I've coded an algorithm that can enter multiple trades based on several signals from different indicators. I am now trying to figure out how to close half of my open positions (partial close) when they've reached a target profit range and then add a trailing stop on the other half of those particular trades that have been closed.
Any ideas on how this can be done? The main issue I'm having is when it comes to writing a code that allows the EA to detect when an open order is the other half of a previously closed order.
Any advice will be much appreciated.
//CLOSING ORDER LOOP
int ticket = OrderTicket();
double orderlots = OrderLots();
for (int b = OrdersTotal() -1 ; b >=0 ;b--)
{
if (!OrderSelect (b, SELECT_BY_POS, MODE_TRADES))continue;
if (OrderSymbol() == Symbol() && OrderType() <= OP_SELL && NormalizeDouble(orderlots,1) == NormalizeDouble(lots, 1))
//CHECK PARTIAL CLOSE
{
if(OrderType() == OP_BUY && (Bid - OrderOpenPrice()) >= TakeProfit)
{
if(!OrderClose(ticket, orderlots/2, Bid, 3, Blue))
return;
}
else
{
if(OrderType() == OP_SELL && (OrderOpenPrice() - Ask) >= (TakeProfit))
{
if(!OrderClose(ticket, orderlots/2, Ask, 3, Blue))
{
bool answer = OrderSelect (b, SELECT_BY_POS, MODE_TRADES);
ticket = OrderTicket();
orderlots = OrderLots();
}
return;
//REDECLARE TICKETS
}
}
if (OrderSymbol() == Symbol() && OrderType() <= OP_SELL && NormalizeDouble(orderlots,1) != NormalizeDouble(lots, 1))continue;
//END
{
if(OrderType() == OP_BUY)
{
if(Bid - OrderOpenPrice() > stopLossATR)
{
if(!OrderModify(ticket, OrderOpenPrice(),Bid + (stopLossATR), OrderTakeProfit(), 0 , Green))
return;
}
}
else
if(OrderType() == OP_SELL)
{
if(OrderOpenPrice() - Ask > (stopLossATR))
{
if(!OrderModify(ticket, OrderOpenPrice(), Ask + (stopLossATR), OrderTakeProfit(), 0 , Green))
return;
if (!OrderSelect (b, SELECT_BY_POS, MODE_TRADES))continue;
}
}
}
}
}
I have a problem where I need to store floating points in sorted fashion. But while retrieving I want to use a different mechanism. Current storing in a std::set by overloading < operator
bool operator < (const SpanStruct_Y &b) const
{
CommonTypes::Point p1 = mobIdentifier.startPoint, p2 = mobIdentifier.startPoint;
CommonTypes::Point p3 = b.mobIdentifier.startPoint, p4 = b.mobIdentifier.startPoint;
editPoint(p1, span, 270.0f + spanOrientation);
editPoint(p2, span, 90.0f + spanOrientation);
editPoint(p3, b.span, 270.0f + b.spanOrientation);
editPoint(p4, b.span, 90.0f + b.spanOrientation);
return cmpPoint_Y(cmpPointR_Y(p1,p2),cmpPointR_Y(p3,p4));
}
But while retrieving data I want to use say different values in the identifier. [mobileId and 2 index values]
struct uniqMobIdentifier
{
CommonTypes::Point startPoint;
uint16_t mobileId;
uint16_t indexXInSpan;
uint16_t indexYInSpan;
uniqMobIdentifier()
{
startPoint = CommonTypes::Point();
mobileId = DUMMY_ID;
indexXInSpan = DUMMY_ID;
indexYInSpan = DUMMY_ID;
}
uniqMobIdentifier(CommonTypes::Point p, uint16_t mobId, uint16_t x, uint16_t y)
{
startPoint = p;
mobileId = mobId;
indexXInSpan = x;
indexYInSpan = y;
}
uniqMobIdentifier& operator =(const uniqMobIdentifier& a)
{
startPoint = a.startPoint;
mobileId = a.mobileId;
indexXInSpan = a.indexXInSpan;
indexYInSpan = a.indexYInSpan;
return *this;
}
bool operator==(const uniqMobIdentifier& a) const
{
return (((std::abs(startPoint.m_xCoordinate - a.startPoint.m_xCoordinate) < POINT_DISTANCE_TOLERANCE)
|| (std::abs(startPoint.m_yCoordinate - a.startPoint.m_yCoordinate) < POINT_DISTANCE_TOLERANCE))
&& mobileId == a.mobileId && indexXInSpan == a.indexXInSpan && indexYInSpan == indexYInSpan);
}
bool operator!=(const uniqMobIdentifier& a) const
{
return (/*((std::abs(startPoint.m_xCoordinate - a.startPoint.m_xCoordinate) > POINT_DISTANCE_TOLERANCE)
|| (std::abs(startPoint.m_yCoordinate - a.startPoint.m_yCoordinate) > POINT_DISTANCE_TOLERANCE))
&& */
!(mobileId == a.mobileId) && !(indexXInSpan == a.indexXInSpan) && !(indexYInSpan == indexYInSpan));
}
};
Is it possible to do so? Is there a better way of doing it?
You either have to do a linear search on retrieval, or else have a container that stores data in multiple ways and then allows you to retrieve by any of them. (Which may default to retrieving them in a particular order.)
If your data manipulation is going to get more and more complex over time, you may want to consider embedding SQLite.
I'm using LigGdx to make a game, it looks like a RPG game. When enemy is in alert state, it have to follow the player, but it can move only forward, backward, left and right, and also have to divert objects when it collides, searching for the best way to reach the player, i'm newbie on game development, and my algorithm may be completely wrong, so, I really need help...
private void enemyInAlert(Enemy enemy, float delta) {
detectDirection(enemy);
enemyWalking(enemy, delta);
if (Math.round(getDistanceXofLastPosition(enemy)) == 0 && Math.round(getDistanceYofLastPosition(enemy)) == 0) {
enemy.setState(States.IDLE);
lastPosition = null;
}
}
private void detectDirection(Enemy enemy) {
float diff = getDistanceXofLastPosition(enemy) - getDistanceYofLastPosition(enemy);
if (diff < 0) {
getDirectionX(enemy);
} else if (diff > 0) {
getDirectionY(enemy);
}
}
private void getDirectionY(Enemy enemy) {
int enemyY = Math.round(enemy.getY());
int lastPositionY = Math.round(lastPosition.getY());
if (enemyY < lastPositionY && enemy.isDirectionBlocked(Direction.FORWARD) == false) { //Enemy needs to go forward
enemy.setDirection(Direction.FORWARD);
enemy.blockDirection(Direction.BACKWARD);
} else if (enemyY > lastPositionY && enemy.isDirectionBlocked(Direction.FORWARD) == false) { //Enemy needs to go backward
enemy.setDirection(Direction.BACKWARD);
enemy.blockDirection(Direction.FORWARD);
} else { //Enemy needs to change direction
if (enemy.isDirectionBlocked(Direction.LEFT) == false || enemy.isDirectionBlocked(Direction.LEFT) == false) {
enemy.blockDirection(Direction.BACKWARD);
enemy.blockDirection(Direction.FORWARD);
getDirectionX(enemy);
} else {
sortRandomDirection(enemy);
}
}
}
private void getDirectionX(Enemy enemy) {
int enemyX = Math.round(enemy.getX());
int lastPositionX = Math.round(lastPosition.getX());
if (enemyX < lastPositionX && enemy.isDirectionBlocked(Direction.RIGHT) == false) { //Enemy needs to go right
enemy.setDirection(Direction.RIGHT);
enemy.blockDirection(Direction.LEFT);
} else if (enemyX > lastPositionX && enemy.isDirectionBlocked(Direction.LEFT) == false) {
enemy.setDirection(Direction.LEFT);
enemy.blockDirection(Direction.RIGHT);
} else { //Enemy needs to change direction
if (enemy.isDirectionBlocked(Direction.FORWARD) == false && enemy.isDirectionBlocked(Direction.BACKWARD) == false) {
enemy.blockDirection(Direction.LEFT);
enemy.blockDirection(Direction.RIGHT);
getDirectionY(enemy);
} else {
sortRandomDirection(enemy);
}
}
}
I'm accepting suggestions, I can change all the code, no mercy... Sorry for the bad English :D
Thanks!!
Edit: now, I'm trying to use A*, or something like it. :D ... my code:
private void calculateRoute(Enemy enemy) {
int lowerPath = getDistanceXofLastPosition(enemy.getBounds()) + getDistanceYofLastPosition(enemy.getBounds());
path = new ArrayList<Rectangle>();
Rectangle finalRect = new Rectangle(enemy.getBounds());
List<Rectangle> openList = new ArrayList<Rectangle>();
while (getDistanceXofLastPosition(finalRect) > 0 || getDistanceYofLastPosition(finalRect) > 0) {
for (int i = -1; i < 2; i+= 1) {
outerloop:
for (int j = -1; j < 2; j+= 1) {
Rectangle temp = new Rectangle(finalRect);
temp.offSet(i, j);
if (openList.contains(temp)) {
continue;
}
if ((i == -1 && j == -1) || (i == 1 && j == -1) || (i == 0 && j == 0) || (i == 1 && j == -1) || (i == 1 && j == 1)) {
continue;
}
for (Collider collider : colliders) {
if (collider.isSolid() && Utils.detectCollision(temp, collider.getBounds())) {
continue outerloop;
}
}
openList.add(temp);
}
}
int lowerDistance = Integer.MAX_VALUE;
for (Rectangle rect : openList) {
int distance = getDistanceXofLastPosition(rect) + getDistanceYofLastPosition(rect);
distance = distance + lowerPath;
if (distance < lowerDistance) {
lowerDistance = distance;
finalRect = rect;
}
}
path.add(new Rectangle(finalRect));
}
}
but is very slow, what I can do to increase performance?
You may want to look into A* .
You can easily convert your map into a graph, using each tile as a vertex, and having said vertex connected with an edge to all the other vertices close to it. The cost associated with the edge may be variable, for instance, moving i tile on a river could cost more than moving one tile in a plane.
Then you can use a path search algorithm to find the best path from one point to another. Using this algorithm will have 2 downsides :
It has an high computational cost
It always finds the optimal solution, making your bot smarter than the average player :)
If computational and storage cost are indeed a problem, you can resort to one of A*'s cousins, such as
IDA* for cheaper memory requirements, iterate over the depth of the solutions
SMA* bounds the amount of memory the algorithm can use
I recently came in contact with this interesting problem. You are given a string containing just the characters '(', ')', '{', '}', '[' and ']', for example, "[{()}]", you need to write a function which will check validity of such an input string, function may be like this:
bool isValid(char* s);
these brackets have to close in the correct order, for example "()" and "()[]{}" are all valid but "(]", "([)]" and "{{{{" are not!
I came out with following O(n) time and O(n) space complexity solution, which works fine:
Maintain a stack of characters.
Whenever you find opening braces '(', '{' OR '[' push it on the stack.
Whenever you find closing braces ')', '}' OR ']' , check if top of stack is corresponding opening bracket, if yes, then pop the stack, else break the loop and return false.
Repeat steps 2 - 3 until end of the string.
This works, but can we optimize it for space, may be constant extra space, I understand that time complexity cannot be less than O(n) as we have to look at every character.
So my question is can we solve this problem in O(1) space?
With reference to the excellent answer from Matthieu M., here is an implementation in C# that seems to work beautifully.
/// <summary>
/// Checks to see if brackets are well formed.
/// Passes "Valid parentheses" challenge on www.codeeval.com,
/// which is a programming challenge site much like www.projecteuler.net.
/// </summary>
/// <param name="input">Input string, consisting of nothing but various types of brackets.</param>
/// <returns>True if brackets are well formed, false if not.</returns>
static bool IsWellFormedBrackets(string input)
{
string previous = "";
while (input.Length != previous.Length)
{
previous = input;
input = input
.Replace("()", String.Empty)
.Replace("[]", String.Empty)
.Replace("{}", String.Empty);
}
return (input.Length == 0);
}
Essentially, all it does is remove pairs of brackets until there are none left to remove; if there is anything left the brackets are not well formed.
Examples of well formed brackets:
()[]
{()[]}
Example of malformed brackets:
([)]
{()[}]
Actually, there's a deterministic log-space algorithm due to Ritchie and Springsteel: http://dx.doi.org/10.1016/S0019-9958(72)90205-7 (paywalled, sorry not online). Since we need log bits to index the string, this is space-optimal.
If you're willing to accept one-sided error, then there's an algorithm that uses n polylog(n) time and polylog(n) space: http://www.eccc.uni-trier.de/report/2009/119/
If the input is read-only, I don't think we can do O(1) space. It is a well known fact that any O(1) space decidable language is regular (i.e writeable as a regular expression). The set of strings you have is not a regular language.
Of course, this is about a Turing Machine. I would expect it to be true for fixed word RAM machines too.
Edit: Although simple, this algorithm is actually O(n^2) in terms of character comparisons. To demonstrate it, one can simply generate a string as '(' * n + ')' * n.
I have a simple, though perhaps erroneous idea, that I will submit to your criticisms.
It's a destructive algorithm, which means that if you ever need the string it would not help (since you would need to copy it down).
Otherwise, the algorithm work with a simple index within the current string.
The idea is to remove pairs one after the others:
([{}()])
([()])
([])
()
empty -> OK
It is based on the simple fact that if we have matching pairs, then at least one is of the form () without any pair character in between.
Algorithm:
i := 0
Find a matching pair from i. If none is found, then the string is not valid. If one is found, let i be the index of the first character.
Remove [i:i+1] from the string
If i is at the end of the string, and the string is not empty, it's a failure.
If [i-1:i] is a matching pair, i := i-1 and back to 3.
Else, back to 1.
The algorithm is O(n) in complexity because:
each iteration of the loop removes 2 characters from the string
the step 2., which is linear, is naturally bound (i cannot grow indefinitely)
And it's O(1) in space because only the index is required.
Of course, if you can't afford to destroy the string, then you'll have to copy it, and that's O(n) in space so no real benefit there!
Unless, of course, I am deeply mistaken somewhere... and perhaps someone could use the original idea (there is a pair somewhere) to better effect.
I doubt you'll find a better solution, since even if you use internal functions to regexp or count occurrences, they still have a O(...) cost. I'd say your solution is the best :)
To optimize for space you could do some run-length encoding on your stack, but I doubt it would gain you very much, except in cases like {{{{{{{{{{}}}}}}}}}}.
http://www.sureinterview.com/shwqst/112007
It is natural to solve this problem with a stack.
If only '(' and ')' are used, the stack is not necessary. We just need to maintain a counter for the unmatched left '('. The expression is valid if the counter is always non-negative during the match and is zero at the end.
In general case, although the stack is still necessary, the depth of the stack can be reduced by using a counter for unmatched braces.
This is an working java code where I filter out the brackets from the string expression and then check the well formedness by replacing wellformed braces by nulls
Sample input = (a+{b+c}-[d-e])+[f]-[g] FilterBrackets will output = ({}[])[][] Then I check for wellformedness.
Comments welcome.
public class ParanString {
public static void main(String[] args) {
String s = FilterBrackets("(a+{b+c}-[d-e])[][]");
while ((s.length()!=0) && (s.contains("[]")||s.contains("()")||s.contains("{}")))
{
//System.out.println(s.length());
//System.out.println(s);
s = s.replace("[]", "");
s = s.replace("()", "");
s = s.replace("{}", "");
}
if(s.length()==0)
{
System.out.println("Well Formed");
}
else
{
System.out.println("Not Well Formed");
}
}
public static String FilterBrackets(String str)
{
int len=str.length();
char arr[] = str.toCharArray();
String filter = "";
for (int i = 0; i < len; i++)
{
if ((arr[i]=='(') || (arr[i]==')') || (arr[i]=='[') || (arr[i]==']') || (arr[i]=='{') || (arr[i]=='}'))
{
filter=filter+arr[i];
}
}
return filter;
}
}
The following modification of Sbusidan's answer is O(n2) time complex but O(log n) space simple.
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
char opposite(char bracket) {
switch(bracket) {
case '[':
return ']';
case '(':
return ')';
}
}
bool is_balanced(int length, char *s) {
int depth, target_depth, index;
char target_bracket;
if(length % 2 != 0) {
return false;
}
for(target_depth = length/2; target_depth > 0; target_depth--) {
depth=0;
for(index = 0; index < length; index++) {
switch(s[index]) {
case '(':
case '[':
depth++;
if(depth == target_depth) target_bracket = opposite(s[index]);
break;
case ')':
case ']':
if(depth == 0) return false;
if(depth == target_depth && s[index] != target_bracket) return false;
depth--;
break;
}
}
}
}
void main(char* argv[]) {
char input[] = "([)[(])]";
char *balanced = is_balanced(strlen(input), input) ? "balanced" : "imbalanced";
printf("%s is %s.\n", input, balanced);
}
If you can overwrite the input string (not reasonable in the use cases I envision, but what the heck...) you can do it in constant space, though I believe the time requirement goes up to O(n2).
Like this:
string s = input
char c = null
int i=0
do
if s[i] isAOpenChar()
c = s[i]
else if
c = isACloseChar()
if closeMatchesOpen(s[i],c)
erase s[i]
while s[--i] != c ;
erase s[i]
c == null
i = 0; // Not optimal! It would be better to back up until you find an opening character
else
return fail
end if
while (s[++i] != EOS)
if c==null
return pass
else
return fail
The essence of this is to use the early part of the input as the stack.
I know I'm a little late to this party; it's also my very first post on StackOverflow.
But when I looked through the answers, I thought I might be able to come up with a better solution.
So my solution is to use a few pointers.
It doesn't even have to use any RAM storage, as registers can be used for this.
I have not tested the code; it's written it on the fly.
You'll need to fix my typos, and debug it, but I believe you'll get the idea.
Memory usage: Only the CPU registers in most cases.
CPU usage: It depends, but approximately twice the time it takes to read the string.
Modifies memory: No.
b: string beginning, e: string end.
l: left position, r: right position.
c: char, m: match char
if r reaches the end of the string, we have a success.
l goes backwards from r towards b.
Whenever r meets a new start kind, set l = r.
when l reaches b, we're done with the block; jump to beginning of next block.
const char *chk(const char *b, int len) /* option 2: remove int len */
{
char c, m;
const char *l, *r;
e = &b[len]; /* option 2: remove. */
l = b;
r = b;
while(r < e) /* option 2: change to while(1) */
{
c = *r++;
/* option 2: if(0 == c) break; */
if('(' == c || '{' == c || '[' == c)
{
l = r;
}
else if(')' == c || ']' == c || '}' == c)
{
/* find 'previous' starting brace */
m = 0;
while(l > b && '(' != m && '[' != m && '{' != m)
{
m = *--l;
}
/* now check if we have the correct one: */
if(((m & 1) + 1 + m) != c) /* cryptic: convert starting kind to ending kind and match with c */
{
return(r - 1); /* point to error */
}
if(l <= b) /* did we reach the beginning of this block ? */
{
b = r; /* set new beginning to 'head' */
l = b; /* obsolete: make left is in range. */
}
}
}
m = 0;
while(l > b && '(' != m && '[' != m && '{' != m)
{
m = *--l;
}
return(m ? l : NULL); /* NULL-pointer for OK */
}
After thinking about this approach for a while, I realized that it will not work as it is right now.
The problem will be that if you have "[()()]", it'll fail when reaching the ']'.
But instead of deleting the proposed solution, I'll leave it here, as it's actually not impossible to make it work, it does require some modification, though.
/**
*
* #author madhusudan
*/
public class Main {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
new Main().validateBraces("()()()()(((((())))))()()()()()()()()");
// TODO code application logic here
}
/**
* #Use this method to validate braces
*/
public void validateBraces(String teststr)
{
StringBuffer teststr1=new StringBuffer(teststr);
int ind=-1;
for(int i=0;i<teststr1.length();)
{
if(teststr1.length()<1)
break;
char ch=teststr1.charAt(0);
if(isClose(ch))
break;
else if(isOpen(ch))
{
ind=teststr1.indexOf(")", i);
if(ind==-1)
break;
teststr1=teststr1.deleteCharAt(ind).deleteCharAt(i);
}
else if(isClose(ch))
{
teststr1=deleteOpenBraces(teststr1,0,i);
}
}
if(teststr1.length()>0)
{
System.out.println("Invalid");
}else
{
System.out.println("Valid");
}
}
public boolean isOpen(char ch)
{
if("(".equals(Character.toString(ch)))
{
return true;
}else
return false;
}
public boolean isClose(char ch)
{
if(")".equals(Character.toString(ch)))
{
return true;
}else
return false;
}
public StringBuffer deleteOpenBraces(StringBuffer str,int start,int end)
{
char ar[]=str.toString().toCharArray();
for(int i=start;i<end;i++)
{
if("(".equals(ar[i]))
str=str.deleteCharAt(i).deleteCharAt(end);
break;
}
return str;
}
}
Instead of putting braces into the stack, you could use two pointers to check the characters of the string. one start from the beginning of the string and the other start from end of the string. something like
bool isValid(char* s) {
start = find_first_brace(s);
end = find_last_brace(s);
while (start <= end) {
if (!IsPair(start,end)) return false;
// move the pointer forward until reach a brace
start = find_next_brace(start);
// move the pointer backward until reach a brace
end = find_prev_brace(end);
}
return true;
}
Note that there are some corner case not handled.
I think that you can implement an O(n) algorithm. Simply you have to initialise an counter variable for each type: curly, square and normal brackets. After than you should iterate the string and should increase the coresponding counter if the bracket is opened, otherwise to decrease it. If the counter is negative return false. AfterI think that you can implement an O(n) algorithm. Simply you have to initialise an counter variable for each type: curly, square and normal brackets. After than you should iterate the string and should increase the coresponding counter if the bracket is opened, otherwise to decrease it. If the counter is negative return false. After you count all brackets, you should check if all counters are zero. In that case, the string is valid and you should return true.
You could provide the value and check if its a valid one, it would print YES otherwise it would print NO
static void Main(string[] args)
{
string value = "(((([{[(}]}]))))";
List<string> jj = new List<string>();
if (!(value.Length % 2 == 0))
{
Console.WriteLine("NO");
}
else
{
bool isValid = true;
List<string> items = new List<string>();
for (int i = 0; i < value.Length; i++)
{
string item = value.Substring(i, 1);
if (item == "(" || item == "{" || item == "[")
{
items.Add(item);
}
else
{
string openItem = items[items.Count - 1];
if (((item == ")" && openItem == "(")) || (item == "}" && openItem == "{") || (item == "]" && openItem == "["))
{
items.RemoveAt(items.Count - 1);
}
else
{
isValid = false;
break;
}
}
}
if (isValid)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("NO");
}
}
Console.ReadKey();
}
var verify = function(text)
{
var symbolsArray = ['[]', '()', '<>'];
var symbolReg = function(n)
{
var reg = [];
for (var i = 0; i < symbolsArray.length; i++) {
reg.push('\\' + symbolsArray[i][n]);
}
return new RegExp('(' + reg.join('|') + ')','g');
};
// openReg matches '(', '[' and '<' and return true or false
var openReg = symbolReg(0);
// closeReg matches ')', ']' and '>' and return true or false
var closeReg = symbolReg(1);
// nestTest matches openSymbol+anyChar+closeSymbol
// and returns an obj with the match str and it's start index
var nestTest = function(symbols, text)
{
var open = symbols[0]
, close = symbols[1]
, reg = new RegExp('(\\' + open + ')([\\s\\S])*(\\' + close + ')','g')
, test = reg.exec(text);
if (test) return {
start: test.index,
str: test[0]
};
else return false;
};
var recursiveCheck = function(text)
{
var i, nestTests = [], test, symbols;
// nestTest with each symbol
for (i = 0; i < symbolsArray.length; i++)
{
symbols = symbolsArray[i];
test = nestTest(symbols, text);
if (test) nestTests.push(test);
}
// sort tests by start index
nestTests.sort(function(a, b)
{
return a.start - b.start;
});
if (nestTests.length)
{
// build nest data: calculate match end index
for (i = 0; i < nestTests.length; i++)
{
test = nestTests[i];
var end = test.start + ( (test.str) ? test.str.length : 0 );
nestTests[i].end = end;
var last = (nestTests[i + 1]) ? nestTests[i + 1].index : text.length;
nestTests[i].pos = text.substring(end, last);
}
for (i = 0; i < nestTests.length; i++)
{
test = nestTests[i];
// recursive checks what's after the nest
if (test.pos.length && !recursiveCheck(test.pos)) return false;
// recursive checks what's in the nest
if (test.str.length) {
test.str = test.str.substring(1, test.str.length - 1);
return recursiveCheck(test.str);
} else return true;
}
} else {
// if no nests then check for orphan symbols
var closeTest = closeReg.test(text);
var openTest = openReg.test(text);
return !(closeTest || openTest);
}
};
return recursiveCheck(text);
};
Using c# OOPS programming... Small and simple solution
Console.WriteLine("Enter the string");
string str = Console.ReadLine();
int length = str.Length;
if (length % 2 == 0)
{
while (length > 0 && str.Length > 0)
{
for (int i = 0; i < str.Length; i++)
{
if (i + 1 < str.Length)
{
switch (str[i])
{
case '{':
if (str[i + 1] == '}')
str = str.Remove(i, 2);
break;
case '(':
if (str[i + 1] == ')')
str = str.Remove(i, 2);
break;
case '[':
if (str[i + 1] == ']')
str = str.Remove(i, 2);
break;
}
}
}
length--;
}
if(str.Length > 0)
Console.WriteLine("Invalid input");
else
Console.WriteLine("Valid input");
}
else
Console.WriteLine("Invalid input");
Console.ReadKey();
This is my solution to the problem.
O(n) is the complexity of time without complexity of space.
Code in C.
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
bool checkBraket(char *s)
{
int curly = 0, rounded = 0, squre = 0;
int i = 0;
char ch = s[0];
while (ch != '\0')
{
if (ch == '{') curly++;
if (ch == '}') {
if (curly == 0) {
return false;
} else {
curly--; }
}
if (ch == '[') squre++;
if (ch == ']') {
if (squre == 0) {
return false;
} else {
squre--;
}
}
if (ch == '(') rounded++;
if (ch == ')') {
if (rounded == 0) {
return false;
} else {
rounded--;
}
}
i++;
ch = s[i];
}
if (curly == 0 && rounded == 0 && squre == 0){
return true;
}
else {
return false;
}
}
void main()
{
char mystring[] = "{{{{{[(())}}]}}}";
int answer = checkBraket(mystring);
printf("my answer is %d\n", answer);
return;
}