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Home/ Questions/Q 757857
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T15:24:43+00:00 2026-05-14T15:24:43+00:00

I’ve written software in the past that uses a stack to check for balanced

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I’ve written software in the past that uses a stack to check for balanced equations, but now I’m asked to write a similar algorithm recursively to check for properly nested brackets and parenthesis.

Good examples: () [] ()
([]()[])

Bad examples: ( (] ([)]

Suppose my function is called: isBalanced.

Should each pass evaluate a smaller substring (until reaching a base case of 2 left)? Or, should I always evaluate the full string and move indices inward?

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  1. Editorial Team
    Editorial Team
    2026-05-14T15:24:43+00:00Added an answer on May 14, 2026 at 3:24 pm

    There are many ways to do this, but the simplest algorithm is to simply process forward left to right, passing the stack as a parameter

    FUNCTION isBalanced(String input, String stack) : boolean
      IF isEmpty(input)
        RETURN isEmpty(stack)
      ELSE IF isOpen(firstChar(input))
        RETURN isBalanced(allButFirst(input), stack + firstChar(input))
      ELSE IF isClose(firstChar(input))
        RETURN NOT isEmpty(stack) AND isMatching(firstChar(input), lastChar(stack))
          AND isBalanced(allButFirst(input), allButLast(stack))
      ELSE
        ERROR "Invalid character"
    

    Here it is implemented in Java. Note that I’ve switched it now so that the stack pushes in front instead of at the back of the string, for convenience. I’ve also modified it so that it just skips non-parenthesis symbols instead of reporting it as an error.

    static String open  = "([<{";
    static String close = ")]>}";
    
    static boolean isOpen(char ch) {
        return open.indexOf(ch) != -1;
    }
    static boolean isClose(char ch) {
        return close.indexOf(ch) != -1;
    }
    static boolean isMatching(char chOpen, char chClose) {
        return open.indexOf(chOpen) == close.indexOf(chClose);
    }
    
    static boolean isBalanced(String input, String stack) {
        return
            input.isEmpty() ?
                stack.isEmpty()
            : isOpen(input.charAt(0)) ?
                isBalanced(input.substring(1), input.charAt(0) + stack)
            : isClose(input.charAt(0)) ?
                !stack.isEmpty() && isMatching(stack.charAt(0), input.charAt(0))
                  && isBalanced(input.substring(1), stack.substring(1))
            : isBalanced(input.substring(1), stack);
    }
    

    Test harness:

        String[] tests = {
            "()[]<>{}",
            "(<",
            "]}",
            "()<",
            "(][)",
            "{(X)[XY]}",
        };
        for (String s : tests) {
            System.out.println(s + " = " + isBalanced(s, ""));
        }
    

    Output:

    ()[]<>{} = true
    (< = false
    ]} = false
    ()< = false
    (][) = false
    {(X)[XY]} = true
    
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