Understanding the Problem
We are given a string that may contain a mix of characters, including parentheses (
and )
. Our goal is to determine the maximum nesting depth of parentheses in the string.
But what exactly does "nesting depth" mean? It refers to the number of parentheses that are open at the same time during the reading of the string. Think of it like keeping track of how deep inside layers of parentheses we are.
For example, in the expression ((a + b))
, when we reach the innermost a + b
, we are two layers deep — that's a nesting depth of 2.
Step-by-Step Solution with Example
Step 1: Initialize two variables
We need to track:
currentDepth
: How many parentheses are currently open.
maxDepth
: The maximum value that currentDepth
reaches throughout the string.
We start both at 0.
Step 2: Traverse the string character by character
Loop through the string. For each character:
- If it's an opening bracket
(
, we increase currentDepth
by 1.
- Update
maxDepth
if currentDepth
exceeds it.
- If it's a closing bracket
)
, we decrease currentDepth
by 1.
- Ignore all other characters.
Step 3: Example Walkthrough
Let’s take the example string: (1+(2*3)+((8)/4))+1
We scan from left to right:
Char ( → currentDepth = 1 → maxDepth = 1
Char 1 → skip
Char + → skip
Char ( → currentDepth = 2 → maxDepth = 2
Char 2 → skip
Char * → skip
Char 3 → skip
Char ) → currentDepth = 1
Char + → skip
Char ( → currentDepth = 2
Char ( → currentDepth = 3 → maxDepth = 3 ✅
Char 8 → skip
Char ) → currentDepth = 2
Char / → skip
Char 4 → skip
Char ) → currentDepth = 1
Char ) → currentDepth = 0
Char + → skip
Char 1 → skip
Final maxDepth = 3. So, the maximum nesting is 3.
Edge Cases
- Empty String: No parentheses → nesting depth is
0
.
- No parentheses at all: For example,
"abc+123"
→ depth is 0
.
- Only one pair:
"(x)"
→ depth is 1
.
- Deep nesting:
"((((x))))"
→ depth is 4
.
- Invalid or mismatched parentheses: If we assume valid input only, we don’t handle errors. But ideally, if we encounter a closing
)
without a matching open (
, we should either ignore or throw an error based on problem constraints.
Finally
The beauty of this problem is its simplicity. You don't need a complex data structure like a stack — just a simple counter to track the current level of nesting. It's a classic example of how a linear pass through a string with just a few variables can solve what seems like a tricky problem.
The overall time complexity is O(n), where n is the length of the string, and the space complexity is O(1).
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