Xor using bitwise operations - python-3.x

Say I have a fn taking two ints A,B
Let binary rep of A be a0,a1,a2....an, and that of B be b0,b1,b2...bn .
I wish to return this ((a0 * b0) ^ (a1 * b1) ^ ..... ^ (an * bn)).
But the challenge is to achieve this without bit conversions i.e. using integers. How can I achieve this?
PS: I know A & B gives me a number. When this number is converted to binary and its elements are xorred amongst each other, I would get my answer. But I do not wish to convert the anded result to binary using bin() for faster computation.
def find( int A,int B):
multiply= A & B
list= bin(multiply)[2:] #(this step I wish to avoid cuz the product can be super large and the binary string is long)
list = [int(d) for d in list]
innerprod = (reduce(lambda i, j: int(i) ^ int(j), list))
return innerprod


First bitwise "and" (&) the numbers to get a number whose bits are the bits of the input numbers, multiplied respectively.
You can use Kernighan's algorithm to count the number of bits that are set to 1 (see link for description), in this number.
Then, mod 2 the result, because XOR is just flipping the result every time a bit set to 1 is encountered (so, an even number of 1's XOR'd together is 0, and an odd number will be 1).
Example:
7 is '111' and 5 is '101'
7 & 5 is '101' (bitwise "and" operation)
Two bits in '101' are set to 1 (so count_set_bits returns 2)
2 modulo 2 is 0.
(1 * 1) ^ (1 * 0) ^ (1 * 1) is 0
def count_set_bits(n):
count = 0
while n:
n &= (n-1)
count += 1
return count
def compute_answer(a, b):
return count_set_bits(a & b) % 2
print(compute_answer(7, 5)) # 0
print(compute_answer(37, 3)) # 1

Related

Removing a specific digit from a number that was specified by the user

I tried to make a Python program that removes specific digit from a number, example a = 12025 k = 2 result is 105, however none of the guides helped me do that, can anybody help me with that?

Conversion to string does not seem elegant.
As pseudo-code:
number without digit (number, digit)
if number == digit
0
else if number < 10
number
else if number % 10 == digit
number without digit (number / 10, digit)
else
number without digit (number / 10, digit) * 10 + (number % 10)
Where / is integer division, truncating the remainder, and % is the modulo, remainder.
So it is a matter of recursion.

You have to convert into str type, then remove the occurrencies, and go back to int
int(str(a).replace(str(k),''))

a = 12025
k = 2
print(int(str(a).replace(str(k), '')))

If you want to use math rather than converting to and from string you can do
a = 12025
k = 2
result = 0
exp = 0
while a:
a, remainder = divmod(a, 10)
if remainder != k:
result = result + 10**exp * remainder
exp += 1

How do I convert a 4 digit number into individual digits?

I need to write logic to break down a 4 digit number into individual digits.
On a reply here at SO to a question regarding 3 digits, someone gave the math below:
int first = 321/100;
int second = (321/10)-first*10;
int third = (321/1)-first*100-second*10;
Can someone help me?
Thank you in advance!

Well, using the sample you found, we can quite easily infer a code for you.
The first line says int first = 321/100;, which returns 3 (integer division is the euclidian one). 3 is indeed the first integer in 321 so that's a good thing. However, we have a 4 digit number, let's try replacing 100 with 1000:
int first = 4321/1000;
This does return 4 !
Let's try adapting the rest of your code (plus I put your four digit number in the variable entry).
int entry = 4321;
int first = entry/1000;
int second = entry/100 - first*10;
int third = entry/10 - first*100 - second*10;
int fourth = entry - first*1000 - second*100 - third*10;
second will be entry/100 (43) minus first*10 (40), so we're okay.
third is then 432 - 400 - 30 which turns to 2. This also works till fourth.
For more-than-four digits, you may want to use a for-loop and maybe some modulos though.

This snip of code counts the number of digits input from the user
then breaks down the digits one by one:
PRINT "Enter value";
INPUT V#
X# = V#
DO
IF V# < 1 THEN
EXIT DO
END IF
D = D + 1
V# = INT(V#) / 10
LOOP
PRINT "Digits:"; D
FOR L = D - 1 TO 0 STEP -1
M = INT(X# / 10 ^ L)
PRINT M;
X# = X# - M * 10 ^ L
NEXT
END

Given two strings, how do I find number of reoccurences of one in another?

For example, s1='abc', s2='kokoabckokabckoab'.
Output should be 3. (number of times s1 appears in s2).
Not allowed to use for or strfind. Can only use reshape,repmat,size.
I thought of reshaping s2, so it would contain all of the possible strings of 3s:
s2 =
kok
oko
koa
oab
.... etc
But I'm having troubles from here..

Assuming you have your matrix reshaped into the format you have in your post, you can replicate s1 and stack the string such that it has as many rows as there are in the reshaped s2 matrix, then do an equality operator. Rows that consist of all 1s means that we have found a match and so you would simply search for those rows where the total sum is equal to the total length of s1. Referring back to my post on dividing up a string into overlapping substrings, we can decompose your string into what you have posted in your question like so:
%// Define s1 and s2 here
s1 = 'abc';
len = length(s1);
s2 = 'kokoabckokabckoab';
%// Hankel starts here
c = (1 : len).';
r = (len : length(s2)).';
nr = length(r);
nc = length(c);
x = [ c; r((2:nr)') ]; %-- build vector of user data
cidx = (1:nc)';
ridx = 0:(nr-1);
H = cidx(:,ones(nr,1)) + ridx(ones(nc,1),:); % Hankel subscripts
ind = x(H); % actual data
%// End Hankel script
%// Now get our data
subseqs = s2(ind.');
%// Case where string length is 1
if len == 1
subseqs = subseqs.';
end
subseqs contains the matrix of overlapping characters that you have alluded to in your post. You've noticed a small bug where if the length of the string is 1, then the algorithm won't work. You need to make sure that the reshaped substring matrix consists of a single column vector. If we ran the above code without checking the length of s1, we would get a row vector, and so simply transpose the result if this is the case.
Now, simply replicate s1 for as many times as we have rows in subseqs so that all of these strings get stacked into a 2D matrix. After, do an equality operator.
eqs = subseqs == repmat(s1, size(subseqs,1), 1);
Now, find the column-wise sum and see which elements are equal to the length of your string. This will produce a single column vector where 1 indicates that we have found a match, and zero otherwise:
sum(eqs, 2) == len
ans =
0
0
0
0
1
0
0
0
0
0
1
0
0
0
0
Finally, to add up how many times the substring matched, you just have to add up all elements in this vector:
out = sum(sum(eqs, 2) == len)
out =
2
As such, we have two instances where abc is found in your string.

Here is another one,
s1='abc';
s2='bkcokbacaabcsoabckokabckoabc';
[a,b] = ismember(s2,s1);
b = [0 0 b 0 0];
a1=circshift(b,[0 -1]);
a2=circshift(b,[0 -2]);
sum((b==1)&(a1==2)&(a2==3))
It gives 3 for your input and 4 for my example, and it seems to work well if ismember is okey.

Just for the fun of it: this can be done with nlfilter from the Image Processing Toolbox (I just discovered this function today and am eager to apply it!):
ds1 = double(s1);
ds2 = double(s2);
result = sum(nlfilter(ds2, [1 numel(ds1)], #(x) all(x==ds1)));

Input as many characters as possible in notepad with fewest keyboard typings [duplicate]

This is an interview question from google. I am not able to solve it by myself. Can somebody shed some light?
Write a program to print the sequence of keystrokes such that it generates the maximum number of character 'A's. You are allowed to use only 4 keys: A, Ctrl+A, Ctrl+C and Ctrl+V. Only N keystrokes are allowed. All Ctrl+ characters are considered as one keystroke, so Ctrl+A is one keystroke.
For example, the sequence A, Ctrl+A, Ctrl+C, Ctrl+V generates two A's in 4 keystrokes.
Ctrl+A is Select All
Ctrl+C is Copy
Ctrl+V is Paste
I did some mathematics. For any N, using x numbers of A's , one Ctrl+A, one Ctrl+C and y Ctrl+V, we can generate max ((N-1)/2)2 number of A's. For some N > M, it is better to use as many Ctrl+A's, Ctrl+C and Ctrl+V sequences as it doubles the number of A's.
The sequence Ctrl+A, Ctrl+V, Ctrl+C will not overwrite the existing selection. It will append the copied selection to selected one.

There's a dynamic programming solution. We start off knowing 0 keys can make us 0 A's. Then we iterate through for i up to n, doing two things: pressing A once and pressing select all + copy followed by paste j times (actually j-i-1 below; note the trick here: the contents are still in the clipboard, so we can paste it multiple times without copying each time). We only have to consider up to 4 consecutive pastes, since select, copy, paste x 5 is equivalent to select, copy, paste, select, copy, paste and the latter is better since it leaves us with more in the clipboard. Once we've reached n, we have the desired result.
The complexity might appear to be O(N), but since the numbers grow at an exponential rate it is actually O(N2) due to the complexity of multiplying the large numbers. Below is a Python implementation. It takes about 0.5 seconds to calculate for N=50,000.
def max_chars(n):
dp = [0] * (n+1)
for i in xrange(n):
dp[i+1] = max(dp[i+1], dp[i]+1) # press a
for j in xrange(i+3, min(i+7, n+1)):
dp[j] = max(dp[j], dp[i]*(j-i-1)) # press select all, copy, paste x (j-i-1)
return dp[n]
In the code, j represents the total number of keys pressed after our new sequence of keypresses. We already have i keypresses at this stage, and 2 new keypresses go to select-all and copy. Therefore we're hitting paste j-i-2 times. Since pasting adds to the existing sequence of dp[i] A's, we need to add 1 making it j-i-1. This explains the j-i-1 in the 2nd-last line.
Here are some results (n => number of A's):
7 => 9
8 => 12
9 => 16
10 => 20
100 => 1391569403904
1,000 => 3268160001953743683783272702066311903448533894049486008426303248121757146615064636953144900245
174442911064952028008546304
50,000 => a very large number!
I agree with #SB that you should always state your assumptions: Mine is that you don't need to paste twice to double the number of characters. This gets the answer for 7, so unless my solution is wrong the assumption must be right.
In case someone wonders why I'm not checking sequences of the form Ctrl+A, Ctrl+C, A, Ctrl+V: The end result will always be the same as A, Ctrl+A, Ctrl+C, Ctrl+V which I do consider.

By using marcog's solution I found a pattern that starts at n=16. To illustrate this here are the keystrokes for n=24 up to n=29, I replaced ^A with S (select), ^C with C (copy), and ^V with P (paste) for readability:
24: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P
4 * 4 * 4 * 4 * 4 = 1024
25: A,A,A,A,S,C,P,P,P,S,C,P,P,S,C,P,P,S,C,P,P,S,C,P,P
4 * 4 * 3 * 3 * 3 * 3 = 1296
26: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,S,C,P,P,S,C,P,P
4 * 4 * 4 * 3 * 3 * 3 = 1728
27: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,S,C,P,P
4 * 4 * 4 * 4 * 3 * 3 = 2304
28: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P
4 * 4 * 4 * 4 * 4 * 3 = 3072
29: A,A,A,A,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P,S,C,P,P,P
4 * 4 * 4 * 4 * 4 * 4 = 4096
After an initial 4 As, the ideal pattern is to select, copy, paste, paste, paste and repeat. This will multiply the number of As by 4 every 5 keystrokes. If this 5 keystroke pattern cannot consume the remaining keystrokes on its own some number of 4 keystroke patterns (SCPP) consume the final keystrokes, replacing SCPPP (or removing one of the pastes) as necessary. The 4 keystroke patterns multiply the total by 3 every 4 keystrokes.
Using this pattern here is some Python code that gets the same results as marcog's solution, but is O(1) edit: This is actually O(log n) due to exponentiation, thanks to IVlad for pointing that out.
def max_chars(n):
if n <= 15:
return (0, 1, 2, 3, 4, 5, 6, 9, 12, 16, 20, 27, 36, 48, 64, 81)[n]
e3 = (4 - n) % 5
e4 = n // 5 - e3
return 4 * (4 ** e4) * (3 ** e3)
Calculating e3:
There are always between 0 and 4 SCPP patterns at the end of the keystroke list, for n % 5 == 4 there are 4, n % 5 == 1 there are 3, n % 5 == 2 there are 2, n % 5 == 3 there are 1, and n % 5 == 4 there are 0. This can be simplified to (4 - n) % 5.
Calculating e4:
The total number of patterns increases by 1 whenever n % 5 == 0, as it turns out this number increases to exactly n / 5. Using floor division we can get the total number of patterns, the total number for e4 is the total number of patterns minus e3. For those unfamiliar with Python, // is the future-proof notation for floor division.

Here's how I would approach it:
assume CtrlA = select all
assume CtrlC = copy selection
assume CtrlV = paste copied selection
given some text, it takes 4 keystrokes to duplicate it:
CtrlA to select it all
CtrlC to copy it
CtrlV to paste (this will paste over the selection - STATE YOUR ASSUMPTIONS)
CtrlV to paste again which doubles it.
From there, you can consider doing 4 or 5 A's, then looping through the above. Note that doing ctrl + a, c, v, v will grow your text exponentially as you loop through. If remaining strokes < 4, just keep doing a CtrlV
The key to interviews # places like Google is to state your assumptions, and communicate your thinking. they want to know how you solve problems.

It's solveable in O(1): Like with the Fibonacci numbers, there is a formula to calculate the number of printed As (and the sequence of keystrokes):
1) We can simplify the problem description:
Having only [A],[C-a]+[C-c],[C-v] and an empty copy-paste-buffer
equals
having only [C-a]+[C-c],[C-v] and "A" in the copy-paste-buffer.
2) We can describe the sequence of keystrokes as a string of N chars out of {'*','V','v'}, where 'v' means [C-v] and '*' means [C-a] and 'V' means [C-c]. Example: "vvvv*Vvvvv*Vvvv"
The length of that string still equals N.
The product of the lengths of the Vv-words in that string equals the number of produced As.
3) Given a fixed length N for that string and a fixed number K of words, the outcome will be maximal iff all words have nearly equal lengths. Their pair-wise difference is not more than ±1.
Now, what is the optimal number K, if N is given?
4) Suppose, we want to increase the number of words by appending one single word of length L, then we have to reduce L+1 times any previous word by one 'v'. Example: "…*Vvvv*Vvvv*Vvvv*Vvvv" -> "…*Vvv*Vvv*Vvv*Vvv*Vvv"
Now, what is the optimal word length L?
(5*5*5*5*5) < (4*4*4*4*4)*4 , (4*4*4*4) > (3*3*3*3)*3
=> Optimal is L=4.
5) Suppose, we have a sufficient large N to generate a string with many words of length 4, but a few keystrokes are left; how should we use them?
If there are 5 or more left: Append another word with length 4.
If there are 0 left: Done.
If there are 4 left: We could either
a) append one word with length 3: 4*4*4*4*3=768.
b) or increase 4 words to lenght 5: 5*5*5*5=625. => Appending one word is better.
If there are 3 left: We could either
a) or append one word with length 3 by adjusting the previus word from length 4 to 3: 4*4*4*2=128 < 4*4*3*3=144.
b) increase 3 words to lenght 5: 5*5*5=125. => Appending one word is better.
If there are 2 left: We could either
a) or append one word with length 3 by adjusting the previus two words from length 4 to 3: 4*4*1=16 < 3*3*3=27.
b) increase 2 words to lenght 5: 5*5=25. => Appending one word is better.
If there is 1 left: We could either
a) or append one word with length 3 by adjusting the previus three words from length 4 to 3: 4*4*4*0=0 < 3*3*3*3=81.
b) increase one word to lenght 5: 4*4*5=80. => Appending one word is better.
6) Now, what if we don't have a "sufficient large N" to use the rules in 5)? We have to stick with plan b), if possible!
The strings for small N are:
1:"v", 2:"vv", 3:"vvv", 4:"vvvv"
5:"vvvvv" → 5 (plan b)
6:"vvvvvv" → 6 (plan b)
7:"vvv*Vvv" → 9 (plan a)
8:"vvvv*Vvv" → 12 (plan a)
9:"vvvv*Vvvv" → 16
10:"vvvv*Vvvvv" → 20 (plan b)
11:"vvv*Vvv*Vvv" → 29 (plan a)
12:"vvvv*Vvv*Vvv" → 36 (plan a)
13:"vvvv*Vvvv*Vvv" → 48 (plan a)
14:"vvvv*Vvvv*Vvvv" → 64
15:"vvv*Vvv*Vvv*Vvv" → 81 (plan a)
…
7) Now, what is the optimal number K of words in a string of length N?
If N < 7 then K=1 else if 6 < N < 11 then K=2 ; otherwise: K=ceil((N+1)/5)
Written in C/C++/Java: int K = (N<7)?(1) : (N<11)?(2) : ((N+5)/5);
And if N > 10, then the number of words with length 3 will be: K*5-1-N. With this, we can calculate the number of printed As:
If N > 10, the number of As will be: 4^{N+1-4K}·3^{5K-N-1}

Using CtrlA + CtrlC + CtrlV is an advantage only after 4 'A's.
So I would do something like this (in pseudo-BASIC-code, since you haven't specified any proper language):
// We should not use the clipboard for the first four A's:
FOR I IN 1 TO MIN(N, 4)
PRINT 'CLICK A'
NEXT
LET N1 = N - 4
// Generates the maximum number of pastes allowed:
FOR I IN 1 TO (N1 DIV 3) DO
PRINT 'CTRL-A'
PRINT 'CTRL-C'
PRINT 'CTRL-V'
LET N1 = N1 - 3
NEXT
// If we still have same keystrokes left, let's use them with simple CTRL-Vs
FOR I IN N1 TO N
PRINT 'CTRL-V'
NEXT
Edit
Back to using a single CtrlV in the main loop.
Added some comments to explain what I'm trying to do here.
Fixed an issue with the "first four A's" block.

It takes 3 keystrokes to double your number of As. It only makes sense to start doubling when you have 3 or more As already printed. You want your last allowed keystroke to be a CtrlV to make sure you are doubling the biggest number you can, so in order to align it we will fill in any extra keystrokes after the first three As at the beginning with more As.
for (i = 3 + n%3; i>0 && n>0; n--, i--) {
print("a");
}
for (; n>0; n = n-3) {
print("ctrl-a");
print("ctrl-c");
print("ctrl-v");
}
Edit:
This is terrible, I completely got ahead of myself and didn't consider multiple pastes for each copy.
Edit 2:
I believe pasting 3 times is optimal, when you have enough keystrokes to do it. In 5 keystrokes you multiply your number of As by 4. This is better than multiplying by 3 using 4 keystrokes and better than multiplying by 5 using 6 keystrokes. I compared this by giving each method the same number of keystrokes, enough so they each would finish a cycle at the same time (60), letting the 3-multiplier do 15 cycles, the 4-multiplier do 12 cycles, and the 5-multiplier do 10 cycles. 3^15 = 14,348,907, 4^12=16,777,216, and 5^10=9,765,625. If there are only 4 keystrokes left, doing a 3-multiplier is better than pasting 4 more times, essentially making the previous 4 multiplier become an 8-multiplier. If there are only 3 keystrokes left, a 2-multiplier is best.

Assume you have x characters in the clipboard and x characters in the text area; let's call it "state x".
Let's press "Paste" a few times (i denote it by m-1 for convenience), then "Select-all" and "Copy"; after this sequence, we get to "state m*x".
Here, we wasted a total of m+1 keystrokes.
So the asymptotic growth is (at least) something like f^n, where f = m^(1/(m+1)).
I believe it's the maximum possible asymptotic growth, though i cannot prove it (yet).
Trying various values of m shows that the maximum for f is obtained for m=4.
Let's use the following algorithm:
Press A a few times
Press Select-all
Press Copy
Repeat a few times:
Press Paste
Press Paste
Press Paste
Press Select-all
Press Copy
While any keystrokes left:
Press Paste
(not sure it's the optimal one).
The number of times to press A at the beginning is 3: if you press it 4 times, you miss the opportunity to double the number of A's in 3 more keystrokes.
The number of times to press Paste at the end is no more than 5: if you have 6 or more keystrokes left, you can use Paste, Paste, Paste, Select-all, Copy, Paste instead.
So, we get the following algorithm:
If (less than 6 keystrokes - special case)
While (any keystrokes left)
A
Else
First 5 keystrokes: A, A, A, Select-all, Copy
While (more than 5 keystrokes left)
Paste, Paste, Paste, Select-all, Copy
While (any keystrokes left)
Paste
(not sure it's the optimal one). The number of characters after executing this is something like
3 * pow(4, floor((n - 6) / 5)) * (2 + (n - 1) % 5).
Sample values: 1,2,3,4,5,6,9,12,15,18,24,36,48,60,72,96,144,192,240,288,...

What follows uses the OP's second edit that pasting does not replace existing text.
Notice a few things:
^A and ^C can be considered a single action that takes two keystrokes, since it never makes sense to do them individually. In fact, we can replace all instances of ^A^C with ^K^V, where ^K is a one-key "cut" operation (let's abbreviate it X). We shall see that dealing with ^K is much nicer than the two-cost ^A^C.
Let's assume that an 'A' starts in the clipboard. Then ^V (let's abbreviate it Y) is strictly superior to A and we can drop the latter from all consideration. (In the actual problem, if the clipboard starts empty, in what follows we'll just replace Y with A instead of ^V up until the first X.)
Every reasonable keystroke sequence can thus be interpreted as a group of Ys separated by Xs, for example YYYXYXYYXY. Denote by V(s) the number of 'A's produced by the sequence s. Then V(nXm) = V(n)*V(m), because X essentially replaces every Y in m with V(n) 'A's.
The copy-paste problem is thus isomorphic to the following problem: "using m+1 numbers which sum to N-m, maximimze their product." For example, when N=6, the answer is m=1 and the numbers (2,3). 6 = 2*3 = V(YYXYYY) = V(AA^A^C^V^V) (or V(YYYXYY) = V(AAA^A^C^V). )
We can make a few observations:
For a fixed value of m, the numbers to choose are ceil( (N-m)/(m+1) ) and floor( (N-m)/(m+1) ) (in whatever combination makes the sum work out; more specifically you will need (N-m) % (m+1) ceils and the rest floors). This is because, for a < b, (a+1)*(b-1) >= a*b.
Unfortunately I don't see an easy way to find the value of m. If this were my interview I would propose two solutions at this point:
Option 1. Loop over all possible m. An O(n log n) solution.
C++ code:
long long ipow(int a, int b)
{
long long val=1;
long long mul=a;
while(b>0)
{
if(b%2)
val *= mul;
mul *= mul;
b/=2;
}
return val;
}
long long trym(int N, int m)
{
int floor = (N-m)/(m+1);
int ceil = 1+floor;
int numceils = (N-m)%(m+1);
return ipow(floor, m+1-numceils) * ipow(ceil, numceils);
}
long long maxAs(int N)
{
long long maxval=0;
for(int m=0; m<N; m++)
{
maxval = std::max(maxval, trym(N,m));
}
return maxval;
}
Option 2. Allow m to attain non-integer values and find its optimal value by taking the derivative of [(N-m)/(m+1)]^m with respect to m and solving for its root. There is no analytic solution, but the root can be found using e.g. Newton's method. Then use the floor and ceiling of that root for the value of m, and choose whichever is best.

public int dp(int n)
{
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = i + 1;
for (int i = 2; i < n - 3; i++)
{
int numchars = arr[i] * 2;
int j = i + 3;
arr[j] = Math.max(arr[j], numchars);
while (j < n - 1)
{
numchars = numchars + arr[i];
arr[++j] = Math.max(arr[j], numchars);
}
}
return arr[n - 1];
}

Here is my approach and solution with code below.
Approach:
There are three distinct operations that can be performed.
Keystroke A - Outputs one character 'A'
Keystroke (Ctrl-A) + (Ctrl-C) - Outputs nothing essentially. These two keystrokes can be combined into one operation because each of these keystrokes individually make no sense. Also, this keystroke sets up the output for the next paste operation.
Keystroke (Ctrl-V) - Output for this keystroke really depends on the previous (second) operation and hence we would need to account for that in our code.
Now given the three distinct operations and their respective outputs, we have to run through all the permutations of these operations.
Assumption:
Now, some version of this problem states that the sequence of keystrokes, Ctrl+A -> Ctrl+C -> Ctrl+V, overwrite the highlighted selection. To factor in this assumption, only one line of code needs to be added to the solution below where the printed variable in case 2 is set to 0
case 2:
//Ctrl-A and then Ctrl-C
if((count+2) < maxKeys)
{
pOutput = printed;
//comment the below statement to NOT factor
//in the assumption described above
printed = 0;
}
For this solution
The code below will print a couple of sequences and the last sequence is the correct answer for any given N. e.g. for N=11 this will be the correct sequence
With the assumption
A, A, A, A, A, C, S, V, V, V, V, :20:
Without the assumption
A, A, A, C, S, V, V, C, S, V, V, :27:
I have decided to retain the assumption for this solution.
Keystroke Legend:
'A' - A
'C' - Ctrl+A
'S' - Ctrl+C
'V' - Ctrl+V
Code:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void maxAprinted(int count, int maxKeys, int op, int printed, int pOutput, int *maxPrinted, char *seqArray)
{
if(count > maxKeys)
return;
if(count == maxKeys)
{
if((*maxPrinted) < printed)
{
//new sequence found which is an improvement over last sequence
(*maxPrinted) = printed;
printf("\n");
int i;
for(i=0; i<maxKeys; i++)
printf(" %c,",seqArray[i]);
}
return;
}
switch(op)
{
case 1:
//A keystroke
printed++;
seqArray[count] = 'A';
count++;
break;
case 2:
//Ctrl-A and then Ctrl-C
if((count+2) < maxKeys)
{
pOutput = printed;
//comment the below statement to NOT factor
//in the assumption described above
printed = 0;
}
seqArray[count] = 'C';
count++;
seqArray[count] = 'S';
count++;
break;
case 3:
//Ctrl-V
printed = printed + pOutput;
seqArray[count] = 'V';
count++;
break;
}
maxAprinted(count, maxKeys, 1, printed, pOutput, maxPrinted, seqArray);
maxAprinted(count, maxKeys, 2, printed, pOutput, maxPrinted, seqArray);
maxAprinted(count, maxKeys, 3, printed, pOutput, maxPrinted, seqArray);
}
int main()
{
const int keyStrokes = 11;
//this array stores the sequence of keystrokes
char *sequence;
sequence = (char*)malloc(sizeof(char)*(keyStrokes + 1));
//stores the max count for As printed for a sqeuence
//updated in the recursive call.
int printedAs = 0;
maxAprinted(0, keyStrokes, 1, 0, 0, &printedAs, sequence);
printf(" :%d:", printedAs);
return 0;
}

Using the tricks mentioned in answers above, Mathematically, Solution can be explained in one equation as,
4 + 4^[(N-4)/5] + ((N-4)%5)*4^[(N-4)/5].
where [] is greatest integer factor

There is a trade-off between printing m-A's manually, then using Ctrl+A, Ctrl+C, and N-m-2 Ctrl+V. The best solution is in the middle. If max key strokes = 10, the best solution is typing 5 A's or 4 A's.
try using this Look at this http://www.geeksforgeeks.org/how-to-print-maximum-number-of-a-using-given-four-keys/ and maybe optimize a bit looking for the results around the mid point.

Here is my solution with dynamic programming, without a nested loop, and which also prints the actual characters that you'd need to type:
N = 52
count = [0] * N
res = [[]] * N
clipboard = [0] * N
def maybe_update(i, new_count, new_res, new_clipboard):
if new_count > count[i] or (
new_count == count[i] and new_clipboard > clipboard[i]):
count[i] = new_count
res[i] = new_res
clipboard[i] = new_clipboard
for i in range(1, N):
# First option: type 'A'.
# Using list concatenation for 'res' to avoid O(n^2) string concatenation.
maybe_update(i, count[i - 1] + 1, res[i - 1] + ['A'], clipboard[i - 1])
# Second option: type 'CTRL+V'.
maybe_update(i, count[i - 1] + clipboard[i - 1], res[i - 1] + ['v'],
clipboard[i - 1])
# Third option: type 'CTRL+A, CTRL+C, CTRL+V'.
# Assumption: CTRL+V always appends.
if i >= 3:
maybe_update(i, 2 * count[i - 3], res[i - 3] + ['acv'], count[i - 3])
for i in range(N):
print '%2d %7d %6d %-52s' % (i, count[i], clipboard[i], ''.join(res[i]))
This is the output ('a' means 'CTRL+A', etc.)
0 0 0
1 1 0 A
2 2 0 AA
3 3 0 AAA
4 4 0 AAAA
5 5 0 AAAAA
6 6 3 AAAacv
7 9 3 AAAacvv
8 12 3 AAAacvvv
9 15 3 AAAacvvvv
10 18 9 AAAacvvacv
11 27 9 AAAacvvacvv
12 36 9 AAAacvvacvvv
13 45 9 AAAacvvacvvvv
14 54 27 AAAacvvacvvacv
15 81 27 AAAacvvacvvacvv
16 108 27 AAAacvvacvvacvvv
17 135 27 AAAacvvacvvacvvvv
18 162 81 AAAacvvacvvacvvacv
19 243 81 AAAacvvacvvacvvacvv
20 324 81 AAAacvvacvvacvvacvvv
21 405 81 AAAacvvacvvacvvacvvvv
22 486 243 AAAacvvacvvacvvacvvacv
23 729 243 AAAacvvacvvacvvacvvacvv
24 972 243 AAAacvvacvvacvvacvvacvvv
25 1215 243 AAAacvvacvvacvvacvvacvvvv
26 1458 729 AAAacvvacvvacvvacvvacvvacv
27 2187 729 AAAacvvacvvacvvacvvacvvacvv
28 2916 729 AAAacvvacvvacvvacvvacvvacvvv
29 3645 729 AAAacvvacvvacvvacvvacvvacvvvv
30 4374 2187 AAAacvvacvvacvvacvvacvvacvvacv
31 6561 2187 AAAacvvacvvacvvacvvacvvacvvacvv
32 8748 2187 AAAacvvacvvacvvacvvacvvacvvacvvv
33 10935 2187 AAAacvvacvvacvvacvvacvvacvvacvvvv
34 13122 6561 AAAacvvacvvacvvacvvacvvacvvacvvacv
35 19683 6561 AAAacvvacvvacvvacvvacvvacvvacvvacvv
36 26244 6561 AAAacvvacvvacvvacvvacvvacvvacvvacvvv
37 32805 6561 AAAacvvacvvacvvacvvacvvacvvacvvacvvvv
38 39366 19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacv
39 59049 19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvv
40 78732 19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvv
41 98415 19683 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvvv
42 118098 59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacv
43 177147 59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvv
44 236196 59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvv
45 295245 59049 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvvv
46 354294 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacv
47 531441 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvv
48 708588 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvv
49 885735 177147 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvvv
50 1062882 531441 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacv
51 1594323 531441 AAAacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvvacvv

If N key Strokes are allowed, then the result is N-3.
A's -> N-3
CTRL+A -> Selecting those N Characters :+1
CTRL+C -> Copying those N Characters :+1
Ctrl+V -> Pasting the N Characters. :+1
i.e., (Since we have selected the whole characters using CTRL+A) Replacing these existing N-3 characters with the copied N-3 Characters(which is overriding the same characters) and the result is N-3.

String permutations rank + data structure

The problem at hand is:
Given a string. Tell its rank among all its permutations sorted
lexicographically.
The question can be attempted mathematically, but I was wondering if there was some other algorithmic method to calculate it ?
Also if we have to store all the string permutations rankwise , how can we generate them efficiently (and what would be the complexity) . What would be a good data structure for storing the permutations and which is also efficient for retrieval?
EDIT
Thanks for the detailed answers on the permutations generation part, could someone also suggest a good data structure? I have only been able to think of trie tree.

There is an O(n|Σ|) algorithm to find the rank of a string of length n in the list of its permutations. Here, Σ is the alphabet.
Algorithm
Every permutation which is ranked below s can be written uniquely in the form pcx; where:
p is a proper prefix of s
c is a character ranked below the character appearing just after p in s. And c is also a character occurring in the part of s not included in p.
x is any permutation of the remaining characters occurring in s; i.e. not included in p or c.
We can count the permutations included in each of these classes by iterating through each prefix of s in increasing order of length, while maintaining the frequency of the characters appearing in the remaining part of s, as well as the number of permutations x represents. The details are left to the reader.
This is assuming the arithmetic operations involved take constant time; which it wont; since the numbers involved can have nlog|Σ| digits. With this consideration, the algorithm will run in O(n2 log|Σ| log(nlog|Σ|)). Since we can add, subtract, multiply and divide two d-digit numbers in O(dlogd).
C++ Implementation
typedef long long int lli;
lli rank(string s){
int n = s.length();
vector<lli> factorial(n+1,1);
for(int i = 1; i <= n; i++)
factorial[i] = i * factorial[i-1];
vector<int> freq(26);
lli den = 1;
lli ret = 0;
for(int i = n-1; i >= 0; i--){
int si = s[i]-'a';
freq[si]++;
den *= freq[si];
for(int c = 0; c < si; c++)
if(freq[c] > 0)
ret += factorial[n-i-1] / (den / freq[c]);
}
return ret + 1;
}

This is similar to the quickselect algorithm. In an unsorted array of integers, find the index of some particular array element. The partition element would be the given string.
Edit:
Actually it is similar to partition method done in QuickSort. The given string is the partition element.Once all permutations are generated, the complexity to find the rank for strings with length k would be O(nk). You can generate string permutations using recursion and store them in a linked list. You can pass this linked list to the partition method.
Here's the java code to generate all String permutations:
private static int generateStringPermutations(String name,int currIndex) {
int sum = 0;
for(int j=name.length()-1;j>=0;j--) {
for(int i=j-1;((i<j) && (i>currIndex));i--) {
String swappedString = swapCharsInString(name,i,j);
list.add(swappedString);
//System.out.println(swappedString);
sum++;
sum = sum + generateStringPermutations(swappedString,i);
}
}
return sum;
}
Edit:
Generating all permutations is costly. If a string contains distinct characters, the rank can be determined without generating all permutations. Here's the link.
This can be extended for cases where there are repeating characters.
Instead of x * (n-1)! which is for distinct cases mentioned as in the link,
For repeating characters it will be:
if there is 1 character which is repeating twice,
x* (n-1)!/2!
Let's take an example. For string abca the combinations are:
aabc,aacb,abac,abca,acab,acba,baac,baca,bcaa,caab,caba,cbaa (in sorted order)
Total combinations = 4!/2! = 12
if we want to find rank of 'bcaa' then we know all strings starting with 'a' are before which is 3! = 6.
Note that because 'a' is the starting character, the remaining characters are a,b,c and there are no repetitions so it is 3!. We also know strings starting with 'ba' will be before which is 2! = 2 so it's rank is 9.
Another example. If we want to find the rank of 'caba':
All strings starting with a are before = 6.
All strings starting with b are before = 3!/2! = 3 (Because once we choose b, we are left with a,a,c and because there are repetitions it is 3!/2!.
All strings starting with caa will be before which is 1
So the final rank is 11.

From GeeksforGeeks:
Given a string, find its rank among all its permutations sorted
lexicographically. For example, rank of “abc” is 1, rank of “acb” is
2, and rank of “cba” is 6.
For simplicity, let us assume that the string does not contain any
duplicated characters.
One simple solution is to initialize rank as 1, generate all
permutations in lexicographic order. After generating a permutation,
check if the generated permutation is same as given string, if same,
then return rank, if not, then increment the rank by 1. The time
complexity of this solution will be exponential in worst case.
Following is an efficient solution.
Let the given string be “STRING”. In the input string, ‘S’ is the
first character. There are total 6 characters and 4 of them are
smaller than ‘S’. So there can be 4 * 5! smaller strings where first
character is smaller than ‘S’, like following
R X X X X X I X X X X X N X X X X X G X X X X X
Now let us Fix S’ and find the smaller strings staring with ‘S’.
Repeat the same process for T, rank is 4*5! + 4*4! +…
Now fix T and repeat the same process for R, rank is 4*5! + 4*4! +
3*3! +…
Now fix R and repeat the same process for I, rank is 4*5! + 4*4! +
3*3! + 1*2! +…
Now fix I and repeat the same process for N, rank is 4*5! + 4*4! +
3*3! + 1*2! + 1*1! +…
Now fix N and repeat the same process for G, rank is 4*5! + 4*4 + 3*3!
+ 1*2! + 1*1! + 0*0!
Rank = 4*5! + 4*4! + 3*3! + 1*2! + 1*1! + 0*0! = 597
Since the value of rank starts from 1, the final rank = 1 + 597 = 598

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