Problem

Given a string s containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

  1. Open brackets must be closed by the same type of brackets.
  2. Open brackets must be closed in the correct order.
  3. Every close bracket has a corresponding open bracket of the same type.

Examples

Example 1:

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Input: s = "()"
Output: true

Example 2:

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

Example 3:

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Input: s = "(]"
Output: false

More Examples

Follow up

  • Only One Type of Parenthesis, Say ()

Solution

A typical problem which can be solved by using a stack data structure.

Version1 - When only 1 parenthesis type is present

Method 1 - Using Stack

Why Stack?

Since the last bracket that is opened must also be the first one to be closed, it makes sense to use a data structure that uses the Last In, First Out (LIFO) principle. Therefore, a stack is a good choice here.

Code

 
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public static boolean isValid(String str) {
	Map<Character, Character> parenthes = new HashMap<>() {{
		put('(', ')');
		put('{', '}');
		put('[', ']');
	}};

	Stack<Character> stack = new Stack<>();
	for (char c : s.toCharArray()) {
		if (parenthese.containsKey(c)) {
			stack.push(c);
		} else {
			if (stack.isEmpty()) {
				return false;
			}
			char cKey = stack.pop();

			char cVal = parenthese.get(cKey);

			if (cVal != c) {
				return false;
			}

		}
	}

	return stack.isEmpty();
}
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We can use counter to reduce space. See method 2 with only 1 type of parenthesis.

Dry Run

Lets see algo in action.

Example 1

Consider the following expression {([])}

Example 2

Example 3

Example 4

Method 2 - Using Counter When Only 1 Type of Parenthesis Present

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Boolean: IsProperlyNested(String: expression)
	Integer: counter = 0
	For Each ch In expression
		If (ch == '(') Then counter = counter + 1
			Else If (ch == ')') Then
				counter = counter - 1
			If (counter < 0) Then Return False
		End If
	Next ch
	If (counter == 0) Then Return True
	 Else Return False
IsProperlyNested