241. Different Ways to Add Parentheses - Detailed Explanation

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Problem Statement

Description:
You are given a string expression that contains numbers and operators (+, -, *). The task is to return all possible results from computing the expression in all possible ways by adding parentheses in different positions. Each way of adding parentheses may yield a different result.

Example 1:

Input: "2-1-1"
Output: [0, 2]
Explanation:
- Grouping as ((2-1)-1) gives 0.
- Grouping as (2-(1-1)) gives 2.

Example 2:

Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation:
There are several ways to add parentheses; for example:
- Grouping as (2*(3-(4*5))) gives -34.
- Grouping as ((2*3)-(4*5)) gives -14.
- Grouping as ((2*(3-4))*5) gives -10.
- Grouping as (2*((3-4)*5)) gives -10.
- Grouping as (((2*3)-4)*5) gives 10.

Constraints:

The input string is non-empty.
The expression contains only digits and the operators '+', '-', and '*'.
The numbers can be multiple digits.

Intuition and Hints

Divide and Conquer:
The problem naturally lends itself to a divide-and-conquer approach. For every operator in the expression, consider it as a potential split point. Compute all possible results for the left and right parts, and then combine these results using the operator.
Recursive Evaluation:
If the expression does not contain any operators (i.e., it’s just a number), simply return that number.
Memoization:
Since many sub-expressions may be evaluated repeatedly, caching (memoization) is beneficial to avoid duplicate computations and improve performance.
Handling Multi-digit Numbers:
Make sure to parse multi-digit numbers correctly when the expression does not contain an operator.

Approaches

Approach 1: Recursive Divide and Conquer (Without Memoization)

Explanation:

Base Case:
- If the expression is a number (no operator present), convert it to an integer and return it as a single-element list.
Recursive Case:
- Iterate over the string; when an operator is encountered:
  - Split the expression into two parts: left and right.
  - Recursively compute all possible results for the left sub-expression.
  - Recursively compute all possible results for the right sub-expression.
  - Combine each result from the left with each result from the right using the current operator.
Return the Combined Results:
- Collect and return all the possible outcomes from these combinations.

Python Code (Recursive Divide and Conquer):

Python3

. . . .

Java Code (Recursive Divide and Conquer):

Java

. . . .

Approach 2: Recursive Divide and Conquer with Memoization

Explanation:

To avoid recalculating the results for the same sub-expression multiple times:

Memoization Map:
Use a hash map (or dictionary) to cache the results of sub-expressions.
Check Cache:
Before computing the result for an expression, check if it’s already in the cache.
Store and Return:
After computing the results for an expression, store it in the cache. Then, return the cached results.

This improvement reduces redundant computation and speeds up the algorithm considerably.

Python Code (Recursive with Memoization):

Python3

. . . .

Java Code (Recursive with Memoization):

Java

. . . .

Complexity Analysis

Time Complexity:
Without memoization, the algorithm may explore many redundant sub-expressions, resulting in exponential time complexity in the worst case.
With memoization, each unique sub-expression is computed only once. The number of unique sub-expressions is bounded by O(n²) (for an expression of length n), making the solution much more efficient.
Space Complexity:
The space required for recursion and memoization is O(n²) in the worst case, where n is the length of the expression.

Common Pitfalls & Edge Cases

Handling Multi-digit Numbers:
Ensure that the code correctly treats multi-digit numbers as a single operand.
Empty or Invalid Expressions:
The input is assumed to be valid per the problem constraints, but always consider error-checking in production code.
Memoization Cache:
Failing to use memoization might lead to significant performance issues for larger expressions.
Operator Precedence:
The problem asks for all possible results from different parenthesizations, so standard operator precedence does not apply.