Solution to Nesting by codility

9 Replies to “Solution to Nesting by codility”

1. C# solution:

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   [crayon-66e6a9de4960a546865143  ]public int solution(string S) {         // write your code in C# 5.0 with .NET 4.5 (Mono)         int result = 0;         foreach(char element in S){             if(element == '(')             {                 result += 1;             }             else             {                 result -= 1;                 if(result < 0) return 0;             }         }         return result == 0 ? 1:0;     }

   [/crayon]

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2. Adding a Java Solution 100/100:

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   public static final int BALANCED = 1;     public static final int UNBALANCED = 0;     public int solution(String S) {         if (S.length() % 2 != 0) {             return UNBALANCED;         }         int counter = 0, increment = 1;         for (char c : S.toCharArray()) {             if (c == '(') {                 increment = increment > 0 ? increment : increment * -1;             } else if (c == ')') {                 increment = increment < 0 ? increment : increment * -1;             }             counter += increment;             if (counter < 0) {                 return UNBALANCED;             }         }         return counter == 0 ? BALANCED : UNBALANCED;     }

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3. I am wondering how it passed test case )))(((

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   • It does pass. For the first iteration, parentheses -= 1 and return 0.

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4. 1 2 3 4 5 6 7 8 9 10 11

   def solution(S):     stack = []     for p in S:         if p == '(':             stack.append(p)         else:             try:                 stack.pop()             except Exception as e:                 return 0     return 1 if not stack else 0

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5. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

   def solution(S):     # write your code in Python 2.7     queue = []     for elem in S:         if (elem in ["(","{","["]) :             queue.append(elem)         else:             if (len(queue) == 0):                 return 0             if (elem in [")","}","]"]):                 elem_open = queue.pop()                 if (elem_open == "(" and elem == ")" or                     elem_open == "[" and elem == "]" or                     elem_open == "{" and elem == "}"):     #                    print elem_open+"n"                         continue                 else:                     return 0;     if len(queue) == 0:         return 1     else:         return 0

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6. 100% in javascript

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   function solution(S) {     // write your code in JavaScript (Node.js 8.9.4)           var sum = 0;       for (let i = 0; i < S.length; i++) {                  // console.log(A[i]);         if (S[i] =='(') {             sum ++;         } else {             sum --;         }           if (sum < 0) {             return 0;         }     }          if (sum === 0) {         return 1;     } else {         return 0;     } }

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7. To save time, if the length of the string is odd you can return 0 as it’s impossible to create an even number of pairs.

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   • This is a great catch! That said, in some languages, getting the string length is O(N).

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