We have:
n1 number of {} brackets ,
n2 number of () brackets ,
n3 number of [] brackets ,
How many different valid combination of these brackets we can have?
What I thought: I wrote a brute force code in java (which comes in the following) and counted all possible combinations, I know it's the worst solution possible,
(the code is for general case in which we can have different types of brackets)
Any mathematical approach ?
Note 1: valid combination is defined as usual, e.g. {{()}} : valid , {(}){} : invalid
Note 2: let's assume that we have 2 pairs of {} , 1 pair of () and 1 pair of [], the number of valid combinations would be 168 and the number of all possible (valid & invalid) combinations would be 840
static void paranthesis_combination(char[] open , char[] close , int[] arr){
int l = 0;
for (int i = 0 ; i < arr.length ; i++)
l += arr[i];
l *= 2;
paranthesis_combination_sub(open , close , arr , new int[arr.length] , new int[arr.length], new StringBuilder(), l);
System.out.println(paran_count + " : " + valid_paran_count);
return;
}
static void paranthesis_combination_sub(char[] open , char[] close, int[] arr , int[] open_so_far , int[] close_so_far, StringBuilder strbld , int l){
if (strbld.length() == l && valid_paran(open , close , strbld)){
System.out.println(new String(strbld));
valid_paran_count++;
return;
}
for (int i = 0 ; i < open.length ; i++){
if (open_so_far[i] < arr[i]){
strbld.append(open[i]);
open_so_far[i]++;
paranthesis_combination_sub(open , close, arr , open_so_far , close_so_far, strbld , l);
open_so_far[i]--;
strbld.deleteCharAt(strbld.length() -1 );
}
}
for (int i = 0 ; i < open.length ; i++){
if (close_so_far[i] < open_so_far[i]){
strbld.append(close[i]);
close_so_far[i]++;
paranthesis_combination_sub(open , close, arr , open_so_far , close_so_far, strbld , l);
close_so_far[i]--;
strbld.deleteCharAt(strbld.length() -1 );
}
}
return;
}
Cn is the nth Catalan number, C(2n,n)/(n+1), and gives the number of valid strings of length 2n that use only (). So if we change all [] and {} into (), there would be Cn1+n2+n3 ways. Then there are C(n1+n2+n3,n1) ways to change n1 () back to {}, and C(n2+n3,n3) ways to change the remaining () into []. Putting that all together, there are C(2n1+2n2+2n3,n1+n2+n3)C(n1+n2+n3,n1)C(n2+n3,n3)/(n1+n2+n3+1) ways.
As a check, when n1=2, n2=n3=1, we have C(8,4)C(4,2)C(2,1)/5=168.
In general, infinitely. However I assume, that you meant to find how many combinations are there provided limited string length. For simplicity lets assume that the limit is an even number. Then, lets create an initial string:
(((...()...))) with length equal to the limit.
Then, we can switch any instance of () pair with [] or {} parenthesis. However, if we change an opening brace, then we ought to change the matching closing brace. So, we can look only at the opening braces, or at pairs. For each parenthesis pair we have 4 options:
leave it unchanged
change it to []
change it to {}
remove it
So, for each of (l/2) objects we choose one of four labels, which gives:
4^(l/2) possibilities.
EDIT: this assumes only "concentric" parenthesis strings (contained in each other), as you've suggested in your edit. Intuitively however, a valid combination is also: ()[]{} - this solution does not take this into account.
Related
So I am doing this question of EDIT DISTANCE and before going to DP approach I am trying to solve this question in recursive manner and I am facing some logical error, please help....
Here is my code -
class Solution {
public int minDistance(String word1, String word2) {
int n=word1.length();
int m=word2.length();
if(m<n)
return Solve(word1,word2,n,m);
else
return Solve(word2,word1,m,n);
}
private int Solve(String word1,String word2,int n,int m){
if(n==0||m==0)
return Math.abs(n-m);
if(word1.charAt(n-1)==word2.charAt(m-1))
return 0+Solve(word1,word2,n-1,m-1);
else{
//insert
int insert = 1+Solve(word1,word2,n-1,m);
//replace
int replace = 1+Solve(word1,word2,n-1,m-1);
//delete
int delete = 1+Solve(word1,word2,n-1,m);
int max1 = Math.min(insert,replace);
return Math.min(max1,delete);
}
}
}
here I am checking the last element of both the strings if both the characters are equal then simple moving both string to n-1 and m-1 resp.
Else
Now I am having 3 cases of insertion , deletion and replace ,and between these 3 I have to find minima.
If I am replacing the character then simply I moved the character to n-1 & m-1.
If I am inserting the character from my logic I think I should insert the character at the last of smaller length string and move the pointer to n-1 and m
To delete the element I think I should delete the element from the larger length String that's why I move pointer to n-1 and m but I think I am making mistake here please help.
Leetcode is giving me wrong answer for word1 = "plasma" and word2 = "altruism".
The problem is that the recursive expression for the insert-case is the same as for the delete-case.
Reasoning further, it turns out the one for the insert-case is wrong. In that case we choose to resolve the letter in word2 (at index m-1) through insertion, so it should not be considered any more during the recursive process. On the other hand the considered letter in word1 could still be matched with another letter in word2, so that letter should still be considered during the recursive process.
That means that m should be decremented, not n.
So change:
int insert = 1+Solve(word1,word2,n-1,m);
to:
int insert = 1+Solve(word1,word2,n,m-1);
...and it will work. Then remains to add the memoization for getting a good efficiency.
Python clean DP based solution,
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
return self.edit_distance(word1, word2)
#cache
def edit_distance(self, s, t):
# Edge conditions
if len(s) == 0:
return len(t)
if len(t) == 0:
return len(s)
# If 1st char matches
if s[0] == t[0]:
return self.edit_distance(s[1:], t[1:])
else:
return min(
1 + self.edit_distance(s[1:], t), # delete
1 + self.edit_distance(s, t[1:]), # insert
1 + self.edit_distance(s[1:], t[1:]) # replace
)
The problem is as such:
given an array of N numbers, find two numbers in the array such that they will have a range(max - min) value of K.
for example:
input:
5 3
25 9 1 6 8
output:
9 6
So far, what i've tried is first sorting the array and then finding two complementary numbers using a nested loop. However, because this is a sort of brute force method, I don't think it is as efficient as other possible ways.
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(), k = sc.nextInt();
int[] arr = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
Arrays.sort(arr);
int count = 0;
int a, b;
for(int i = 0; i < n; i++) {
for(int j = i; j < n; j++) {
if(Math.max(arr[i], arr[j]) - Math.min(arr[i], arr[j]) == k) {
a = arr[i];
b = arr[j];
}
}
}
System.out.println(a + " " + b);
}
}
Much appreciated if the solution was in code (any language).
Here is code in Python 3 that solves your problem. This should be easy to understand, even if you do not know Python.
This routine uses your idea of sorting the array, but I use two variables left and right (which define two places in the array) where each makes just one pass through the array. So other than the sort, the time efficiency of my code is O(N). The sort makes the entire routine O(N log N). This is better than your code, which is O(N^2).
I never use the inputted value of N, since Python can easily handle the actual size of the array. I add a sentinel value to the end of the array to make the inner short loops simpler and quicker. This involves another pass through the array to calculate the sentinel value, but this adds little to the running time. It is possible to reduce the number of array accesses, at the cost of a few more lines of code--I'll leave that to you. I added input prompts to aid my testing--you can remove those to make my results closer to what you seem to want. My code prints the larger of the two numbers first, then the smaller, which matches your sample output. But you may have wanted the order of the two numbers to match the order in the original, un-sorted array--if that is the case, I'll let you handle that as well (I see multiple ways to do that).
# Get input
N, K = [int(s) for s in input('Input N and K: ').split()]
arr = [int(s) for s in input('Input the array: ').split()]
arr.sort()
sentinel = max(arr) + K + 2
arr.append(sentinel)
left = right = 0
while arr[right] < sentinel:
# Move the right index until the difference is too large
while arr[right] - arr[left] < K:
right += 1
# Move the left index until the difference is too small
while arr[right] - arr[left] > K:
left += 1
# Check if we are done
if arr[right] - arr[left] == K:
print(arr[right], arr[left])
break
I'm trying to improve my recursion skill(reading a written recursion function) by looking at examples. However, I can easily get the logic of recursions without local variables. In below example, I can't understand how the total variables work. How should I think a recursive function to read and write by using local variables? I'm thinking it like stack go-hit-back. By the way, I wrote the example without variables. I tried to write just countThrees(n / 10); instead of total = total + countThrees(n / 10); but it doesn't work.
with total variable:
int countThrees(int n) {
if (n == 0) { return 0; }
int lastDigit = n % 10;
int total = 0;
total = total + countThrees(n / 10);
if (lastDigit == 3) {
total = total + 1;
}
return total;
}
simplified version
int countThrees(int x)
{
if (x / 10 == 0) return 0;
if (x % 10 == 3)
return 1 + countThrees(x / 10);
return countThrees(x / 10);
}
In both case, you have to use a stack indeed, but when there are local variables, you need more space in the stack as you need to put every local variables inside. In all cases, the line number from where you jump in a new is also store.
So, in your second algorithme, if x = 13, the stack will store "line 4" in the first step, and "line 4; line 3" in the second one, in the third step you don't add anything to the stack because there is not new recursion call. At the end of this step, you read the stack (it's a First in, Last out stack) to know where you have to go and you remove "line 3" from the stack, and so.
In your first algorithme, the only difference is that you have to add the locale variable in the stack. So, at the end of the second step, it looks like "Total = 0, line 4; Total = 0, line 4".
I hope to be clear enough.
The first condition should read:
if (x == 0) return 0;
Otherwise the single 3 would yield 0.
And in functional style the entire code reduces to:
return x == 0 ? 0
: countThrees(x / 10) + (x % 10 == 3 ? 1 : 0);
On the local variables:
int countThrees(int n) {
if (n == 0) {
return 0;
}
// Let an alter ego do the other digits:
int total = countThrees(n / 10);
// Do this digit:
int lastDigit = n % 10;
if (lastDigit == 3) {
++total;
}
return total;
}
The original code was a bit undecided, when or what to do, like adding to total after having it initialized with 0.
By declaring the variable at the first usage, things become more clear.
For instance the absolute laziness: first letting the recursive instances calculate the total of the other digits, and only then doing the last digit oneself.
Using a variable lastDigit with only one usage is not wrong; it explains what is happening: you inspect the last digit.
Preincrement operator ++x; is x += 1; is x = x + 1;.
One could have done it (recursive call and own work) the other way around, so it probably says something about the writer's psychological preferences
The stack usage: yes total before the recursive call is an extra variable on the stack. Irrelevant for numbers. Also a smart compiler could see that total is a result.
On the usage of variables: they can be stateful, and hence are useful for turning recursion into iteration. For that tail recursion is easiest: the recursion happening last.
int countThrees(int n) {
int total = 0;
while (n != 0) {
int digit = n % 10;
if (digit == 3) {
++total;
}
n /= 10; // Divide by 10
}
return total;
}
The question
Given a MiniZinc array of strings:
int: numStats;
set of int: Stats = 1..numStats;
array[Stats] of string: statNames;
... with data loaded from a MiniZinc data file:
numStats = 3;
statNames = ["HEALTH", "ARMOR", "MANA"];
How can one look up the index of a specific string in the array? For example, that ARMOR is located at position 2.
The context
I need to find an optimal selection of items with regard to some constraints on their stats. This information is stored in a 2D array declared as follows:
int: numItems;
set of int: Items = 1..numItems;
array[Items, Stats] of float: itemStats;
So in order to write a constraint on, say, the minimum amount of ARMOR obtained through the selected items, I need to know that ARMOR has index 2 in the inner array.
Since the data file is generated by an external program, and the number and order of stats are dynamic, I cannot hardcode the indices in the constraints.
One solution (that won't work in my case)
The MiniZinc tutorial uses an interesting trick to achieve something similar:
set of int: Colors = 1..3;
int: red = 1;
int: yellow = 2;
int: blue = 3;
array[Colors] of string: name = ["red", "yellow", "blue"];
var Colors: x;
constraint x != red;
output [ name[fix(x)] ];
Unfortunately, as variable declarations are not allowed in MiniZinc data files, this trick won't work in my case.
You can write your own custom function to get the index of a string within a string array:
function int: getIndexOfString(string: str,
array[int] of string: string_array) =
sum( [ if str = string_array[i]
then i
else 0 endif
| i in index_set(string_array) ]
);
In this function I create an array of integers where the integer at position i either equals the index of str if string_array[i]=str and 0 otherwise. For instance, for your sample string array ["HEALTH", "ARMOR", "MANA"] and str ARMOR the resulting int array will be [0,2,0].
This is why I can simply sum over the int array to get the index of the string. If the string does not occur, the return value is 0, which is fine since indices in MiniZinc start with 1 by default.
Here is how you can call the function above for your first example:
int: numStats;
set of int: Stats = 1..numStats;
array[Stats] of string: statNames;
numStats = 3;
statNames = ["HEALTH", "ARMOR", "MANA"];
var int: indexOfArmor;
constraint
indexOfArmor = getIndexOfString("ARMOR",statNames);
solve satisfy;
Note however that the function above is limited and has some flaws. First, if you have multiple occurrences of the string in the array, then you will receive an invalid index (the sum of all indices where str occurred). Also, if you have your own index set for your string array (say (2..6)), then you will need to adapt the function.
Another, cleaner option is to write a function that uses a recursive helper function:
% main function
function int: index_of(string: elem, array[int] of string: elements) =
let {
int: index = length(elements);
} in % calls the helper function with the last index
get_index(elem, elements, index)
;
% recursive helper function
function int: get_index(string: elem, array[int] of string: elements, int: index) =
if index == 0
then -1 % the element was not found (base case of recursion)
elseif elements[index] == elem
then index % the element was found
else
get_index(elem, elements, index - 1) % continue searching
endif
;
The helper function iterates recursively over the array, starting from the last element, and when it finds the element, it returns the index. If the element was not found in the array, then -1 is returned. Alternatively, you can also throw an assertion following the suggestion of Patrick Trentin by replacing then -1 with then assert(false, "unknown element: " + elem).
An example of calling this function:
set of int: Customers = 1..5;
array[Customers] of string: ids = ["a-1", "a-2", "a-3", "a-4", "a-5"];
var int: index = index_of("a-3", ids);
var int: unknown_index = index_of("x-3", ids);
where index will be assigned 3 and unknown_index will be -1.
An alternative approach to that presented by Andrea Rendl-Pitrey, is the following one:
array[int] of string: statNames = array1d(10..12, ["HEALTH", "ARMOR", "MANA"]);
var int: indexOfArmor =
sum([i | i in index_set(statNames) where statNames[i] = "ARMOR"]);
solve satisfy;
output [
"indexOfArmor=", show(indexOfArmor), "\n",
];
which outputs:
~$ mzn2fzn example.mzn ; flatzinc example.fzn
indexOfArmor = 11;
----------
note: that var can be dropped from the declaration of indexOfArmor, since the index can be statically computed. I kept it here only for output purposes.
A better solution is to declare a new predicate:
predicate index_of_str_in_array(var int: idx,
string: str,
array[int] of string: arr) =
assert(
not exists(i in index_set(arr), j in index_set(arr))
(i != j /\ arr[i] = str /\ arr[j] = str),
"input string occurs at multiple locations",
assert(
exists(i in index_set(arr))
(arr[i] = str),
"input string does not occur in the input array",
exists(i in index_set(arr))
(arr[i] = str /\ i = idx)
));
which enforces both of the following conditions:
str occurs at least once in arr
str does not occur multiple times in arr
e.g
predicate index_of_str_in_array(var int: idx,
string: str,
array[int] of string: arr) =
...
array[10..13] of string: statNames =
array1d(10..13, ["HEALTH", "ARMOR", "MANA", "ATTACK"]);
var int: indexOfArmor;
constraint index_of_str_in_array(indexOfArmor, "ARMOR", statNames);
solve satisfy;
output [
"indexOfArmor=", show(indexOfArmor), "\n",
];
outputs
~$ mzn2fzn example.mzn ; flatzinc example.fzn
indexOfArmor = 11;
----------
If one changes statNames in the following way
array[10..13] of string: statNames =
array1d(10..13, ["HEALTH", "ARMOR", "MANA", "ARMOR"]);
then mzn2fzn detects an assertion violation:
~$ mzn2fzn example.mzn ; flatzinc example.fzn
MiniZinc: evaluation error:
example.mzn:24:
in call 'index_of_str_in_array'
example.mzn:4:
in call 'assert'
Assertion failed: input string occurs at multiple locations
flatzinc:
example.fzn: cannot open input file: No such file
A similar result would be obtained by searching for the index of a string that does not occur in the array. This condition can of course be removed if not necessary.
DISCLAIMER: older versions of mzn2fzn don't seem to check that the declared index-set of an array of strings variable matches the index-set of an array of strings literal that is being assigned to it. This rule is enforced on newer versions, as it is also valid for other data types.
According to this other post on Stackoverflow there is no way of converting strings to integers in MiniZinc, only the other way around. You need to first pre process your data in some other language and turn it into integers. You can however turn those integers into string once you are done in MiniZinc.
You can however load MiniZinc files instead of data files if you would like. Use the include syntax to include any .mzn file.
I'm learning how to do recursion, and I want to make sure that I'm doing it correctly. I just finished a question on codingbat that reads like this:
Given a non-negative int n, return the count of the occurrences of 7
as a digit, so for example 717 yields 2. (no loops). Note that mod (%)
by 10 yields the rightmost digit (126 % 10 is 6), while divide (/) by
10 removes the rightmost digit (126 / 10 is 12).
count7(717) → 2
count7(7) → 1
count7(123) → 0
And my solution, which worked, looks like this:
public int count7(int n) {
int count = 0;
if(n < 7) {
return count;
} else {
int divided = n / 10;
if(n % 10 == 7) count++;
return count + count7(divided);
}
}
Even though my solution passed, I want to make sure that I'm working through these recursion problems correctly. Should I have a counter sitting outside the if/else statement? If not, why? If not, how would you solve it instead.
The more self-contained, the better - and your answer is self-contained. And it has the two requisites for correct recursion:
Provide a "stopper"
public int count7(int n) {
int count = 0;
if(n < 7) {
return count;
If n is less than 7, return 0, since n clearly contains no 7s.
Otherwise, assume the problem is solved for some smaller number
} else {
int divided = n / 10;
if(n % 10 == 7) count++;
return count + count7(divided);
}
}
Remove the rightmost digit and assume that the problem is solved for what's left. That's the recursion count7(divided). Meanwhile, what about that rightmost digit? If it is 7, that needs to go into our ultimate answer, so add it in.
So far, so good.
Critique
Your structure is misleading. count at the start actually does nothing. You could have written this:
public int count7(int n) {
if(n < 7) {
return 0;
In that case there is also no need for your count++. We will add 1 if this is a 7 and not if it isn't:
} else {
int divided = n / 10;
if(n % 10 == 7) return 1 + count7(divided);
return count7(divided);
}
}
Observe that that is your answer - but it is more "honest" than what you wrote. There was nothing wrong with how you were recursing, but the presentation, I'm suggesting, can be clearer and less crabbed. Your code should read like a verbal description of the approach you are taking: "If this number is less than 7, return 0. Otherwise, pull off the last digit and recurse on what's left, adding 1 only if the number we pulled off is a 7."
There are recursion problems where you might generate a "count" value of some sort and pass it into the recursion call, but this is not one of them. Thus the whole count variable thing is just a red herring.