Buses

Moderate

0/80

Average time to solve is 15m

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Problem statement

You are given a vector of 'N' integers denoting the number of buses that can be boarded from the i-th position. The bus stops only at stops whose number is a multiple of the bus stop number from which the bus originates. You need to find the number of buses originating from each bus stop from 1 to 'N'.

For example:

If 'N' = 4 and the given vector is: [1 3 4 3].

1 bus can be boarded from the first bus stop which means that 1 bus originates from the first bus stop.

3 buses can be boarded from the second bus stop which means that (3 - 1 = 2) buses originate from the second bus stop. This is because the bus originating from the first stop will stop at the second stop as well.

4 buses can be boarded from the third bus stop which means that (4-1 = 3) buses originate from the third bus stop. This is because the bus originating from the first stop will stop at the third stop as well.

3 buses can be boarded from the fourth bus stop which means that (3-3 = 0) buses originate from the fourth bus stop. This is because the buses originating from the first and second stop will stop at the fourth stop as well.

So the final vector would be: [1 2 3 0].

Note:

The given vector uses 1-based indexing.

Detailed explanation ( Input/output format, Notes, Images )

Input Format:

The first line of input contains a single integer 'T', representing the number of test cases or queries to be run. 
Then the 'T' test cases follow.

The first line of each test case contains a single integer 'N' representing the length of the vector.

The second line of each test case contains 'N' space-separated integers denoting the elements of the given vector.

Output Format:

For each test case, print 'N' integers denoting the number of buses originating from each bus stop from 1 to 'N'.

Note:

You are not required to print the expected output, it has already been taken care of. Just implement the function.

Constraints:

1 ≤ T ≤ 50
1 ≤ N ≤ 10^4
1 ≤ Ai ≤ 10^6

Time Limit : 1 sec

Sample Input 1:

Sample Output 1:

1 2 3 0 
1 1 2 2 4 
4 0 0

Explanation For Sample Input 1:

For the first test case, the explanation is already given.

For the second test case, the given vector is: [1 2 3 4 5].
1 bus can be boarded from the first bus stop which means that 1 bus originates from the first bus stop.
2 buses can be boarded from the second bus stop which means that (2-1 = 1) buses originate from the second bus stop. This is because the bus originating from the first stop will stop at the second stop as well.
3 buses can be boarded from the third bus stop which means that (3-1 = 2) buses originate from the third bus stop. This is because the bus originating from the first stop will stop at the third stop as well.
4 buses can be boarded from the fourth bus stop which means that (4-1-1 = 1) buses originate from the fourth bus stop. This is because the buses originating from the first and second stop will stop at the fourth stop as well.
5 buses can be boarded from the fifth bus stop which means that (5-1 = 4) buses originate from the fifth bus stop. This is because the bus originating from the first stop will stop at the fifth stop as well.
So the final vector would be: [1 1 2 2 4]

For the third test case, the given vector is: [4 4 4].
4 buses can be boarded from the first bus stop which means that 4 buses originate from the first bus stop.
4 buses can be boarded from the second bus stop which means that (4-4 = 0) buses originate from the second bus stop. This is because the bus originating from the first stop will stop at the second stop as well.
4 buses can be boarded from the third bus stop which means that (4-4 = 0) buses originate from the third bus stop. This is because the bus originating from the first stop will stop at the third stop as well.
So the final vector would be: [4 0 0]

Sample Input 2:

Sample Output 2

5 
5 1 
1 2 5 5

Hint 1

Hint 2

Try to iterate on all the number of array.

Approaches (2)

Approach 1

Approach 2

Brute Force

Use two nested loops for finding the answer.
Use the outer loop for finding the final solution and the inner loop for subtracting the buses which do not start from that place.
For each bus stop in the inner loop which is a divisor of the outer loop, subtract it from the number of buses in the inner loop.
Go on doing this, till we traverse the whole array/list.

Time Complexity

O(N^2) Where N is the size of array/list.

As we are iterating on all the elements of an array/list using two nested loops. Hence the time complexity is O(N^2).

Space Complexity

O(1)

As there is no extra space required.

(100% EXP penalty)

Buses

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