Got it, Rakesh 🚀 Let’s turn this into a neat blog post. I’ll explain the problem, walk through the approach, give a complexity analysis, and finally embed your visualization iframe.
Here’s a polished draft for your blog:
Generate Parentheses — Explained with Backtracking & Visualization
One of the classic problems in coding interviews (especially LeetCode #22) is "Generate Parentheses". It’s a great way to understand backtracking, recursion, and how constraints guide the search space.
Problem Statement
Given n
pairs of parentheses, generate all possible combinations of well-formed parentheses.
Example:
Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
Approach
This problem is a textbook case for backtracking.
We want to build strings step by step, only adding characters when they keep the parentheses valid.
Rules for Valid Parentheses:
- We can add an opening parenthesis
"("
as long as we haven’t used up alln
. - We can add a closing parenthesis
")"
only if the number of closing brackets is less than the number of opening ones so far (to avoid invalid strings like")("
).
We use a stack to build the current sequence and a recursive function to explore all possibilities.
Code Implementation
/**
* @param {number} n
* @return {string[]}
*/
var generateParenthesis = function (n) {
let stack = [];
let res = [];
const backtrack = (openN, closedN) => {
// Base case: if both counts reach n, we found a valid string
if (openN === n && closedN === n) {
res.push(stack.join(""));
return;
}
// Add "(" if we still have openings left
if (openN < n) {
stack.push("(");
backtrack(openN + 1, closedN);
stack.pop(); // backtrack
}
// Add ")" if it's valid
if (openN > closedN) {
stack.push(")");
backtrack(openN, closedN + 1);
stack.pop(); // backtrack
}
};
backtrack(0, 0);
return res;
};
Complexity Analysis
-
Time Complexity:
Each valid sequence has length
2n
. The number of valid sequences is the nth Catalan number:
$$
C_n = \frac{1}{n+1} \binom{2n}{n}
$$
So, the complexity is O(Cn × n) (since joining each sequence takes O(n)
time).
-
Space Complexity:
- The recursion depth is at most
2n
. - We use a stack of size
2n
in the worst case. - Result storage takes
O(Cn × n)
.
- The recursion depth is at most
So overall space complexity is O(Cn × n).
Visualization 🎨
Here’s an interactive visualization of the backtracking tree:
Final Thoughts
This problem is a perfect introduction to backtracking. It teaches how to explore a search space while pruning invalid paths early. Once you master this, you’ll notice a similar structure in problems like N-Queens, Combinations, and Subsets.
👉 Tip: When stuck in a backtracking problem, always ask:
- “What choices do I have at this step?”
- “What constraints stop me from making a choice?”
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