Terms Of AP
Ayush is given a number ‘X’. He has been told that he has to find the first ‘X’ terms of the series 3 * ‘N’ + 2, which are not multiples of 4. Help Ayush to find it as he has not been able to answer.

Example: Given an ‘X’ = 4. The output array/list which must be passed to Ayush will be [ 5, 11, 14, 17 ].

Input Format:

The first line contains a single integer ‘T’ representing the number of test cases. The first line of each test case will contain one integer, ‘X’ that denotes the number of terms he has to answer.

Output Format:

For each test case, return a vector with the first ‘X’ integer of the series 3 * ‘N’ + 2, which is not multiple of 4. Output for every test case will be printed in a separate line.

Note:

You don’t need to print anything; It has already been taken care of.

Constraints:

1

Question

Terms Of AP

Ayush is given a number ‘X’. He has been told that he has to find the first ‘X’ terms of the series 3 * ‘N’ + 2, which are not multiples of 4. Help Ayush to find it as he has not been able to answer.

Example: Given an ‘X’ = 4. The output array/list which must be passed to Ayush will be [ 5, 11, 14, 17 ].

Input Format:

The first line contains a single integer ‘T’ representing the number of test cases. 

The first line of each test case will contain one integer, ‘X’ that denotes the number of terms he has to answer.

Output Format:

For each test case, return a vector with the first ‘X’ integer of the series 3 * ‘N’ + 2, which is not multiple of 4.

Output for every test case will be printed in a separate line.

Note:

You don’t need to print anything; It has already been taken care of.

Constraints:

1 <= T <= 10^2
1 <= X <= 10^5

Time Limit: 1 sec

Accepted Answer

Approach :
1) Declare a temporary array/list variable ‘ANS’ in which we store our answer.

2) Declare a temporary variable ‘GOT’ that will store the total number of elements we obtained until now, which are acceptable.

3) Declare a temporary variable ‘CURRENT’ that will store the current number of series 3 * ‘N’ + 2 and initialize with the first number of series 5.

4) Run a loop while ‘GOT’ is not equal to ‘X’:
4.1) If ‘CURRENT’ is not divisible by 4, we will append the value at the end of ‘ANS’ and increment the value of ‘GOT’ by 1.
4.2) Increment the value of ‘CURRENT’ by three as the next value of the series.

5) Finally, return ‘ANS’.

TC : O(N), where N is the number of elements we have to find in the series.
SC : O(N)

Accepted Answer

Brute Force

The basic idea of this approach is to iterate through all the elements obtained in series 3 * ‘N’ + 2. We will see if the number is divisible by 4 or not. If it is acceptable, then we will store it in our final answer array/list. We will run the process until we don’t get the first ‘X’ number of elements.

Here is the algorithm:

Declare a temporary array/list variable ‘ANS’ in which we store our answer.
Declare a temporary variable ‘GOT’ that will store the total number of elements we obtained until now, which are acceptable.
Declare a temporary variable ‘CURRENT’ that will store the current number of series 3 * ‘N’ + 2 and initialize with the first number of series 5.
Run a loop while ‘GOT’ is not equal to ‘X’:
- If ‘CURRENT’ is not divisible by 4, we will append the value at the end of ‘ANS’ and increment the value of ‘GOT’ by 1.
- Increment the value of ‘CURRENT’ by three as the next value of the series.
Finally, return ‘ANS’.

Space Complexity: O(n)Explanation:

O(N), where N is the number of elements we have to find in the series.

Because we are using an array/list of size ‘N’to find our answer. Hence the space complexity is O(N).

Time Complexity: O(n)Explanation:

O(N), where N is the number of elements we have to find in the series.

Because we will iterate the loop until the number of elements in the array/list is not equal to ‘X’. The number of elements rejected is approximately one-third of the elements accepted. Thus, the time complexity is O(N).

Python (3.5)

'''

    Time Complexity : O(N)
    Space Complexity : O(N)

    Where N is the number of elements we have to find of the A.P. series.

'''

def termsOfAP(x):
    
    # Declaring a vector to store all the elements which are to be returned.
    ans = [0 for i in range(x)]
    
    # Declaring variable to have a look at number of elements in 'ANS' vector.
    got = 0 
    
    # Pointing to current number of series 3N + 2.
    curr = 5
    
    # Running a loop until total number of elements in 'ANS' is not equal to 'X'.
    while(got != x):
        
        # If curr value of series is not divisible by 4 then we will append it at
        # the end of 'ANS' vector and increment the value of 'GOT' by 1.
        if curr % 4 != 0:
            ans[got] = curr
            got += 1
            
        # Getting the next value of series by incrementing by 3.
        curr += 3
        
    # Finally we will return our 'ANS' vector.
    return ans

Java (SE 1.8)

/*

    Time Complexity : O(N)
    Space Complexity : O(N)

    Where N is the number of elements we have to find of the A.P. series.

*/


public class Solution {

    public static int[] termsOfAP(int x) {
    
        // Declaring a vector to store all the elements which are to be returned.
        int ans[] = new int[x];
        
        // Declaring variable to have a look at number of elements in 'ANS' vector.
        int got = 0;
        
        // Pointing to current number of series 3N + 2.
        int curr = 5;
        
        // Running a loop until total number of elements in 'ANS' is not equal to 'X'.
        while (got != x) {
            
            // If curr value of series is not divisible by 4 then we will append it at
            // the end of 'ANS' vector and increment the value of 'GOT' by 1.
            if (curr % 4 != 0) {
                ans[got] = curr;
                got++;
            }
            
            // Getting the next value of series by incrementing by 3.
            curr += 3;
        }
        
        // Finally we will return our 'ANS' vector.
        return ans;
    }
}

C++ (g++ 5.4)

/*

    Time Complexity : O(N)
    Space Complexity : O(N)

    Where N is the number of elements we have to find of the A.P. series.

*/

#include 

vector < int > termsOfAP(int x) {
    
    // Declaring a vector to store all the elements which are to be returned.
    vector < int > ans;
    
    // Declaring variable to have a look at number of elements in 'ANS' vector.
    int got = 0;
    
    // Pointing to current number of series 3N + 2.
    int curr = 5;
    
    // Running a loop until total number of elements in 'ANS' is not equal to 'X'.
    while (got != x) {
        
        // If curr value of series is not divisible by 4 then we will append it at
        // the end of 'ANS' vector and increment the value of 'GOT' by 1.
        if (curr % 4 != 0) {
            ans.push_back(curr);
            got++;
        }
        
        // Getting the next value of series by incrementing by 3.
        curr += 3;
    }
    
    // Finally we will return our 'ANS' vector.
    return ans;
}

Ayush is given a number ‘X’. He has been told that he has to find the first ‘X’ terms of the series 3 * ‘N’ + 2, which are not multiples of 4. Help Ayush to find it as he has not been able to answer.

Example: Given an ‘X’ = 4. The output array/list which must be passed to Ayush will be [ 5, 11, 14, 17 ].

Input Format:

Output Format:

Note:

Constraints:

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