How do you solve Different Ways to Add Parentheses on LeetCode?

Different Ways to Add Parentheses (LeetCode #241) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n^3) time complexity and O(n) space complexity. Key insight: Recursive breakdown of expressions leads to overlapping subproblems.

What is the brute force solution for Different Ways to Add Parentheses?

The brute force approach for Different Ways to Add Parentheses is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves evaluating every possible way to group the numbers and operators in the expression. This means recursively splitting the expression at each operator and calculating all possible results for each split. The optimal Optimal Solution approach improves this to O(n^3).

What are the interview tips for Different Ways to Add Parentheses?

Explain your thought process clearly as you code. Discuss the time and space complexity of your solution. Consider edge cases and how they affect your solution.

What are common mistakes when solving Different Ways to Add Parentheses?

Not using memoization in recursive solutions. Failing to handle base cases correctly.

#241

Different Ways to Add Parentheses

Medium

MathStringDynamic ProgrammingRecursionMemoizationRecursionMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n^3)
Space	O(n)	O(n)

💡

Intuition

Time O(n^3)Space O(n)

The optimal solution uses memoization to store results of previously computed expressions, avoiding redundant calculations. This reduces the time complexity significantly by ensuring each unique expression is computed only once.

⚙️

Algorithm

3 steps

1Step 1: Create a memoization dictionary to store results for expressions already computed.
2Step 2: For each operator in the expression, split it and recursively compute results for the left and right parts, checking the memoization dictionary first.
3Step 3: Combine results from the left and right parts using the operator and store the results in the memoization dictionary.

solution.py20 lines

1def diffWaysToCompute(expression, memo={}):
2    if expression in memo:
3        return memo[expression]
4    if expression.isdigit():
5        return [int(expression)]
6    results = []
7    for i, char in enumerate(expression):
8        if char in '+-*':
9            left = diffWaysToCompute(expression[:i], memo)
10            right = diffWaysToCompute(expression[i + 1:], memo)
11            for l in left:
12                for r in right:
13                    if char == '+':
14                        results.append(l + r)
15                    elif char == '-':
16                        results.append(l - r)
17                    elif char == '*':
18                        results.append(l * r)
19    memo[expression] = results
20    return results

ℹ

Complexity note: The time complexity is O(n^3) due to the recursive calls and the memoization lookup. The space complexity is O(n) due to the memoization storage.

1Recursive breakdown of expressions leads to overlapping subproblems.
2Memoization can significantly reduce redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.