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

Balanced Brackets

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1621 Discussions

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  • h2400100036
     + 0 comments

    char brackets[3][2] = { {'['. ']'}, {'(', ')'}, {'{', '}'}, }; char* isBalanced(char* s) { char* stack = calloc(sizeof(char),strlen(s)); int tos = 0; char b; while((b = *s)){ for(int i = 0; i < 3; i++) { if(brackets[i][0] ==b) { stack[tos++] = b; } else if(brackets[i][1] = b) { char left_b = stack[--tos]; if(brackets[i][O] == b) { char left_b = stack[--tos]; if(brackets[i][O] == b){ stack[tos++] = b; } else if((brackets[i][1] == b) { char left_b = stack[--tos]; if(brackets[i][O] != left_b){ return "NO"; } } } ++s; }free(stack); returntosv == 0 ?"YES" : "NO"; }

  • kuro_ece
     + 0 comments

    in C:

    char brackets[3][2] = {
    {'[', ']'},
    {'(', ')'},
    {'{', '}'},
 };
 
char* isBalanced(char* s) {
    char* stack = calloc(sizeof(char), strlen(s));
    int tos = 0;
    char b;
    
    while((b = *s)) {
        for(int i = 0; i < 3; i++) {
            if(brackets[i][0] == b) {
                stack[tos++] = b;
                
            } else if(brackets[i][1] == b) {
                char left_b = stack[--tos];
                if(brackets[i][0] != left_b) {
                    return "NO";
                }
            }
        }
        ++s;
    }
    free(stack);
    return tos == 0 ? "YES" : "NO";
}

  • wilson155079
     + 0 comments
    public static String isBalanced(String s) {
    // Use ArrayDeque instead of Stack for better performance
    Deque<Character> stack = new ArrayDeque<>();

    for (char ch : s.toCharArray()) {
        // Push opening brackets onto the stack
        switch (ch) {
            case '(':
            case '[':
            case '{':
                stack.push(ch);
                break;
            default: 
                // If the stack is empty or brackets don't match, return "NO"
                if (stack.isEmpty() //
                    || !isMatchingPair(stack.pop(), ch)) {
                    return "NO";
                }
                break;
        }
    }
    // If the stack is empty, the string is balanced
    return stack.isEmpty() ? "YES" : "NO";
}

// Helper method to check if brackets match
private static boolean isMatchingPair(char open, char close) {
    return (open == '(' && close == ')') //
        || (open == '[' && close == ']') //
        || (open == '{' && close == '}');
}

  • illogicalsleepi1
     + 0 comments
    def isBalanced(s):
    matching_pairs = {')': '(', '}': '{', ']': '['}
    stack = []
    
    for char in s:
        if char in "({[":
            stack.append(char)
        elif char in ")}]":
            if not stack or stack[-1] != matching_pairs[char]:
                return "NO"
            stack.pop()
    
    return "YES" if not stack else "NO"

  • saha15_5764
     + 0 comments

    This problem Test case 4 may be some issue