How to print maximum number of A’s using given four keys

This is a famous interview question asked in Google, Paytm and many other company interviews. 
Below is the problem statement.

Imagine you have a special keyboard with the following keys: 
Key 1:  Prints 'A' on screen
Key 2: (Ctrl-A): Select screen
Key 3: (Ctrl-C): Copy selection to buffer
Key 4: (Ctrl-V): Print buffer on screen appending it
                 after what has already been printed. 

If you can only press the keyboard for N times (with the above four
keys), write a program to produce maximum numbers of A's. That is to
say, the input parameter is N (No. of keys that you can press), the 
output is M (No. of As that you can produce).

Examples:

Input:  N = 3
Output: 3
We can at most get 3 A's on screen by pressing 
following key sequence.
A, A, A

Input:  N = 7
Output: 9
We can at most get 9 A's on screen by pressing 
following key sequence.
A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V

Input:  N = 11
Output: 27
We can at most get 27 A's on screen by pressing 
following key sequence.
A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V, Ctrl A, 
Ctrl C, Ctrl V, Ctrl V

Below are few important points to note.
a) For N < 7, the output is N itself. 
b) Ctrl V can be used multiple times to print current buffer (See last two examples above). The idea is to compute the optimal string length for N keystrokes by using a simple insight. The sequence of N keystrokes which produces an optimal string length will end with a suffix of Ctrl-A, a Ctrl-C, followed by only Ctrl-V’s . (For N > 6)
The task is to find out the break=point after which we get the above suffix of keystrokes. Definition of a breakpoint is that instance after which we need to only press Ctrl-A, Ctrl-C once and the only Ctrl-V’s afterward to generate the optimal length. If we loop from N-3 to 1 and choose each of these values for the break-point, and compute that optimal string they would produce. Once the loop ends, we will have the maximum of the optimal lengths for various breakpoints, thereby giving us the optimal length for N keystrokes.

Below is implementation based on above idea.  

C++14

/* A recursive C++ program to print maximum number of A's using
following four keys */
#include <bits/stdc++.h>
using namespace std;

// A recursive function that returns the optimal length string
// for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;

    // Initialize result
    int max = 0;

    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to loop from N-3 keystrokes
    // back to 1 keystroke to find a breakpoint 'b' after which we
    // will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way.
    int b;
    for (b = N - 3; b >= 1; b--) {

        // If the breakpoint is s at b'th keystroke then
        // the optimal string would have length
        // (n-b-1)*screen[b-1];
        int curr = (N - b - 1) * findoptimal(b);
        if (curr > max)
            max = curr;
    }
    return max;
}

// Driver program
int main()
{
    int N;

    // for the rest of the array we will reply on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        cout << "Maximum Number of A's with " << N <<
        " keystrokes is " << findoptimal(N) << endl;
}

// This code is contributed by shubhamsingh10

C

/* A recursive C program to print maximum number of A's using
   following four keys */
#include <stdio.h>

// A recursive function that returns the optimal length string
// for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;

    // Initialize result
    int max = 0;

    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to loop from N-3 keystrokes
    // back to 1 keystroke to find a breakpoint 'b' after which we
    // will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way.
    int b;
    for (b = N - 3; b >= 1; b--) {
        // If the breakpoint is s at b'th keystroke then
        // the optimal string would have length
        // (n-b-1)*screen[b-1];
        int curr = (N - b - 1) * findoptimal(b);
        if (curr > max)
            max = curr;
    }
    return max;
}

// Driver program
int main()
{
    int N;

    // for the rest of the array we will reply on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        printf("Maximum Number of A's with %d keystrokes is %d\n",
               N, findoptimal(N));
}

Java

/* A recursive Java program to print
  maximum number of A's using
  following four keys */
import java.io.*;

class GFG {

    // A recursive function that returns
    // the optimal length string  for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length is N
        // when N is smaller than 7
        if (N <= 6)
            return N;

        // Initialize result
        int max = 0;

        // TRY ALL POSSIBLE BREAK-POINTS
        // For any keystroke N, we need to
        // loop from N-3 keystrokes back to
        // 1 keystroke to find a breakpoint
        // 'b' after which we will have Ctrl-A,
        // Ctrl-C and then only Ctrl-V all the way.
        int b;
        for (b = N - 3; b >= 1; b--) {
            // If the breakpoint is s at b'th
            // keystroke then the optimal string
            // would have length
            // (n-b-1)*screen[b-1];
            int curr = (N - b - 1) * findoptimal(b);
            if (curr > max)
                max = curr;
        }
        return max;
    }

    // Driver program
    public static void main(String[] args)
    {
        int N;

        // for the rest of the array we
        // will reply on the previous
        // entries to compute new ones
        for (N = 1; N <= 20; N++)
            System.out.println("Maximum Number of A's with keystrokes is " + N + findoptimal(N));
    }
}

// This code is contributed by vt_m.

Python3

# A recursive Python3 program to print maximum
# number of A's using following four keys

# A recursive function that returns
# the optimal length string for N keystrokes

def findoptimal(N):

    # The optimal string length is
    # N when N is smaller than
    if N<= 6:
        return N

    # Initialize result
    maxi = 0

    # TRY ALL POSSIBLE BREAK-POINTS
    # For any keystroke N, we need
    # to loop from N-3 keystrokes
    # back to 1 keystroke to find
    # a breakpoint 'b' after which we
    # will have Ctrl-A, Ctrl-C and then
    # only Ctrl-V all the way.
    for b in range(N-3, 0, -1):
        curr =(N-b-1)*findoptimal(b)
        if curr>maxi:
            maxi = curr

    return maxi
# Driver program
if __name__=='__main__':


# for the rest of the array we will
# reply on the previous
# entries to compute new ones
    for n in range(1, 21):
        print('Maximum Number of As with ', n, 'keystrokes is ', findoptimal(n))

# this code is contributed by sahilshelangia

C#

/* A recursive C# program
to print maximum number
of A's using following
four keys */
using System;

class GFG {
    // A recursive function that
    // returns the optimal length
    // string for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length
        // is N when N is smaller than 7
        if (N <= 6)
            return N;

        // Initialize result
        int max = 0;

        // TRY ALL POSSIBLE BREAK-POINTS
        // For any keystroke N, we need
        // to loop from N-3 keystrokes
        // back to 1 keystroke to find
        // a breakpoint 'b' after which
        // we will have Ctrl-A, Ctrl-C
        // and then only Ctrl-V all the way.
        int b;
        for (b = N - 3; b >= 1; b--) {
            // If the breakpoint is s
            // at b'th keystroke then
            // the optimal string would
            // have length (n-b-1)*screen[b-1];
            int curr = (N - b - 1) * findoptimal(b);
            if (curr > max)
                max = curr;
        }
        return max;
    }

    // Driver code
    static void Main()
    {
        int N;

        // for the rest of the array
        // we will reply on the
        // previous entries to compute
        // new ones
        for (N = 1; N <= 20; N++)
            Console.WriteLine("Maximum Number of A's with " + N + " keystrokes is " + findoptimal(N));
    }
}

// This code is contributed by Sam007

PHP

<?php
/* A recursive PHP program to print maximum number of A's using
following four keys */

// A recursive function that returns the optimal
//  length string for N keystrokes
function findoptimal($N)
{
    // The optimal string length is
    // N when N is smaller than 7
    if ($N <= 6)
        return $N;

    // Initialize result
    $max = 0;

    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to loop from N-3 keystrokes
    // back to 1 keystroke to find a breakpoint 'b' after which we
    // will have Ctrl-A, Ctrl-C and then only Ctrl-V all the way.
    $b;
    for ($b = $N - 3; $b >= 1; $b -= 1)
    {
            // If the breakpoint is s at b'th keystroke then
            // the optimal string would have length
            // (n-b-1)*screen[b-1];
            $curr = ($N - $b - 1) * findoptimal($b);
            if ($curr > $max)
                $max = $curr;
    }
    return $max;
}

// Driver code
$N;

// for the rest of the array we will reply on the previous
// entries to compute new ones
for ($N = 1; $N <= 20; $N += 1)
    echo("Maximum Number of A's with"
        .$N."keystrokes is ".findoptimal($N)."\n");

// This code is contributed by aman neekhara
?>

Javascript

<script>

// A recursive Javascript program to print
// maximum number of A's using
// following four keys

// A recursive function that returns the
// optimal length string  for N keystrokes
function findoptimal(N)
{

    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;

    // Initialize result
    let max = 0;

    // TRY ALL POSSIBLE BREAK-POINTS
    // For any keystroke N, we need to
    // loop from N-3 keystrokes back to
    // 1 keystroke to find a breakpoint
    // 'b' after which we will have Ctrl-A,
    // Ctrl-C and then only Ctrl-V all the way.
    let b;

    for(b = N - 3; b >= 1; b--)
    {

        // If the breakpoint is s at b'th
        // keystroke then the optimal string
        // would have length
        // (n-b-1)*screen[b-1];
        let curr = (N - b - 1) * findoptimal(b);

        if (curr > max)
            max = curr;
    }
    return max;
}

// Driver code
let N;

// For the rest of the array we
// will reply on the previous
// entries to compute new ones
for(N = 1; N <= 20; N++)
    document.write("Maximum Number of A's with "+
                   N + " keystrokes is " +
                   findoptimal(N) + "<br>");

// This code is contributed by avanitrachhadiya2155

</script>

Output:

Maximum Number of A's with 1 keystrokes is 1
Maximum Number of A's with 2 keystrokes is 2
Maximum Number of A's with 3 keystrokes is 3
Maximum Number of A's with 4 keystrokes is 4
Maximum Number of A's with 5 keystrokes is 5
Maximum Number of A's with 6 keystrokes is 6
Maximum Number of A's with 7 keystrokes is 9
Maximum Number of A's with 8 keystrokes is 12
Maximum Number of A's with 9 keystrokes is 16
Maximum Number of A's with 10 keystrokes is 20
Maximum Number of A's with 11 keystrokes is 27
Maximum Number of A's with 12 keystrokes is 36
Maximum Number of A's with 13 keystrokes is 48
Maximum Number of A's with 14 keystrokes is 64
Maximum Number of A's with 15 keystrokes is 81
Maximum Number of A's with 16 keystrokes is 108
Maximum Number of A's with 17 keystrokes is 144
Maximum Number of A's with 18 keystrokes is 192
Maximum Number of A's with 19 keystrokes is 256
Maximum Number of A's with 20 keystrokes is 324

The above function computes the same subproblems again and again. Recomputations of same subproblems can be avoided by storing the solutions to subproblems and solving problems in a bottom-up manner. 

Below is Dynamic Programming based C implementation where an auxiliary array screen[N] is used to store result of subproblems. 

C++

/* A Dynamic Programming based C program to find maximum number of A's
that can be printed using four keys */
#include <iostream>
using namespace std;


// this function returns the optimal length string for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;

    // An array to store result of subproblems
    int screen[N];

    int b; // To pick a breakpoint

    // Initializing the optimal lengths array for until 6 input
    // strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;

    // Solve all subproblems in bottom manner
    for (n = 7; n <= N; n++) {
        // Initialize length of optimal string for n keystrokes
        screen[n - 1] = 0;

        // For any keystroke n, we need to loop from n-3 keystrokes
        // back to 1 keystroke to find a breakpoint 'b' after which we
        // will have ctrl-a, ctrl-c and then only ctrl-v all the way.
        for (b = n - 3; b >= 1; b--) {
            // if the breakpoint is at b'th keystroke then
            // the optimal string would have length
            // (n-b-1)*screen[b-1];
            int curr = (n - b - 1) * screen[b - 1];
            if (curr > screen[n - 1])
                screen[n - 1] = curr;
        }
    }

    return screen[N - 1];
}

// Driver program
int main()
{
    int N;

    // for the rest of the array we will reply on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        cout << "Maximum Number of A's with "<<N<<" keystrokes is "
        << findoptimal(N) << endl;
}

//This is contributed by shubhamsingh10

C

/* A Dynamic Programming based C program to find maximum number of A's
   that can be printed using four keys */
#include <stdio.h>

// this function returns the optimal length string for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;

    // An array to store result of subproblems
    int screen[N];

    int b; // To pick a breakpoint

    // Initializing the optimal lengths array for until 6 input
    // strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;

    // Solve all subproblems in bottom manner
    for (n = 7; n <= N; n++) {
        // Initialize length of optimal string for n keystrokes
        screen[n - 1] = 0;

        // For any keystroke n, we need to loop from n-3 keystrokes
        // back to 1 keystroke to find a breakpoint 'b' after which we
        // will have ctrl-a, ctrl-c and then only ctrl-v all the way.
        for (b = n - 3; b >= 1; b--) {
            // if the breakpoint is at b'th keystroke then
            // the optimal string would have length
            // (n-b-1)*screen[b-1];
            int curr = (n - b - 1) * screen[b - 1];
            if (curr > screen[n - 1])
                screen[n - 1] = curr;
        }
    }

    return screen[N - 1];
}

// Driver program
int main()
{
    int N;

    // for the rest of the array we will reply on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        printf("Maximum Number of A's with %d keystrokes is %d\n",
               N, findoptimal(N));
}

Java

/* A Dynamic Programming based C
   program to find maximum number
   of A's that can be printed using
   four keys */
import java.io.*;

class GFG {

    // this function returns the optimal
    // length string for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length is N
        // when N is smaller than 7
        if (N <= 6)
            return N;

        // An array to store result
        // of subproblems
        int screen[] = new int[N];

        int b; // To pick a breakpoint

        // Initializing the optimal lengths
        // array for until 6 input strokes
        int n;
        for (n = 1; n <= 6; n++)
            screen[n - 1] = n;

        // Solve all subproblems in bottom manner
        for (n = 7; n <= N; n++) {
            // Initialize length of optimal
            // string for n keystrokes
            screen[n - 1] = 0;

            // For any keystroke n, we need
            // to loop from n-3 keystrokes
            // back to 1 keystroke to find
            // a breakpoint 'b' after which we
            // will have ctrl-a, ctrl-c and
            // then only ctrl-v all the way.
            for (b = n - 3; b >= 1; b--) {
                // if the breakpoint is
                // at b'th keystroke then
                // the optimal string would
                // have length
                // (n-b-1)*screen[b-1];
                int curr = (n - b - 1) * screen[b - 1];
                if (curr > screen[n - 1])
                    screen[n - 1] = curr;
            }
        }

        return screen[N - 1];
    }

    // Driver program
    public static void main(String[] args)
    {
        int N;

        // for the rest of the array we will reply on the previous
        // entries to compute new ones
        for (N = 1; N <= 20; N++)
            System.out.println("Maximum Number of A's with keystrokes is " + N + findoptimal(N));
    }
}

//

Python3

# A Dynamic Programming based Python program
# to find maximum number of A's
# that can be printed using four keys

# this function returns the optimal
# length string for N keystrokes
def findoptimal(N):

    # The optimal string length is
    # N when N is smaller than 7
    if (N <= 6):
        return N

    # An array to store result of
    # subproblems
    screen = [0]*N

    # Initializing the optimal lengths
    # array for until 6 input
    # strokes.

    for n in range(1, 7):
        screen[n-1] = n

    # Solve all subproblems in bottom manner
    for n in range(7, N + 1):

        # Initialize length of optimal
        # string for n keystrokes
        screen[n-1] = 0

        # For any keystroke n, we need to
        # loop from n-3 keystrokes
        # back to 1 keystroke to find a breakpoint
        # 'b' after which we
        # will have ctrl-a, ctrl-c and then only
        # ctrl-v all the way.
        for b in range(n-3, 0, -1):

            # if the breakpoint is at b'th keystroke then
            # the optimal string would have length
            # (n-b-1)*screen[b-1];
            curr = (n-b-1)*screen[b-1]
            if (curr > screen[n-1]):
                screen[n-1] = curr

    return screen[N-1]

# Driver program
if __name__ == "__main__":

    # for the rest of the array we
    # will reply on the previous
    # entries to compute new ones
    for N in range(1, 21):
        print("Maximum Number of A's with ", N, " keystrokes is ",
                findoptimal(N))

# this code is contributed by
# ChitraNayal

C#

/* A Dynamic Programming based C#
program to find maximum number
of A's that can be printed using
four keys */
using System;
public class GFG {

    // this function returns the optimal
    // length string for N keystrokes
    static int findoptimal(int N)
    {
        // The optimal string length is N
        // when N is smaller than 7
        if (N <= 6)
            return N;

        // An array to store result
        // of subproblems
        int[] screen = new int[N];

        int b; // To pick a breakpoint

        // Initializing the optimal lengths
        // array for until 6 input strokes
        int n;
        for (n = 1; n <= 6; n++)
            screen[n - 1] = n;

        // Solve all subproblems in bottom manner
        for (n = 7; n <= N; n++) {
            // Initialize length of optimal
            // string for n keystrokes
            screen[n - 1] = 0;

            // For any keystroke n, we need
            // to loop from n-3 keystrokes
            // back to 1 keystroke to find
            // a breakpoint 'b' after which we
            // will have ctrl-a, ctrl-c and
            // then only ctrl-v all the way.
            for (b = n - 3; b >= 1; b--) {
                // if the breakpoint is
                // at b'th keystroke then
                // the optimal string would
                // have length
                // (n-b-1)*screen[b-1];
                int curr = (n - b - 1) * screen[b - 1];
                if (curr > screen[n - 1])
                    screen[n - 1] = curr;
            }
        }

        return screen[N - 1];
    }

    // Driver program
    public static void Main(String[] args)
    {
        int N;

        // for the rest of the array we will reply on the previous
        // entries to compute new ones
        for (N = 1; N <= 20; N++)
            Console.WriteLine("Maximum Number of A's with {0} keystrokes is {1}\n",
                              N, findoptimal(N));
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// A Dynamic Programming based javascript
// program to find maximum number
// of A's that can be printed using
// four keys

// This function returns the optimal
// length string for N keystrokes
function findoptimal(N)
{

    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;

    // An array to store result
    // of subproblems
    let screen = new Array(N);
    for(let i = 0; i < N; i++)
    {
        screen[i] = 0;
    }

    // To pick a breakpoint
    let b;

    // Initializing the optimal lengths
    // array for until 6 input strokes
    let n;
    for(n = 1; n <= 6; n++)
        screen[n - 1] = n;

    // Solve all subproblems in bottom manner
    for(n = 7; n <= N; n++)
    {

        // Initialize length of optimal
        // string for n keystrokes
        screen[n - 1] = 0;

        // For any keystroke n, we need
        // to loop from n-3 keystrokes
        // back to 1 keystroke to find
        // a breakpoint 'b' after which we
        // will have ctrl-a, ctrl-c and
        // then only ctrl-v all the way.
        for(b = n - 3; b >= 1; b--)
        {

            // If the breakpoint is
            // at b'th keystroke then
            // the optimal string would
            // have length
            // (n-b-1)*screen[b-1];
            let curr = (n - b - 1) * screen[b - 1];

            if (curr > screen[n - 1])
                screen[n - 1] = curr;
        }
    }
    return screen[N - 1];
}

// Driver code
let N;
for(N = 1; N <= 20; N++)
    document.write("Maximum Number of A's with " +
                   N + " keystrokes is " +
                   findoptimal(N) + "<br>");

// This code is contributed by rag2127

</script>

Output: 

Maximum Number of A's with 1 keystrokes is 1
Maximum Number of A's with 2 keystrokes is 2
Maximum Number of A's with 3 keystrokes is 3
Maximum Number of A's with 4 keystrokes is 4
Maximum Number of A's with 5 keystrokes is 5
Maximum Number of A's with 6 keystrokes is 6
Maximum Number of A's with 7 keystrokes is 9
Maximum Number of A's with 8 keystrokes is 12
Maximum Number of A's with 9 keystrokes is 16
Maximum Number of A's with 10 keystrokes is 20
Maximum Number of A's with 11 keystrokes is 27
Maximum Number of A's with 12 keystrokes is 36
Maximum Number of A's with 13 keystrokes is 48
Maximum Number of A's with 14 keystrokes is 64
Maximum Number of A's with 15 keystrokes is 81
Maximum Number of A's with 16 keystrokes is 108
Maximum Number of A's with 17 keystrokes is 144
Maximum Number of A's with 18 keystrokes is 192
Maximum Number of A's with 19 keystrokes is 256
Maximum Number of A's with 20 keystrokes is 324

Thanks to Gaurav Saxena for providing the above approach to solve this problem.
As the number of A’s become large, the effect of pressing Ctrl-V more than 3 times starts to become insubstantial as compared to just pressing Ctrl-A, Ctrl-C and Ctrl-V again. So, the above code can be made more efficient by checking the effect of pressing Ctrl-V for 1, 2, and 3 times only.

C++

// A Dynamic Programming based C++ program to find maximum
// number of A's that can be printed using four keys

#include <bits/stdc++.h>
using namespace std;

// This function returns the optimal length string for N keystrokes
int findoptimal(int N)
{
    // The optimal string length is N when N is smaller than 7
    if (N <= 6)
        return N;

    // An array to store result of subproblems
    int screen[N];

    int b; // To pick a breakpoint

    // Initializing the optimal lengths array for until 6 input
    // strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;

    // Solve all subproblems in bottom-up manner
    for (n = 7; n <= N; n++) {

        // for any keystroke n, we will need to choose between:-
        // 1. pressing Ctrl-V once after copying the
        // A's obtained by n-3 keystrokes.

        // 2. pressing Ctrl-V twice after copying the A's
        // obtained by n-4 keystrokes.

        // 3. pressing Ctrl-V thrice after copying the A's
        // obtained by n-5 keystrokes.
        screen[n - 1] = max(2 * screen[n - 4],
                            max(3 * screen[n - 5],
4 * screen[n - 6]));
    }

    return screen[N - 1];
}

// Driver program
int main()
{
    int N;

    // for the rest of the array we will reply on the previous
    // entries to compute new ones
    for (N = 1; N <= 20; N++)
        printf("Maximum Number of A's with %d keystrokes is %d\n",
               N, findoptimal(N));
}

Java

// A Dynamic Programming based Java program
// to find maximum number of A's that
// can be printed using four keys
class GFG
{

// This function returns the optimal length
// string for N keystrokes
static int findoptimal(int N)
{
    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;

    // An array to store result of subproblems
    int []screen = new int[N];

    int b; // To pick a breakpoint

    // Initializing the optimal lengths array
    // for until 6 input strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;

    // Solve all subproblems in bottom-up manner
    for (n = 7; n <= N; n++)
    {

        // for any keystroke n, we will need to choose between:-
        // 1. pressing Ctrl-V once after copying the
        // A's obtained by n-3 keystrokes.

        // 2. pressing Ctrl-V twice after copying the A's
        // obtained by n-4 keystrokes.

        // 3. pressing Ctrl-V thrice after copying the A's
        // obtained by n-5 keystrokes.
        screen[n - 1] = Math.max(2 * screen[n - 4],
                        Math.max(3 * screen[n - 5],
                                 4 * screen[n - 6]));
    }
    return screen[N - 1];
}

// Driver Code
public static void main(String[] args)
{
    int N;

    // for the rest of the array we will reply
    // on the previous entries to compute new ones
    for (N = 1; N <= 20; N++)
        System.out.printf("Maximum Number of A's with" +
                               " %d keystrokes is %d\n",
                                     N, findoptimal(N));
    }
}

// This code is contributed by Princi Singh

Python3

# A Dynamic Programming based Python3 program
# to find maximum number of A's
# that can be printed using four keys

# this function returns the optimal
# length string for N keystrokes
def findoptimal(N):

    # The optimal string length is
    # N when N is smaller than 7
    if (N <= 6):
        return N

    # An array to store result of
    # subproblems
    screen = [0] * N

    # Initializing the optimal lengths
    # array for until 6 input
    # strokes.

    for n in range(1, 7):
        screen[n - 1] = n

    # Solve all subproblems in bottom manner
    for n in range(7, N + 1):

        # for any keystroke n, we will need to choose between:-
        # 1. pressing Ctrl-V once after copying the
        # A's obtained by n-3 keystrokes.

        # 2. pressing Ctrl-V twice after copying the A's
        # obtained by n-4 keystrokes.

        # 3. pressing Ctrl-V thrice after copying the A's
        # obtained by n-5 keystrokes.
        screen[n - 1] = max(2 * screen[n - 4],
                        max(3 * screen[n - 5],
                            4 * screen[n - 6]));

    return screen[N - 1]

# Driver Code
if __name__ == "__main__":

    # for the rest of the array we
    # will reply on the previous
    # entries to compute new ones
    for N in range(1, 21):
        print("Maximum Number of A's with ", N,
              " keystrokes is ", findoptimal(N))

# This code is contributed by ashutosh450

C#

// A Dynamic Programming based C# program
// to find maximum number of A's that
// can be printed using four keys
using System;

class GFG
{

// This function returns the optimal length
// string for N keystrokes
static int findoptimal(int N)
{
    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;

    // An array to store result of subproblems
    int []screen = new int[N];

    // Initializing the optimal lengths array
    // for until 6 input strokes.
    int n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;

    // Solve all subproblems in bottom-up manner
    for (n = 7; n <= N; n++)
    {

        // for any keystroke n, we will need to choose between:-
        // 1. pressing Ctrl-V once after copying the
        // A's obtained by n-3 keystrokes.

        // 2. pressing Ctrl-V twice after copying the A's
        // obtained by n-4 keystrokes.

        // 3. pressing Ctrl-V thrice after copying the A's
        // obtained by n-5 keystrokes.
        screen[n - 1] = Math.Max(2 * screen[n - 4],
                        Math.Max(3 * screen[n - 5],
                                4 * screen[n - 6]));
    }
    return screen[N - 1];
}

// Driver Code
public static void Main(String[] args)
{
    int N;

    // for the rest of the array we will reply
    // on the previous entries to compute new ones
    for (N = 1; N <= 20; N++)
        Console.Write("Maximum Number of A's with" +
                            " {0} keystrokes is {1}\n",
                                    N, findoptimal(N));
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// A Dynamic Programming based
// JavaScript program to find maximum
// number of A's that can be printed
// using four keys

// This function returns the optimal
// length string for N keystrokes
function findoptimal(N){
    // The optimal string length is N
    // when N is smaller than 7
    if (N <= 6)
        return N;

    // An array to store result
    // of subproblems
    let screen = [];

    let b; // To pick a breakpoint

    // Initializing the optimal lengths
    // array for until 6 input
    // strokes.
    let n;
    for (n = 1; n <= 6; n++)
        screen[n - 1] = n;

    // Solve all subproblems in bottom-up manner
    for (n = 7; n <= N; n++) {

        // for any keystroke n, we will
        // need to choose between:-
        // 1. pressing Ctrl-V once
        // after copying the
        // A's obtained by n-3 keystrokes.

        // 2. pressing Ctrl-V twice after
        // copying the A's
        // obtained by n-4 keystrokes.

        // 3. pressing Ctrl-V thrice
        // after copying the A's
        // obtained by n-5 keystrokes.
        screen[n - 1] = Math.max(2 * screen[n - 4],
                        Math.max(3 * screen[n - 5],
4 * screen[n - 6]));
    }

    return screen[N - 1];
}

// Driver program
let N;
// for the rest of the array we will reply on the previous
// entries to compute new ones
for (N = 1; N <= 20; N++)
    document.write(
    "Maximum Number of A's with "+ N +" keystrokes is "
    +findoptimal(N)+"<br>"
    );

</script>

Output:

Maximum Number of A's with 1 keystrokes is 1
Maximum Number of A's with 2 keystrokes is 2
Maximum Number of A's with 3 keystrokes is 3
Maximum Number of A's with 4 keystrokes is 4
Maximum Number of A's with 5 keystrokes is 5
Maximum Number of A's with 6 keystrokes is 6
Maximum Number of A's with 7 keystrokes is 9
Maximum Number of A's with 8 keystrokes is 12
Maximum Number of A's with 9 keystrokes is 16
Maximum Number of A's with 10 keystrokes is 20
Maximum Number of A's with 11 keystrokes is 27
Maximum Number of A's with 12 keystrokes is 36
Maximum Number of A's with 13 keystrokes is 48
Maximum Number of A's with 14 keystrokes is 64
Maximum Number of A's with 15 keystrokes is 81
Maximum Number of A's with 16 keystrokes is 108
Maximum Number of A's with 17 keystrokes is 144
Maximum Number of A's with 18 keystrokes is 192
Maximum Number of A's with 19 keystrokes is 256
Maximum Number of A's with 20 keystrokes is 324

Time Complexity: O(N)
Auxiliary Space: O(N)
Thanks to Sahil for providing the above approach to solve this problem.