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++ 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 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 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# 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_ |
<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> |
5
Time complexity: O(N)
Auxiliary space: O(N).