Following works in groovy -
for(def i=0;i<10;i++)
print i
But this which is valid in Java, C++ does not work in groovy -
for(def i=0,j=0;i<10;i++,j++)
print i + ' ' + j
Why? How to make this work?
It will not working as Groovy does not accept multiple expressions in a for loop.
Try this:
[0..10,0..10].transpose().each{ i, j ->
println i + ' ' + j
}
to achieve the same result.
Update to make it more generalized. This update is equivalent to increment with i++, j+=3.
(0..<10).collect{[it, it+3]}.each{ i, j ->
println i + ' ' + j
}
Have you tried this:
for( def ( int i, int j ) = [ 0, 0 ]; i < 10; i++, j++ )
If that doesn't work, it might be failing because of the last part.
C++ has a explicit comma operator, which is how it allows constructs like this.
Java does not have a comma operator, but presumably allows constructs such as this as a hack to the for loop.
If Groovy wont allow this, it's most probably because it doesn't allow this hack.
Related
I would like to know how to increase variables in Nim by 1, alike the (variablename)++ function in other programming languages.
My Code by far:
int i = 1
int j = 1
for i in countup(1, 10):
j = j + 1
echo "number: "
echo i
I want to change the j = j + 1 with the Nim version of: j++
You can use the += operator:
j += 1
You can increment a variable by one in Nim with the inc() function: inc(var_name)
I was asked this question in an interview:
Given a string (1<=|s|<=10^5), check if it is possible to partition it into three palindromes. If there are multiple answers possible, output the one where the cuts are made the earliest. If no answer is possible, print "Impossible".
**Input:**
radarnoonlevel
aabab
abcdefg
**Output:**
radar noon level
a a bab (Notice how a, aba, b is also an answer, but we will output the one with the earliest cuts)
Impossible
I was able to give a brute force solution, running two loops and checking palindrome property for every 3 substrings ( 0-i, i-j, j-end). This was obviously not optimal, but I have not been able to find a better solution since then.
I need a way of checking that if I know the palindrome property of a string, then how removing a character from the start or adding one at the end can give me the property of the new string without having to do the check for the whole string again. I am thinking of using three maps where each character key is mapped to number of occurences but that too doesn't lead me down anything.
Still O(n^2) solution, but you can store the result of palindrome substrings in a table and use that to get to the answer.
vector<string> threePalindromicSubstrings(string word) {
int n = word.size();
vector<vector<bool>> dp (n,vector<bool>(n,false));
for(int i = 0 ; i < n ; ++i)
dp[i][i] = 1;
for(int l = 2 ; l <= n ; ++l){
for(int i = 0 ; i < n - l +1 ; ++i){
int j = i + l - 1;
if(l == 2)
dp[i][j] = (word[i] == word[j]);
else
dp[i][j] = (word[i] == word[j]) && (dp[i+1][j-1]);
}
}
vector<string> ans;
for(int i = 0 ; i < n - 2 ; ++i){
if(dp[0][i]) {
for(int j = i+1 ; j < n - 1 ; ++j){
if(dp[i+1][j] && dp[j+1][n-1]){
ans.push_back(word.substr(0,i + 1));
ans.push_back(word.substr(i+1,j-i));
ans.push_back(word.substr(j+1,n-j));
return ans;
}
}
}
}
if(ans.empty())
ans.push_back("Impossible");
return ans;
}
Given two bit strings, x and y, with x longer than y, I'd like to compute a kind of asymmetric variant of the Levensthein distance between them. Starting with x, I'd like to know the minimum number of deletions and substitutions it takes to turn x into y.
Can I just use the usual Levensthein distance for this, or do I need I need to modify the algorithm somehow? In other words, with the usual set of edits of deletion, substitution, and addition, is it ever beneficial to delete more than the difference in lengths between the two strings and then add some bits back? I suspect the answer is no, but I'm not sure. If I'm wrong, and I do need to modify the definition of Levenshtein distance to disallow deletions, how do I do so?
Finally, I would expect intuitively that I'd get the same distance if I started with y (the shorter string) and only allowed additions and substitutions. Is this right? I've got a sense for what these answers are, I just can't prove them.
If i understand you correctly, I think the answer is yes, the Levenshtein edit distance could be different than an algorithm that only allows deletions and substitutions to the larger string. Because of this, you would need to modify, or create a different algorithm to get your limited version.
Consider the two strings "ABCD" and "ACDEF". The Levenshtein distance is 3 (ABCD->ACD->ACDE->ACDEF). If we start with the longer string, and limit ourselves to deletions and substitutions we must use 4 edits (1 deletion and 3 substitutions. The reason is that strings where deletions are applied to the smaller string to efficiently get to the larger string can't be achieved when starting with the longer string, because it does not have the complimentary insertion operation (since you're disallowing that).
Your last paragraph is true. If the path from shorter to longer uses only insertions and substitutions, then any allowed path can simply be reversed from the longer to the shorter. Substitutions are the same regardless of direction, but the inserts when going from small to large become deletions when reversed.
I haven't tested this thoroughly, but this modification shows the direction I would take, and appears to work with the values I've tested with it. It's written in c#, and follows the psuedo code in the wikipedia entry for Levenshtein distance. There are obvious optimizations that can be made, but I refrained from doing that so it was more obvious what changes I've made from the standard algorithm. An important observation is that (using your constraints) if the strings are the same length, then substitution is the only operation allowed.
static int LevenshteinDistance(string s, string t) {
int i, j;
int m = s.Length;
int n = t.Length;
// for all i and j, d[i,j] will hold the Levenshtein distance between
// the first i characters of s and the first j characters of t;
// note that d has (m+1)*(n+1) values
var d = new int[m + 1, n + 1];
// set each element to zero
// c# creates array already initialized to zero
// source prefixes can be transformed into empty string by
// dropping all characters
for (i = 0; i <= m; i++) d[i, 0] = i;
// target prefixes can be reached from empty source prefix
// by inserting every character
for (j = 0; j <= n; j++) d[0, j] = j;
for (j = 1; j <= n; j++) {
for (i = 1; i <= m; i++) {
if (s[i - 1] == t[j - 1])
d[i, j] = d[i - 1, j - 1]; // no operation required
else {
int del = d[i - 1, j] + 1; // a deletion
int ins = d[i, j - 1] + 1; // an insertion
int sub = d[i - 1, j - 1] + 1; // a substitution
// the next two lines are the modification I've made
//int insDel = (i < j) ? ins : del;
//d[i, j] = (i == j) ? sub : Math.Min(insDel, sub);
// the following 8 lines are a clearer version of the above 2 lines
if (i == j) {
d[i, j] = sub;
} else {
int insDel;
if (i < j) insDel = ins; else insDel = del;
// assign the smaller of insDel or sub
d[i, j] = Math.Min(insDel, sub);
}
}
}
}
return d[m, n];
}
Question: What is y after the following switch statement is executed? Rewrite the code using an if statement.
y = 3; x = 3;
switch (x + 3)
{
case 6: y = 1;
default: y += 1;
}
I'm at the beginning of the C++ hike. I don't know what do with this. It's not working in C++ Visual Studio 2013. I've put it in as is and nothing happens.
I use:
y = 3; x = 3;
switch (x + 3)
{
case 6: y = 1;
default: y += 1;
}
return 0;
}
And nothing happens. I have both the answers but I have no clue how to get them...
y is 2
if (x + 3 == 6)
y = 1;
y += 1;
I'm strictly supposed to be using
#include <iostream>
using namespace std;
int main()
{
return 0;
}
switch basics in C/C++:
switch expression is evaluated once, at the beginning. Then the 1st matching case is executed
after the case clause, the execution falls through to the next clause unless there's a break. Since this typically isn't what was indended, fall-throughs are normally documented with a comment or (if multiple cases refer to the same code) by stacking them together (case 2: case 3: <code>).
I know from Algebra class that with ABC and 123 we can make 216 different permutations for a three letter string, right? (6 x 6 x 6) I'd like to create a console program in C++ that displays ever possible permutation for the example above. The thing is, how would I even begin trying to calculate them. Perhaps:
AAA
BAA
CAA
1BA
2BA
3CA
1AB
2BC
3CA
etc.
This is really hard to ask, but what would I have to do to ensure that I include every permutation? I know there are 216 but I don't know how to actually go about going through all of them.
Any suggestions would be greatly appreciated!!!
If you need a fixed-number strings, you can use N nested loops (three in your case).
string parts = "ABC123";
for (int i = 0 ; i != parts.size() ; i++)
for (int j = 0 ; j != parts.size() ; j++)
for (int k = 0 ; k != parts.size() ; k++)
cout << parts[i] << parts[j] << parts[k] << endl;
If N is not fixed, you would need a more general recursive solution.
It's really easy to do using recursion. Provided you have an array of all six elements, here's java code to do it. I am sure you can translate it to C++ easily.
void getAllCombinations(List<String> output, char[] chrs, String prefix, int length) {
if (prefix.length() == length) {
output.add(prefix);
} else {
for (int i = 0;i < chrs.length;i++) {
getAllCombinations(output, chrs, prefix + chrs[i], length);
}
}
return;
}
This is not perfect, but it should give you the general idea.
Run it with parameters: empty list, array of available characters, empty string and length of desired strings.
With three nested loops (one per character position) iterating over each of the 6 allowed characters it's hard not to see that every possibly combination has a corresponding set of loop indices, and that every set of legal loop indices has a corresponding 3 letter string. And that 1-1 correspondence between loop indices and strings is what you're looking for, I gather.