Minimum time to write characters using insert, delete and copy operation

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We need to write N same characters on a screen and each time we can insert a character, delete the last character and copy and paste all written characters i.e. after copy operation count of total written character will become twice. Now we are given time for insertion, deletion and copying. We need to output minimum time to write N characters on the screen using these operations.

Examples:

Input : N = 9    
        insert time = 1    
        removal time = 2    
        copy time = 1
Output : 5
N character can be written on screen in 5 time units as shown below,
insert a character    
characters = 1  total time = 1
again insert character      
characters = 2  total time = 2
copy characters             
characters = 4  total time = 3
copy characters             
characters = 8  total time = 4
insert character           
characters = 9  total time = 5

We can solve this problem using dynamic programming. We can observe a pattern after solving some examples by hand that for writing each character we have two choices either get it by inserting or get it by copying, whichever takes less time. Now writing relation accordingly,
Let dp[i] be the optimal time to write i characters on screen then,

If i is even then,
   dp[i] = min((dp[i-1] + insert_time), 
               (dp[i/2] + copy_time))
Else (If i is odd)
   dp[i] = min(dp[i-1] + insert_time),
              (dp[(i+1)/2] + copy_time + removal_time)

In the case of odd, removal time is added because when (i+1)/2 characters will be copied one extra character will be on the screen which needs to be removed.

C++

// C++ program to write characters in
// minimum time by inserting, removing
// and copying operation
#include <bits/stdc++.h>
using namespace std;
 
//  method returns minimum time to write
// 'N' characters
int minTimeForWritingChars(int N, int insert,
                       int remove, int copy)
{
    if (N == 0)
       return 0;
    if (N == 1)
       return insert;
 
    //  declare dp array and initialize with zero
    int dp[N + 1];
    memset(dp, 0, sizeof(dp));
 
    // first char will always take insertion time
    dp[1] = insert;
 
    //  loop for 'N' number of times
    for (int i = 2; i <= N; i++)
    {
        /*  if current char count is even then
            choose minimum from result for (i-1)
            chars and time for insertion and
            result for half of chars and time
            for copy  */
        if (i % 2 == 0)
            dp[i] = min(dp[i-1] + insert,
                        dp[i/2] + copy);
 
        /*  if current char count is odd then
            choose minimum from
            result for (i-1) chars and time for
            insertion and
            result for half of chars and time for
            copy and one extra character deletion*/
        else
            dp[i] = min(dp[i-1] + insert,
                        dp[(i+1)/2] + copy + remove);
    }
    return dp[N];
}
 
// Driver code
int main()
{
    int N = 9;
    int insert = 1, remove = 2, copy = 1;
    cout << minTimeForWritingChars(N, insert,
                                remove, copy);
    return 0;
}

Java

// Java program to write characters in 
// minimum time by inserting, removing 
// and copying operation 
 
public class GFG{ 
     
    // method returns minimum time to write 
    // 'N' characters 
    static int minTimeForWritingChars(int N, int insert, 
                                      int remove, int copy) 
    { 
        if (N == 0) 
        return 0; 
        if (N == 1) 
        return insert; 
     
        // declare dp array and initialize with zero 
        int dp[] = new int [N + 1]; 
         
          // first char will always take insertion time
          dp[1] = insert;
     
        // loop for 'N' number of times 
        for (int i = 2; i <= N; i++) 
        { 
            /* if current char count is even then 
                choose minimum from result for (i-1) 
                chars and time for insertion and 
                result for half of chars and time 
                for copy */
            if (i % 2 == 0) 
                dp[i] = Math.min(dp[i-1] + insert, dp[i/2] + copy); 
     
            /* if current char count is odd then 
                choose minimum from 
                result for (i-1) chars and time for 
                insertion and 
                result for half of chars and time for 
                copy and one extra character deletion*/
            else
                dp[i] = Math.min(dp[i-1] + insert, 
                                 dp[(i+1)/2] + copy + remove); 
        } 
        return dp[N]; 
    } 
     
    // Driver code to test above methods 
    public static void main(String []args) 
    { 
        int N = 9; 
        int insert = 1, remove = 2, copy = 1; 
        System.out.println(minTimeForWritingChars(N, insert,remove, copy)); 
    } 
    // This code is contributed by Ryuga 
} 

Python3

# Python3 program to write characters in
# minimum time by inserting, removing
# and copying operation
 
def minTimeForWritingChars(N, insert, 
                           remove, cpy):
     
    # method returns minimum time 
    # to write 'N' characters 
    if N == 0:
        return 0
    if N == 1:
        return insert
 
    # declare dp array and initialize
    # with zero
    dp = [0] * (N + 1)
     
    # first char will always take insertion time
    dp[1] = insert
     
    # loop for 'N' number of times
    for i in range(2, N + 1):
 
        # if current char count is even then
        # choose minimum from result for (i-1)
        # chars and time for insertion and
        # result for half of chars and time
        # for copy
        if i % 2 == 0:
            dp[i] = min(dp[i - 1] + insert,
                        dp[i // 2] + cpy)
 
        # if current char count is odd then
        # choose minimum from
        # result for (i-1) chars and time for
        # insertion and
        # result for half of chars and time for
        # copy and one extra character deletion
        else:
            dp[i] = min(dp[i - 1] + insert, 
                        dp[(i + 1) // 2] +
                        cpy + remove)
 
    return dp[N]
 
# Driver Code
if __name__ == "__main__":
    N = 9
    insert = 1
    remove = 2
    cpy = 1
    print(minTimeForWritingChars(N, insert, 
                                 remove, cpy))
 
# This code is contributed
# by vibhu4agarwal

C#

// C# program to write characters in 
// minimum time by inserting, removing 
// and copying operation
using System; 
 
class GFG 
{ 
    // method returns minimum time to write 
    // 'N' characters 
    static int minTimeForWritingChars(int N, int insert, 
                                        int remove, int copy) 
    { 
        if (N == 0) 
            return 0; 
        if (N == 1) 
            return insert; 
     
        // declare dp array and initialize with zero 
        int[] dp = new int[N + 1]; 
        for(int i = 0; i < N + 1; i++)
            dp[i] = 0;
     
          // first char will always take insertion time
          dp[1] = insert;
       
        // loop for 'N' number of times 
        for (int i = 2; i <= N; i++) 
        { 
             
            /* if current char count is even then 
                choose minimum from result for (i-1) 
                chars and time for insertion and 
                result for half of chars and time 
                for copy */
            if (i % 2 == 0) 
                dp[i] = Math.Min(dp[i - 1] + insert, 
                            dp[i / 2] + copy); 
     
            /* if current char count is odd then 
                choose minimum from 
                result for (i-1) chars and time for 
                insertion and 
                result for half of chars and time for 
                copy and one extra character deletion*/
            else
                dp[i] = Math.Min(dp[i - 1] + insert, 
                            dp[(i + 1) / 2] + copy + remove); 
        } 
        return dp[N]; 
    } 
     
    // Driver code 
    static void Main() 
    { 
        int N = 9; 
        int insert = 1, remove = 2, copy = 1; 
        Console.Write(minTimeForWritingChars(N, insert, 
                                            remove, copy)); 
    }
}
 
//This code is contributed by DrRoot_

Javascript

<script>
    // Javascript program to write characters in
    // minimum time by inserting, removing
    // and copying operation
     
    // method returns minimum time to write
    // 'N' characters
    function minTimeForWritingChars(N, insert, remove, copy)
    {
        if (N == 0)
            return 0;
        if (N == 1)
            return insert;
      
        // declare dp array and initialize with zero
        let dp = new Array(N + 1);
        for(let i = 0; i < N + 1; i++)
            dp[i] = 0;
      
          // first char will always take insertion time
          dp[1] = insert;
        
        // loop for 'N' number of times
        for (let i = 2; i <= N; i++)
        {
              
            /* if current char count is even then
                choose minimum from result for (i-1)
                chars and time for insertion and
                result for half of chars and time
                for copy */
            if (i % 2 == 0)
                dp[i] = Math.min(dp[i - 1] + insert, dp[parseInt(i / 2, 10)] + copy);
      
            /* if current char count is odd then
                choose minimum from
                result for (i-1) chars and time for
                insertion and
                result for half of chars and time for
                copy and one extra character deletion*/
            else
                dp[i] = Math.min(dp[i - 1] + insert,
                            dp[parseInt((i + 1) / 2, 10)] + copy + remove);
        }
        return dp[N];
    }
     
    let N = 9;
    let insert = 1, remove = 2, copy = 1;
    document.write(minTimeForWritingChars(N, insert,
                                         remove, copy));
     
    // This code is contributed by divyeshrabadiya07.
</script>

Output

Time complexity: O(N)
Auxiliary space: O(N).

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