What is the time complexity of Different Ways to Add Parentheses?

The optimal solution has a time complexity of O(4^n/n^(3/2)).

What is the space complexity of Different Ways to Add Parentheses?

The space complexity is O(4^n/n^(3/2)).

What data structures are used to solve Different Ways to Add Parentheses?

This problem uses Math, String, Dynamic Programming, Recursion, Memoization. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Different Ways to Add Parentheses?

Insight 1: The problem can be solved by breaking down the expression recursively at each operator. The results of the left and right sub-expressions are then combined based on the operator. Insight 2: Memoization is crucial for optimizing the recursive solution because the same sub-expressions are evaluated multiple times, leading to exponential time complexity without it.

Which companies ask LeetCode 241. Different Ways to Add Parentheses in interviews?

This problem is frequently asked at DeltaX, Salesforce.

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Different Ways to Add Parentheses - LeetCode 241 Solution

Different Ways to Add Parentheses - Complete Solution Guide

Different Ways to Add Parentheses is LeetCode problem 241, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators . You may return the answer in any order . The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does not exceed 10 4 . Example 1: Input: expression = "2-1-1" Output: [0,2] Explanation: ((2-1)-1) = 0 (2-(1-1)) = 2 Example 2: Input: expression = "2*3-4*5" Output: [-34,-14,-10,-10

Detailed Explanation

The problem asks us to take a string expression containing numbers and operators (+, -, *) and find all possible results by adding parentheses in different ways. The input expression consists of digits and the operators '+', '-', and '*'. The goal is to return a list of all possible numerical results after evaluating the expression with every possible parenthesization. The order of the results doesn't matter. For example, given "2-1-1", we can parenthesize it as "(2-1)-1" which equals 0, or as "2-(1-1)" which equals 2. Therefore, the output should be [0, 2]. The constraints limit the expression length to 20 and the values to be within a 32-bit integer.

Solution Approach

The solution uses a recursive approach with memoization to efficiently explore all possible parenthesizations of the expression. The core idea is to iterate through the expression, and for each operator found, split the expression into two sub-expressions (left and right). Recursively compute all possible results for both the left and right sub-expressions. Then, combine the results by applying the operator to each pair of results from the left and right sides. The memoization technique stores the results of already computed sub-expressions in a hash map to avoid redundant computations, significantly improving the performance.

Step-by-Step Algorithm

Step 1: Implement a recursive function `compute(expr)` that takes a string expression as input.
Step 2: Check if the result for the current expression is already memoized. If yes, return the memoized result.
Step 3: If the expression contains only digits (no operators), convert it to an integer and return a list containing only that integer. This is the base case.
Step 4: Iterate through the expression. When an operator (+, -, *) is encountered at index `i`, split the expression into two sub-expressions: `expr[:i]` (left) and `expr[i+1:]` (right).
Step 5: Recursively call `compute()` on the left and right sub-expressions to get lists of possible results for each side.
Step 6: Iterate through all possible results from the left and right sub-expressions and combine them using the operator at index `i`. Add the combined results to the `results` list.
Step 7: Store the `results` list in the memoization table with the current expression as the key.
Step 8: Return the `results` list.

Key Insights

Insight 1: The problem can be solved by breaking down the expression recursively at each operator. The results of the left and right sub-expressions are then combined based on the operator.
Insight 2: Memoization is crucial for optimizing the recursive solution because the same sub-expressions are evaluated multiple times, leading to exponential time complexity without it.
Insight 3: The base case for the recursion is when the sub-expression is a number (i.e., contains no operators). In this case, the result is simply the integer value of the number.

Complexity Analysis

Time Complexity: O(4^n/n^(3/2))

Space Complexity: O(4^n/n^(3/2))

Topics

This problem involves: Math, String, Dynamic Programming, Recursion, Memoization.

Companies

Asked at: DeltaX, Salesforce.

Different Ways to Add Parentheses | LeetCode 241

Medium

Math String Dynamic Programming Recursion Memoization

Time: O(4^n/n^(3/2))Space: O(4^n/n^(3/2))

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<Integer> diffWaysToCompute(String expression) {
        Map<String, List<Integer>> memo = new HashMap<>();

        return compute(expression, memo);
    }

    private List<Integer> compute(String expr, Map<String, List<Integer>> memo) {
        if (memo.containsKey(expr)) {
            return memo.get(expr);
        }

        if (expr.matches("\\d+")) {
            List<Integer> result = new ArrayList<>();
            result.add(Integer.parseInt(expr));
            return result;
        }

        List<Integer> results = new ArrayList<>();
        for (int i = 0; i < expr.length(); i++) {
            char charAtI = expr.charAt(i);
            if (charAtI == '+' || charAtI == '-' || charAtI == '*') {
                List<Integer> left_results = compute(expr.substring(0, i), memo);
                List<Integer> right_results = compute(expr.substring(i + 1), memo);

                for (int l : left_results) {
                    for (int r : right_results) {
                        if (charAtI == '+') {
                            results.add(l + r);
                        } else if (charAtI == '-') {
                            results.add(l - r);
                        } else {
                            results.add(l * r);
                        }
                    }
                }
            }
        }

        memo.put(expr, results);
        return results;
    }
}

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Problem Info

Description

Problem 241: Different Ways to Add Parentheses - Detailed explanation and solution approaches.

Topics

Math String Dynamic Programming Recursion Memoization

Companies

DeltaXSalesforce

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:241

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Implement a recursive function `compute(expr)` that takes a string expression as input.
2
Step 2: Check if the result for the current expression is already memoized. If yes, return the memoized result.
3
Step 3: If the expression contains only digits (no operators), convert it to an integer and return a list containing only that integer. This is the base case.
4
Step 4: Iterate through the expression. When an operator (+, -, *) is encountered at index `i`, split the expression into two sub-expressions: `expr[:i]` (left) and `expr[i+1:]` (right).
5
Step 5: Recursively call `compute()` on the left and right sub-expressions to get lists of possible results for each side.
6
Step 6: Iterate through all possible results from the left and right sub-expressions and combine them using the operator at index `i`. Add the combined results to the `results` list.
7
Step 7: Store the `results` list in the memoization table with the current expression as the key.
8
Step 8: Return the `results` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(4^n/n^(3/2))

💾

Space Complexity

O(4^n/n^(3/2))

📝

Why?

The recursive nature leads to an exponential number of calls initially. The memoization significantly reduces the number of actual computations by storing and reusing results of subproblems. The time and space complexities are derived from the number of possible ways to parenthesize the expression, which is bounded by the Catalan number.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming (Bottom-up)

Instead of using recursion, a bottom-up dynamic programming approach can be used. We start by computing the results for small sub-expressions and build up to the entire expression. A table stores results for sub-expressions.

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

Trade-offs: While having potentially better theoretical time complexity in some cases, the bottom-up DP approach can be harder to implement and understand compared to the recursive memoization approach. The recursive solution is also naturally suited to the divide-and-conquer paradigm.

✨

Key Insights

1Insight 1: The problem can be solved by breaking down the expression recursively at each operator. The results of the left and right sub-expressions are then combined based on the operator.
2Insight 2: Memoization is crucial for optimizing the recursive solution because the same sub-expressions are evaluated multiple times, leading to exponential time complexity without it.
3Insight 3: The base case for the recursion is when the sub-expression is a number (i.e., contains no operators). In this case, the result is simply the integer value of the number.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Recursion

A function that calls itself to solve smaller versions of the same problem.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Single number input: If the input expression is just a number (e.g., "5"), the solution correctly parses it and returns a list containing only that number.
Edge case 2: Input with only one operator: If the input expression has only one operator, the solution correctly evaluates it and returns the single possible result.
Edge case 3: Empty expression: While the problem statement constraints say that the expression length is at least 1, a robust solution would ideally handle an empty string gracefully, possibly by returning an empty list.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using memoization: Forgetting to memoize leads to exponential time complexity, resulting in a Time Limit Exceeded (TLE) error for larger inputs.

✗Common mistake 2: Incorrect base case: Handling the base case (when the expression is just a number) incorrectly can lead to errors in the overall result.

✗Common mistake 3: Incorrect splitting of the expression: Splitting the expression at the wrong indices or not handling the operator correctly will result in wrong computations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Compiler Design: In compiler design, this algorithm can be applied to parse and evaluate arithmetic expressions in a programming language.

🏦Real-world scenario 2: Financial Modeling: Evaluating complex financial models often involves arithmetic expressions, where the order of operations (parenthesization) can significantly impact the result. This algorithm can be used to explore different scenarios.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 1, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Different Ways to Add Parentheses"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?