On the first row, we write a 0. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10.
Given row N and index K, return the K-th indexed symbol in row N. (The values of K are 1-indexed.) (1 indexed).
Examples:
Input: N = 1, K = 1
Output: 0
Input: N = 2, K = 1
Output: 0
Input: N = 2, K = 2
Output: 1
Input: N = 4, K = 5
Output: 1
Explanation:
row 1: 0
row 2: 01
row 3: 0110
row 4: 01101001
Link to the Problem:
https://leetcode.com/explore/learn/card/recursion-i/253/conclusion/1675/
Solution:
class Solution {
public:
int kthGrammar(int N, int K) {
if(N==0||K==0)
return 0;
string result="0";
string finals;
int i,j;
for(j=0;j<N-1;j++)
{
for(i=0;i<result.length();i++)
{
if(result[i]=='0')
finals.append("01");
else
finals.append("10");
}
result=finals;
}
return result[K-1]-'0';
}
};
Your finals string remains with old contents. Seems you need to clear it at every loop turn.
Anyway, your approach is not suitable for large inputs - so instead of (huge) string generation consider calculation of needed symbol with some math.
def f(n,k):
if n == 1:
return 0
if k<=pow(2,n-2):
return f(n-1,k)
else:
return 1-f(n-1,k-(pow(2,n-2)))
the above is a better soln. written in python but same logic can be used
Related
I need a program that will output this figure:
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
If you add the numbers on both ends, it will print the output beside it (inward). And then you will also add those two sums and print it again inwardly. Another thing, the input should be the largest number (in this case, number 8) It could be larger than 8 like the figure below.
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
1 2 4 8 16 8 4 2 1
In this case the input is 16. And so on. This is my latest program.
#include<iostream>
using namespace std;
int main(){
int i, j, k, b, a, space=10;
for(int i=0;i<=5;i++){
for(k=0;k<space;k++){
cout<<" ";
}
for(j=1;j<=2*i-1;j=j*2){
cout<<j<<" ";
}
space--;
cout<<endl;
}
system("pause");
return 0;
}
Please help me improve this. It's not yet a pyramid. Help me to output the desired figure at least.
To correctly format your pyramid, supposing you're using fixed width characters, you need to know beforehand some information, e.g.:
what is the largest number that you're going to print.
how many numbers have which width.
Since the pyramid is increasing downwards, this information is available when you print the last line.
So what you need to do is to calculate (but not output, of course) the last line first. Say that you want five rows, then the middle number will be 2^(5-1), i.e. 16. So you will have to output 1 2 4 8 16. The column positions will be 0 (beginning), 2 (0 plus length of "1" plus 1 space), 4 (2 plus 1 plus 1 space), 6 (4 plus 1 plus 1), 8, 11 (8 plus length of "16" which is 2, plus 1 space), 13, 15, 17.
At this point you start output of the first line, beginning at column 5, i.e. at position 8.
The second line will start at column 4, i.e. at position 6.
And so on.
Another possibility is to imagine you're filling a table (as if you were generating a HTML table):
- fill it top to bottom
- "explore" every cell size the same way as above, in any order
- generate column positions accordingly
- print the table top to bottom
This requires only one round of calculations, but needs memory storage for the table itself.
A shortcut is to verify what is the largest number you're gonna print, and format all columns with that width. In this case 16 is 2 characters, so you add one space padding and output all columns padded to 3 character width. This may waste unnecessary space.
The latter case can be implemented using cout.width:
int main() {
int line;
// Read input from standard input
cin >> line;
// We output the pyramid by allocating a fixed width to each number.
// This requires to know beforehand which will be the largest number.
// We can observe that at every line, the largest number is 2 to the
// power of that line number: on line 0, the largest number is 2^0
// which is 1, on line 1 it is 2 which is 2^1... on line 4 it is 16
// which is 2^4. So if we have five lines (from 0 to 4), the largest
// number will be 2 to the 4th.
// Now the length of a number in base 10 is given by the logarithm
// base 10 of that number, truncated, plus 1. For example log10 of
// 1000 is exactly 3, and 3+1 is 4 digits. Log10 of 999 is
// 2.9995654... which truncates to 2, 2+1 is 3 and 999 is 3 digits.
// Here our number is 2 to the power of (line-1).
// By the properties of the logarithm
// this is the same as (line-1)*log10(2), and log10(2) is around 0.3.
// So we multiply (line-1) by log10(2), truncate to integer and add 1
// (or vice versa: we add 1 and then assign to width, which is an
// integer, thereby truncating the value to integer.
// But we need to add another 1 for the padding space (we want 1 2 4
// 2 1, not 12421...). So before assigning, we add 2, not 1.
int width = 2+(line-1)*0.30102999566398119521373889472449;
//////////////////////
// TODO: we're gonna output 2*line+1 strings, each 'width' wide.
// So if (2*line+1)*width > 80 we'd better say it and stop, or the
// output will be sorely messed up, since a terminal is only 80 chars
// wide at the most. Which means that N=9 is the maximum number we
// can print out and still be "nice".
// Having not been asked to do this, we proceed instead.
//////////////////////
// For every line that we need to output...
for (int i = 0; i < line; i++) {
// Pad line-i empty spaces
for (int j = 0; j < (line-i); j++) {
// Set the width of the next cout to "width" bytes
cout.width(width);
cout<<" ";
}
int n = 1;
// output the forward sequence: 1, 2, 4... doubling each time
for (int j = 0; j < i; j++) {
cout.width(width);
cout <<n;
n *= 2;
}
// output the top number, which is the next doubling
cout.width(width);
cout <<n;
// output the sequence in reverse. Halve, output, repeat.
for (int j = 0; j < i; j++) {
n /= 2;
cout.width(width);
cout<<n;
}
// Now n is 1 again (not that we care...), and we output newline
cout <<"\n";
}
// Return 0 to signify "no error".
return 0;
}
Check the Code. This will give the desire output .
#include<iostream>
using namespace std;
int main(){
int line = 4;
for (int i =0; i < line; i++){
for(int j = line - i; j >0 ; j --){
cout<<" ";
}
int temp = 1;
for(int k = 0; k < i + 1; k ++){
cout << " "<<temp;
temp = temp *2;
}
temp /=2;
for(int k =0; k < i; k ++){
temp /=2;
cout << " "<<temp;
}
cout <<"\n";
}
return 0;
}
Output:
1
1 2 1
1 2 4 2 1
1 2 4 8 4 2 1
I'm student of second year on CS. On my algorithms and data structures course I've been tasked with following problem:
Input:
2<=r<=20
2<=o<=10
0<=di<=100
Output:
number of combinations
or "NO" if there are none
r is number of integers
di are said integers
o is number of groups
I have to find the number of correct combinations. The correct combination is one where every integer is assigned to some group, none of the groups are empty and the sum of integers in every group is the same:
For an instance:
r = 4;
di = {5, 4, 5, 6}
o = 2;
So the sum of integers in every group should add up to 10:
5 + 4 + 5 + 6 = 20
20 / o = 20 / 2 = 10
So we can make following groups:
{5, 5}, {4, 6}
{5, 5}, {6, 4}
{5, 5}, {4, 6}
{5, 5}, {6, 5}
So as we can see, the every combination is essentialy same as first one.( The order of elements in group doesnt matter.)
So actually we have just one correct combination: {5, 5}, {4, 6}. Which means output is equal to one.
Other examples:
r = 4;
di = {10, 2, 8, 6}
o = 2;
10 + 2 + 8 + 6 = 26;
26 / o = 26 / 2 = 13
There is no way to make such a sum of these integers, so the output is "NO".
I had a following idea of getting this thing done:
struct Input { // holds data
int num; // number of integers
int groups; // number of groups
int sumPerGroup; // sum of integers per group
int *integers; // said integers
};
bool f(bool *t, int s) { // generates binary numbers (right to left"
int i = 0;
while (t[i]) i++;
t[i] = 1;
if (i >= s) return true;
if (!t[i + 1])
for (int j = i - 1; j >= 0; j--)
t[j] = 0;
return false;
}
void solve(Input *input, int &result) {
bool bin[input->num]; // holds generated binary numbers
bool used[input->num]; // integers already used
for (int i = 0; i < input->num; i++) {
bin[i] = 0;
used[i] = 0;
}
int solved = 0;
do {
int sum = 0;
for (int i = 0; i < input->num; i++) { // checking if generated combination gets me nice sum
if (sum > input->sumPerGroup) break;
if (bin[i] && !used[i]) sum += input->integers[i]; // if generated combination wasnt used before, start adding up
if (sum == input->sumPerGroup) { // if its add up as it shoul
for (int j = 0; j < input->num; j++) used[j] = bin[j]; // mark integers as used
solved ++; // and mark group as solved
sum = 0;
}
if (udane == input->groups) { // if the number of solved groups is equal to number of groups
result ++; // it means we found another correct combination
solved = 0;
}
}
} while (!f(bin, input->num)); // as long as I can get more combinations
}
So, the main idea is:
1. I generate combination of some numbers as binary number
2. I check if that combination gets me a nice sum
3. If it does, I mark that up
4. Rinse and repeat.
So for input from first example {5, 4, 5, 6} in 2 groups:
5 4 5 6
-------
0 0 0 0
1 0 0 0
...
1 0 1 0 -> this one is fine, becouse 5 + 5 = 10; I mark it as used
1 1 1 0
...
0 1 0 1 -> another one works (4 + 6 = 10); Marked as used
So far i got myself 2 working groups which is equal to 2 groups - job done, it's a correct combination.
The real problem behind my idea is that I have no way of using some integer once I mark it as "used". This way in more complicated examples I would miss quite alot of correct groups. My question is, what is correct approach to this kind of problem? I've tried recursive approach and it didin't work any better (for the same reason)
Another idea I had is to permutate (std:next_permutate(...) for instance) integers from input each time I mark some group as used, but even on paper that looks silly.
I don't ask you to solve that problem for me, but if you could point any flaws in my reasoning that would be terrific.
Also, not a native speaker. So I'd like to apologise in advance if I butchered any sentence (I know i did).
I am fairly new to C++, and am struggling through a problem that seems to have a solid solution but I just can't seem to find it. I have a contiguous array of ints starting at zero:
int i[6] = { 0, 1, 2, 3, 4, 5 }; // this is actually from an iterator
I would like to partition the array into groups of three. The design is to have two methods, j and k, such that given an i they will return the other two elements from the same group of three. For example:
i j(i) k(i)
0 1 2
1 0 2
2 0 1
3 4 5
4 3 5
5 3 4
The solution seems to involve summing the i with its value mod three and either plus or minus one, but I can't quite seem to work out the logic.
This should work:
int d = i % 3;
int j = i - d + ( d == 0 );
int k = i - d + 2 - ( d == 2 );
or following statement for k could be more readable:
int k = i - d + ( d == 2 ? 1 : 2 );
This should do it:
int j(int i)
{
int div = i / 3;
if (i%3 != 0)
return 3*div;
else
return 3*div+1;
}
int k(int i)
{
int div = i / 3;
if (i%3 != 2)
return 3*div+2;
else
return 3*div+1;
}
Test.
If you want shorter functions:
int j(int i)
{
return i/3*3 + (i%3 ? 0 : 1);
}
int k(int i)
{
return i/3*3 + (i%3-2 ? 2 : 1);
}
Well, first, notice that
j(i) == j(3+i) == j(6+i) == j(9+i) == ...
k(i) == k(3+i) == k(6+i) == k(9+i) == ...
In other words, you only need to find a formula for
j(i), i = 0, 1, 2
k(i), i = 0, 1, 2
and then for the rest of the cases simply plug in i mod 3.
From there, you'll have trouble finding a simple formula because your "rotation" isn't standard. Instead of
i j(i) k(i)
0 1 2
1 2 0
2 0 1
for which the formula would have been
j(i) = (i + 1) % 3
k(i) = (i + 2) % 3
you have
i j(i) k(i)
0 1 2
1 0 1
2 0 2
for which the only formula I can think of at the moment is
j(i) = (i == 0 ? 1 : 0)
k(i) = (i == 1 ? 1 : 2)
If the values of your array (let's call it arr, not i in order to avoid confusion with the index i) do not coincide with their respective index, you have to perform a reverse lookup to figure out their index first. I propose using an std::map<int,size_t> or an std::unordered_map<int,size_t>.
That structure reflects the inverse of arr and you can extra the index for a particular value with its subscript operator or the at member function. From then, you can operate purely on the indices, and use modulo (%) to access the previous and the next element as suggested in the other answers.
consider that
0 -- is the first
1 -- is the second
2 -- is the third
.....
9 -- is the 10th
11 -- is the 11th
what is an efficient algorithm to find the nth palindromic number?
I'm assuming that 0110 is not a palindrome, as it is 110.
I could spend a lot of words on describing, but this table should be enough:
#Digits #Pal. Notes
0 1 "0" only
1 9 x with x = 1..9
2 9 xx with x = 1..9
3 90 xyx with xy = 10..99 (in other words: x = 1..9, y = 0..9)
4 90 xyyx with xy = 10..99
5 900 xyzyx with xyz = 100..999
6 900 and so on...
The (nonzero) palindromes with even number of digits start at p(11) = 11, p(110) = 1001, p(1100) = 100'001,.... They are constructed by taking the index n - 10^L, where L=floor(log10(n)), and append the reversal of this number: p(1101) = 101|101, p(1102) = 102|201, ..., p(1999) = 999|999, etc. This case must be considered for indices n >= 1.1*10^L but n < 2*10^L.
When n >= 2*10^L, we get the palindromes with odd number of digits, which start with p(2) = 1, p(20) = 101, p(200) = 10001 etc., and can be constructed the same way, using again n - 10^L with L=floor(log10(n)), and appending the reversal of that number, now without its last digit: p(21) = 11|1, p(22) = 12|1, ..., p(99) = 89|8, ....
When n < 1.1*10^L, subtract 1 from L to be in the correct setting with n >= 2*10^L for the case of an odd number of digits.
This yields the simple algorithm:
p(n) = { L = logint(n,10);
P = 10^(L - [1 < n < 1.1*10^L]); /* avoid exponent -1 for n=1 */
n -= P;
RETURN( n * 10^L + reverse( n \ 10^[n >= P] ))
}
where [...] is 1 if ... is true, 0 else, and \ is integer division.
(The expression n \ 10^[...] is equivalent to: if ... then n\10 else n.)
(I added the condition n > 1 in the exponent to avoid P = 10^(-1) for n=0. If you use integer types, you don't need this. Another choice it to put max(...,0) as exponent in P, or use if n=1 then return(0) right at the start. Also notice that you don't need L after assigning P, so you could use the same variable for both.)
I'm reading about permutations and I'm interested in ranking/unranking methods.
From the abstract of a paper:
A ranking function for the permutations on n symbols assigns a unique
integer in the range [0, n! - 1] to each of the n! permutations. The corresponding
unranking function is the inverse: given an integer between 0 and n! - 1, the
value of the function is the permutation having this rank.
I made a ranking and an unranking function in C++ using next_permutation. But this isn't practical for n>8. I'm looking for a faster method and factoradics seem to be quite popular.
But I'm not sure if this also works with duplicates. So what would be a good way to rank/unrank permutations with duplicates?
I will cover one half of your question in this answer - 'unranking'. The goal is to find the lexicographically 'K'th permutation of an ordered string [abcd...] efficiently.
We need to understand Factorial Number System (factoradics) for this. A factorial number system uses factorial values instead of powers of numbers (binary system uses powers of 2, decimal uses powers of 10) to denote place-values (or base).
The place values (base) are –
5!= 120 4!= 24 3!=6 2!= 2 1!=1 0!=1 etc..
The digit in the zeroth place is always 0. The digit in the first place (with base = 1!) can be 0 or 1. The digit in the second place (with base 2!) can be 0,1 or 2 and so on. Generally speaking, the digit at nth place can take any value between 0-n.
First few numbers represented as factoradics-
0 -> 0 = 0*0!
1 -> 10 = 1*1! + 0*0!
2 -> 100 = 1*2! + 0*1! + 0*0!
3 -> 110 = 1*2! + 1*1! + 0*0!
4 -> 200 = 2*2! + 0*1! + 0*0!
5 -> 210 = 2*2! + 1*1! + 0*0!
6 -> 1000 = 1*3! + 0*2! + 0*1! + 0*0!
7 -> 1010 = 1*3! + 0*2! + 1*1! + 0*0!
8 -> 1100 = 1*3! + 1*2! + 0*1! + 0*0!
9 -> 1110
10-> 1200
There is a direct relationship between n-th lexicographical permutation of a string and its factoradic representation.
For example, here are the permutations of the string “abcd”.
0 abcd 6 bacd 12 cabd 18 dabc
1 abdc 7 badc 13 cadb 19 dacb
2 acbd 8 bcad 14 cbad 20 dbac
3 acdb 9 bcda 15 cbda 21 dbca
4 adbc 10 bdac 16 cdab 22 dcab
5 adcb 11 bdca 17 cdba 23 dcba
We can see a pattern here, if observed carefully. The first letter changes after every 6-th (3!) permutation. The second letter changes after 2(2!) permutation. The third letter changed after every (1!) permutation and the fourth letter changes after every (0!) permutation. We can use this relation to directly find the n-th permutation.
Once we represent n in factoradic representation, we consider each digit in it and add a character from the given string to the output. If we need to find the 14-th permutation of ‘abcd’. 14 in factoradics -> 2100.
Start with the first digit ->2, String is ‘abcd’. Assuming the index starts at 0, take the element at position 2, from the string and add it to the Output.
Output String
c abd
2 012
The next digit -> 1.String is now ‘abd’. Again, pluck the character at position 1 and add it to the Output.
Output String
cb ad
21 01
Next digit -> 0. String is ‘ad’. Add the character at position 1 to the Output.
Output String
cba d
210 0
Next digit -> 0. String is ‘d’. Add the character at position 0 to the Output.
Output String
cbad ''
2100
To convert a given number to Factorial Number System,successively divide the number by 1,2,3,4,5 and so on until the quotient becomes zero. The reminders at each step forms the factoradic representation.
For eg, to convert 349 to factoradic,
Quotient Reminder Factorial Representation
349/1 349 0 0
349/2 174 1 10
174/3 58 0 010
58/4 14 2 2010
14/5 2 4 42010
2/6 0 2 242010
Factoradic representation of 349 is 242010.
One way is to rank and unrank the choice of indices by a particular group of equal numbers, e.g.,
def choose(n, k):
c = 1
for f in xrange(1, k + 1):
c = (c * (n - f + 1)) // f
return c
def rank_choice(S):
k = len(S)
r = 0
j = k - 1
for n in S:
for i in xrange(j, n):
r += choose(i, j)
j -= 1
return r
def unrank_choice(k, r):
S = []
for j in xrange(k - 1, -1, -1):
n = j
while r >= choose(n, j):
r -= choose(n, j)
n += 1
S.append(n)
return S
def rank_perm(P):
P = list(P)
r = 0
for n in xrange(max(P), -1, -1):
S = []
for i, p in enumerate(P):
if p == n:
S.append(i)
S.reverse()
for i in S:
del P[i]
r *= choose(len(P) + len(S), len(S))
r += rank_choice(S)
return r
def unrank_perm(M, r):
P = []
for n, m in enumerate(M):
S = unrank_choice(m, r % choose(len(P) + m, m))
r //= choose(len(P) + m, m)
S.reverse()
for i in S:
P.insert(i, n)
return tuple(P)
if __name__ == '__main__':
for i in xrange(60):
print rank_perm(unrank_perm([2, 3, 1], i))
For large n-s you need arbitrary precision library like GMP.
this is my previous post for an unranking function written in python, I think it's readable, almost like a pseudocode, there is also some explanation in the comments: Given a list of elements in lexicographical order (i.e. ['a', 'b', 'c', 'd']), find the nth permutation - Average time to solve?
based on this you should be able to figure out the ranking function, it's basically the same logic ;)
Java, from https://github.com/timtiemens/permute/blob/master/src/main/java/permute/PermuteUtil.java (my public domain code, minus the error checking):
public class PermuteUtil {
public <T> List<T> nthPermutation(List<T> original, final BigInteger permutationNumber) {
final int size = original.size();
// the return list:
List<T> ret = new ArrayList<>();
// local mutable copy of the original list:
List<T> numbers = new ArrayList<>(original);
// Our input permutationNumber is [1,N!], but array indexes are [0,N!-1], so subtract one:
BigInteger permNum = permutationNumber.subtract(BigInteger.ONE);
for (int i = 1; i <= size; i++) {
BigInteger factorialNminusI = factorial(size - i);
// casting to integer is ok here, because even though permNum _could_ be big,
// the factorialNminusI is _always_ big
int j = permNum.divide(factorialNminusI).intValue();
permNum = permNum.mod(factorialNminusI);
// remove item at index j, and put it in the return list at the end
T item = numbers.remove(j);
ret.add(item);
}
return ret;
}
}