spectacle not working kde neon

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sudo snap remove spectacle
sudo apt install kde-spectacle

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Source: askubuntu.com

Tags: neon spectacle whatever

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Contributed on Dec 02 2024

Mehrab Islam Arnab

0 Answers Avg Quality 2/10

valid parentheses

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2

let s = '([]{})()'

const isValid = function(s) {
    let stack = []
    let characters = '()[]{}'   // even, odd, even, odd, even, odd

    for (let i = 0; i < s.length; i++) {
        let index = characters.indexOf(s[i])

        if (index % 2 === 0) {
            stack.push(s[i])
        } else if (index % 2 !== 0 && stack[stack.length - 1] === characters[index - 1]) {
            stack.pop()
        } else {
            return false
        }

        console.log(stack)
    }

    return !stack.length
}


console.log(isValid(s))

// [Log]:
[ '(' ]
[ '(', '[' ]
[ '(' ]
[ '(', '{' ]
[ '(' ]
[]
[ '(' ]
[]
true

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Source: Grepper

Tags: javascript j

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Contributed on Nov 27 2022

Mysterious Macaque

0 Answers Avg Quality 2/10

Valid Parentheses

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class Solution:
    def isValid(self, s):
        if s == "":
            return True
        if len(s) < 2:
            return False

        pair_brkts = {
            "{" : "}",
            "(" : ")",
            "[" : "]"
            }
        stack = []

        for i in s:
            if i in pair_brkts:
                stack.append(i)                                #stack i(forward facing brackets) that also exists as keys in our pair_brkts dictionary)
                #print("forward facing brackets", stack)       #to see the contents of your stacked list 
            else:
                if len(stack) == 0 or pair_brkts[stack.pop()] != i:   #if stack list is empty or the value pair of last 
                                                                        #list item isnt same, return False, otherwise break out of loop
                    #print("backward facing brackets", stack)        #print to see whats left of your list after 
                                                                    #looping is over for all your i(i.e brackets)
                    return False
        if len(stack) > 0:                                          #if stack list is not empty at this point, return False, else return True
            return False
        return True

 
        

Task = Solution()
print("1. ", Task.isValid("({[()]})"))
print("2. ", Task.isValid("()[]{}"))
print("3. ", Task.isValid("(]"))

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Source: Grepper

Tags: parentheses pare

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Contributed on May 18 2022

Kingsley Atuba

0 Answers Avg Quality 2/10

Valid parentheses

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1

class Solution:
   def isValid (self, sequence: str):
       '''
       Function to pair if sequence contains valid parenthesis
       :param sequence: Sequence of brackets
       :return: True is sequence is valid else False
       '''
       stack = []
       opening = set('([{')
       closing = set(')]}')
       pair = {')' : '(' , ']' : '[' , '}' : '{'}
       for i in sequence :
           if i in opening :
               stack.append(i)
           if i in closing :
               if not stack :
                   return False
               elif stack.pop() != pair[i] :
                   return False
               else :
                   continue
       if not stack :
           return True
       else :
           return False

if __name__ == '__main__':
   sequence = '{[()]}'
   print(f'Is {sequence} valid ? : {Solution().isValid(sequence)}')
   sequence1 = '{[()]}{]{}}'
   print(f'Is {sequence1} valid ? : {Solution().isValid (sequence1)}')

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Source: favtutor.com

Tags: parentheses pare

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Contributed on Jan 24 2022

Itchy Impala

0 Answers Avg Quality 2/10

valid parentheses

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4

/*
	n = 3
    
    [ '()()()', '()(())', '(())()', '(()())', '((()))' ]
*/

const generateParenthesis = function (n) {
    const res = [];

    const dfs = function (left = n, right = n, combos = '') {
        if (combos.length === 2*n) {
            res.push(combos);
        }
        else if (left <= right) {
            if (left > 0) {
                dfs(left - 1, right, combos + '(');
            }

             if (right > 0) {
                dfs(left, right - 1, combos + ')');
            }
        }
    }

    dfs();

    return res;
}

console.log(generateParenthesis(3));

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Source: Grepper

Tags: javascript j

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Contributed on Nov 28 2023

Mysterious Macaque

0 Answers Avg Quality 2/10

valid parentheses

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0

{ { } [ ] [ [ [ ] ] ] } is VALID expression
          [ [ [ ] ] ]    is VALID sub-expression
  { } [ ]                is VALID sub-expression

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Source: leetcode.com

Tags: java

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Contributed on Mar 06 2022

Hugo Barcelos

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Valid Parentheses

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bool isValid(char * s){

}

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Source: Grepper

Tags: c parentheses

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Contributed on Aug 17 2022

CodeBlack reactor

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Valid Parentheses

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function isValid(s: string): boolean {

};

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Contributed on Aug 17 2022

CodeBlack reactor

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valid parentheses

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Input: s = "()[]{}"
Output: true

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Source: leetcode.com

Tags: javascript parentheses pare

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Contributed on Dec 11 2022

shivam chauhan

0 Answers Avg Quality 2/10

Valid Parentheses

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import scala.collection.mutable.Stack

object ValidParentheses {
  def isValid(s: String): Boolean = {
    val stack = Stack[Char]()
    val map = Map(')' -> '(', '}' -> '{', ']' -> '[')

    for (char <- s) {
      if (map.contains(char)) {
        if (stack.isEmpty || stack.pop() != map(char))
          return false
      } else {
        stack.push(char)
      }
    }

    stack.isEmpty
  }

  def main(args: Array[String]): Unit = {
    // Example usage
    val input1 = "()[]{}" // Valid
    val input2 = "([)]"   // Invalid
    val input3 = "{[]}"   // Valid

    println(s"Is '$input1' valid? ${isValid(input1)}")
    println(s"Is '$input2' valid? ${isValid(input2)}")
    println(s"Is '$input3' valid? ${isValid(input3)}")
  }
}

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Source: chat.openai.com

Tags: parentheses scala

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Contributed on Sep 08 2023

Sumit Rawal

0 Answers Avg Quality 2/10

Valid Parentheses

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/*
  You are given a string s consisting of the following characters: 
  '(', ')', '{', '}', '[' and ']'.

  The input string s is valid if and only if:

  Every open bracket is closed by the same type of close bracket.
  Open brackets are closed in the correct order.
  Every close bracket has a corresponding open bracket of the same type.
  Return true if s is a valid string, and false otherwise.

  Example 1:
  Input: s = "[]"
  Output: true
  
  Example 2:
  Input: s = "([{}])"
  Output: true
  
  Example 3:
  Input: s = "[(])"
  Output: false
  Explanation: The brackets are not closed in the correct order.
*/

	const isValidParentheses = s => {
        let [open, sLeng, isValid] = [[], s.length, true]
        if(!s || sLeng < 2) return false

        const brackets = {
            "[": "]",
            "(": ")",
            "{": "}",
        }

        s.split("").forEach(bracket => {
            if(!brackets[bracket]){
                const lastBracket = open.pop()
                if(brackets[lastBracket] !== bracket){
                    isValid = false
                    return
                }
            }else open.push(bracket)
        })

        return open.length > 0 ? false : isValid
    }
    
    const parentheses = "[(])"
    const result = isValidParentheses(parentheses)
    console.log(result) // false

// With love @kouqhar

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Source: Grepper

Tags: javascript parentheses

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Contributed on Nov 30 2024

kouqhar

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spectacle not working kde neon

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