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  1. Prepare
  2. Data Structures
  3. Stacks
  4. Balanced Brackets
  5. Discussions

Balanced Brackets

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1628 Discussions

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  • awadallayossef
     + 0 comments
    def isBalanced(s):
    closeToOpen = { ")" : "(", "]" : "[", "}" : "{" }
    Stack = []

    for x in s:
        if x in closeToOpen:
            if Stack and Stack[-1] == closeToOpen[x]:
                Stack.pop()
            else:
                return "NO"
        else:
            Stack.append(x)

    return "NO" if len(Stack) > 0 else "YES"

  • jain_sushmit9
     + 0 comments

    C++:

    string isBalanced(string s) { std::map m {{'(',')'}, {'{','}'}, {'[',']'}}; std::stack st;

    if (s.length() %2 != 0)
    return("NO");
for (int i=0;i<s.length();i++)
{
    auto it = m.find(s[i]);
    if(it != m.end())
    {
        st.push(s[i]);
    }
    else
    {
        if(st.empty())
        {
            return("NO");
        }
        if(m[st.top()] == s[i])
            st.pop();
        else
            return("NO");
    }
}
if(st.empty())
    return("YES");
else    
    return("NO");

    }

  • joosx2000
     + 0 comments

    Simple java solution in O(n) 100%:

    public static String isBalanced(String s) {
    Stack<Character> stack = new Stack<>();
    Map<Character, Character> dict = Map.of('{','}', '[',']', '(',')');

    for (char c : s.toCharArray()) {
        if (dict.containsKey(c)) stack.push(dict.get(c));
        else if (stack.isEmpty() || c != stack.pop()) return "NO";
    }

    return stack.isEmpty() ? "YES" : "NO";
}

  • alex1269
     + 0 comments

    Python 3:

    Write your code here

    stack = []
pairs = {
    '[': ']',
    '{': '}',
    '(': ')'
}

for bracket in s:
    if bracket in pairs:
        stack.append(bracket)
    elif not stack or pairs.get(stack.pop(-1)) != bracket:
        return 'NO'

return 'YES' if not stack else 'NO'

  • jono_reshef
     + 0 comments

    Python3

    def isBalanced(s):
    # Valid combos that must exist as pairs
    combos = ["[]", "{}", "()"]
    
    # While there are characters available, attempt to remove combos
    while len(s) > 0:
        new_s = s
        for c in combos:
            new_s = new_s.replace(c, "")
        # If the new_s has not changed length (ie no valid combos) return NO
        if len(new_s) == len(s):
            return "NO"
        s = new_s
    return "YES"