What is the time complexity of Minimize Result by Adding Parentheses to Expression?

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

What data structures are used to solve Minimize Result by Adding Parentheses to Expression?

This problem uses String, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimize Result by Adding Parentheses to Expression?

Insight 1: The problem can be solved by iterating through all possible placements of the parentheses and evaluating the resulting expressions. Insight 2: We need to handle cases where the parentheses are placed at the very beginning or end of the numbers, effectively multiplying the result of the parenthesized expression by 1 if there are no numbers outside the parenthesis on that side.

Which companies ask LeetCode 2232. Minimize Result by Adding Parentheses to Expression in interviews?

This problem is frequently asked at Pinterest, Snap.

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Minimize Result by Adding Parentheses to Expression - LeetCode 2232 Solution

Minimize Result by Adding Parentheses to Expression - Complete Solution Guide

Minimize Result by Adding Parentheses to Expression is LeetCode problem 2232, 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

You are given a 0-indexed string expression of the form "<num1>+<num2>" where <num1> and <num2> represent positive integers. Add a pair of parentheses to expression such that after the addition of parentheses, expression is a valid mathematical expression and evaluates to the smallest possible value. The left parenthesis must be added to the left of '+' and the right parenthesis must be added to the right of '+' . Return expression after adding a pair of parentheses such that expression evaluate

Detailed Explanation

The problem asks us to find the optimal placement of parentheses in a string expression of the form "<num1>+<num2>" such that the resulting expression evaluates to the smallest possible value. The left parenthesis must be placed to the left of the '+' sign, and the right parenthesis must be placed to the right of the '+' sign. The goal is to return the expression with the parentheses inserted in the optimal position. The input expression consists only of digits and a single '+' sign. We need to consider all possible positions for the parentheses and choose the one that yields the minimum value.

Solution Approach

The provided solution uses a brute-force approach. It iterates through all possible positions to place the left parenthesis within the first number and the right parenthesis within the second number. For each combination of parenthesis placement, it calculates the value of the resulting expression. The solution maintains a running minimum value and the corresponding expression string. After iterating through all possibilities, it returns the expression that corresponds to the minimum value found.

Step-by-Step Algorithm

Step 1: Find the index of the '+' character in the input string.
Step 2: Split the input string into two substrings, `num1_str` (left of '+') and `num2_str` (right of '+').
Step 3: Initialize `min_val` to infinity (or a very large number) and `result_expr` to an empty string.
Step 4: Iterate through all possible positions `i` for the left parenthesis in `num1_str` (from 0 to length of `num1_str`).
Step 5: Iterate through all possible positions `j` for the right parenthesis in `num2_str` (from 0 to length of `num2_str` - 1). The index j represents how many characters from the start of num2_str will be inside the parenthesis.
Step 6: For each combination of `i` and `j`, create the substrings `a_str`, `b_str`, `c_str`, and `d_str` representing the parts of the numbers outside and inside the parentheses: a_str is num1_str[:i], b_str is num1_str[i:], c_str is num2_str[:j+1], and d_str is num2_str[j+1:].
Step 7: Convert the substrings to integers. If a substring is empty, treat it as 1 (to simulate multiplication by 1).
Step 8: Calculate the value of the expression `val_a * (val_b + val_c) * val_d`.
Step 9: If the calculated value is less than `min_val`, update `min_val` and `result_expr` with the current value and the corresponding expression string.
Step 10: After iterating through all possible parenthesis placements, return the `result_expr`.

Key Insights

Insight 1: The problem can be solved by iterating through all possible placements of the parentheses and evaluating the resulting expressions.
Insight 2: We need to handle cases where the parentheses are placed at the very beginning or end of the numbers, effectively multiplying the result of the parenthesized expression by 1 if there are no numbers outside the parenthesis on that side.
Insight 3: Integer overflow is mentioned in the prompt and needs to be considered. The problem statement guarantees that intermediate calculations fit in a 32-bit integer. Thus no explicit overflow handling is needed.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

This problem involves: String, Enumeration.

Companies

Asked at: Pinterest, Snap.

Minimize Result by Adding Parentheses to Expression | LeetCode 2232

Medium

String Enumeration

Time: O(n^2)Space: O(1)

Solution Code

Available in:

class Solution {
    public String minimizeResult(String expression) {
        int plus_index = expression.indexOf('+');
        String num1_str = expression.substring(0, plus_index);
        String num2_str = expression.substring(plus_index + 1);
        
        int min_val = Integer.MAX_VALUE;
        String result_expr = "";
        
        for (int i = 0; i < num1_str.length(); i++) {
            for (int j = 0; j < num2_str.length(); j++) {
                
                String a_str = num1_str.substring(0, i);
                String b_str = num1_str.substring(i);
                
                String c_str = num2_str.substring(0, j + 1);
                String d_str = num2_str.substring(j + 1);
                
                int val_a = a_str.isEmpty() ? 1 : Integer.parseInt(a_str);
                int val_b = Integer.parseInt(b_str);
                int val_c = Integer.parseInt(c_str);
                int val_d = d_str.isEmpty() ? 1 : Integer.parseInt(d_str);
                
                int current_val = val_a * (val_b + val_c) * val_d;
                
                if (current_val < min_val) {
                    min_val = current_val;
                    result_expr = a_str + "(" + b_str + "+" + c_str + ")" + d_str;
                }
            }
        }
        
        return result_expr;
    }
}

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Problem Info

Description

Problem 2232: Minimize Result by Adding Parentheses to Expression - Detailed explanation and solution approaches.

Topics

String Enumeration

Companies

PinterestSnap

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2232

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

10 steps

1
Step 1: Find the index of the '+' character in the input string.
2
Step 2: Split the input string into two substrings, `num1_str` (left of '+') and `num2_str` (right of '+').
3
Step 3: Initialize `min_val` to infinity (or a very large number) and `result_expr` to an empty string.
4
Step 4: Iterate through all possible positions `i` for the left parenthesis in `num1_str` (from 0 to length of `num1_str`).
5
Step 5: Iterate through all possible positions `j` for the right parenthesis in `num2_str` (from 0 to length of `num2_str` - 1). The index j represents how many characters from the start of num2_str will be inside the parenthesis.
6
Step 6: For each combination of `i` and `j`, create the substrings `a_str`, `b_str`, `c_str`, and `d_str` representing the parts of the numbers outside and inside the parentheses: a_str is num1_str[:i], b_str is num1_str[i:], c_str is num2_str[:j+1], and d_str is num2_str[j+1:].
7
Step 7: Convert the substrings to integers. If a substring is empty, treat it as 1 (to simulate multiplication by 1).
8
Step 8: Calculate the value of the expression `val_a * (val_b + val_c) * val_d`.
9
Step 9: If the calculated value is less than `min_val`, update `min_val` and `result_expr` with the current value and the corresponding expression string.
10
Step 10: After iterating through all possible parenthesis placements, return the `result_expr`.

💡

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(n^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n^2) because there are two nested loops, each iterating a maximum of n times, where n is the length of the input expression. The space complexity is O(1) because the extra space used by variables like min_val and result_expr is independent of the input size.

📖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

While not immediately obvious, a dynamic programming approach could be considered for more complex variations of this problem where there are multiple '+' signs or more complex expressions. However, given the small size of the input, dynamic programming would likely be overkill here.

Time: Potentially O(n^3) or higher, depending on the specific implementation.Space: O(n^2) or higher, depending on the specific implementation.

Trade-offs: Dynamic programming would likely be less efficient than the brute-force approach for this problem due to the constraints on input size. It would only be advantageous if the constraints were relaxed.

✨

Key Insights

1Insight 1: The problem can be solved by iterating through all possible placements of the parentheses and evaluating the resulting expressions.
2Insight 2: We need to handle cases where the parentheses are placed at the very beginning or end of the numbers, effectively multiplying the result of the parenthesized expression by 1 if there are no numbers outside the parenthesis on that side.
3Insight 3: Integer overflow is mentioned in the prompt and needs to be considered. The problem statement guarantees that intermediate calculations fit in a 32-bit integer. Thus no explicit overflow handling is needed.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: When the left parenthesis is at the beginning of `num1_str` (i = 0) or the right parenthesis is at the end of `num2_str` (j = len(num2_str) -1), the corresponding `a_str` or `d_str` will be empty. The code correctly handles this by treating the empty string as a value of 1, effectively multiplying the parenthesized expression by 1.
Edge case 2: The entire `num1_str` or `num2_str` could be enclosed within parenthesis. The solution handles these cases correctly because it iterates through all possible 'i' and 'j' values including 0 and the length of the strings.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the substring outside the parentheses is empty. This leads to incorrect calculations as it assumes the values outside the parentheses are zero.

✗Common mistake 2: Not correctly parsing the substrings into integers, leading to incorrect evaluation of the expression.

✗Common mistake 3: Incorrect memory management in C. Not allocating enough memory for result_expr or failing to free dynamically allocated strings.

✅

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: This problem demonstrates a simplified form of expression optimization. In compilers, similar techniques (though far more complex) are used to rearrange expressions to minimize execution time or memory usage.

🏦Real-world scenario 2: In financial modeling, optimizing the placement of operations in a formula can improve the speed and accuracy of calculations when dealing with large datasets.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 12, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimize Result by Adding Parentheses to Expression"?

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

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