store the leftmost element in it i.e a[o]
Now run the loop from 0 to second last element of the array
6.a[i]=a[i+1] -> shifts the element towards left for eg. for i=0 a[0]=a[1]; it shifts all the elements towards left
7.now simply store the first element in temp variable to the last position.
steps 4 to 7 for first rotation now while loop will run as per the given number of rotations these steps will be repeated until r becomes 0.
Keep in mind a left rotation of x is equal to (len-x) of righ rotations. Hence, it would be more efficient to the rotation with least amount of "moves".
Left Rotation
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int main() {
} test case#8 is showing "terminated due to timeout"! Any suggestions? i am new to this!
I have the exactly same problem. Did anyone figure out why?
I don't particularly know what test case 8 is exactly, but I would assume that this test case has set the input parameters of n and d to be very large. If you look at the constraints on n, then you will see that n can be as large as 10,000 and d also can be as large as n. With an approach that manually shifts the array one by one a certain amount of times can lead to a lot of work. Possibly shifting all 10,000 elements to the left 10,000 times racks up to a lot of work. You would have to shift something up to 10,000,000,000 times in this scenario. That is a lot of work and would most likely be the reason the test case times out before finishing.
I did solve it! It was proably beacuse of the nested loops! they take more time to run!
which is known as complexity
how did you solve it ?
Its easy man i cant figure out the last test case though !!
1.Input an array of size n 2.Input number of rotations as r 3. r=r%n -> it reduces the number of repeatative rotations thus decreasing the complexity of our code . suppose the size of the array is 3 and number of rotations is also 3 then our loop wont run it will simply print the same elements .
:)
Man i have better version of solution. See my code.
vector array_left_rotation(vector a, int n, int k) { rotate(a.begin(),a.begin()+k,a.end()); return a; }
int main(){ int n; int k; cin >> n >> k; vector a(n); for(int a_i = 0;a_i < n;a_i++){ cin >> a[a_i]; } vector output = array_left_rotation(a, n, k); for(int i = 0; i < n;i++) cout << output[i] << " "; cout << endl; return 0; }
chup.. space dedeta kuch to
Doesn't work for 8th test case
how did u solve??
Keep in mind a left rotation of x is equal to (len-x) of righ rotations. Hence, it would be more efficient to the rotation with least amount of "moves".
Thats true. I got accepted for test case 8 using right rotation. We need to choose :
If #rotations < (size /2) Use Left Rotation If #rotations > (size /2) Use Right Rotation
can you elaborate @crystalcode
just type,
for i=0 to n-1
print a [ ( i + d ) %n ].
thats all..