What is the time complexity of Expression Add Operators?

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

What data structures are used to solve Expression Add Operators?

This problem uses Math, String, Backtracking. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Expression Add Operators?

Insight 1: Backtracking is essential because we need to explore all possible combinations of operators and operands. Insight 2: The current expression value and the last operand value need to be tracked during backtracking to handle multiplication correctly due to operator precedence.

Which companies ask LeetCode 282. Expression Add Operators in interviews?

This problem is frequently asked at Pinterest.

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Expression Add Operators - LeetCode 282 Solution

Expression Add Operators - Complete Solution Guide

Expression Add Operators is LeetCode problem 282, a Hard 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 num that contains only digits and an integer target , return all possibilities to insert the binary operators '+' , '-' , and/or '*' between the digits of num so that the resultant expression evaluates to the target value . Note that operands in the returned expressions should not contain leading zeros. Note that a number can contain multiple digits. Example 1: Input: num = "123", target = 6 Output: ["1*2*3","1+2+3"] Explanation: Both "1*2*3" and "1+2+3" evaluate to 6. Example 2:

Detailed Explanation

The problem asks us to find all possible ways to insert binary operators ('+', '-', '*') between the digits of a given string `num` such that the resulting expression evaluates to a specified `target` value. We need to return a list of strings, where each string represents a valid expression. Importantly, operands in the expressions should not contain leading zeros. For example, if `num` is "123" and `target` is 6, the valid expressions are "1*2*3" and "1+2+3". The constraints specify the length of `num` (1 to 10) and the range of `target` which is -2^31 to 2^31 - 1.

Solution Approach

The solution employs a backtracking algorithm to explore all possible expressions. The `backtrack` function recursively builds expressions by iterating through all possible substrings of the input `num`, considering each substring as an operand. At each step, it tries adding '+', '-', and '*' operators and recursively calls itself to explore further possibilities. The base case for the recursion is when the entire input string `num` has been processed. At this point, the current expression's value is compared with the target value. If they match, the current expression is added to the result list. The multiplication operation requires special handling, as it has precedence over addition and subtraction. To handle this, the algorithm keeps track of the last operand used. When multiplication is applied, the current value is adjusted by subtracting the last operand and adding the product of the last operand and the current operand.

Step-by-Step Algorithm

Step 1: Initialize an empty list `ans` to store the valid expressions.
Step 2: Define a recursive `backtrack` function with the following parameters: `index` (current index in `num`), `path` (current expression string), `current_val` (current value of the expression), and `last_val` (the last operand added/subtracted to handle multiplication).
Step 3: Base case: If `index` reaches the end of the `num` string, check if `current_val` is equal to `target`. If it is, add the `path` to `ans`.
Step 4: Iterate from `index` to the end of `num` to extract substrings as operands.
Step 5: Check for leading zeros in the operand substring. If leading zeros exist (e.g., "05"), skip this operand.
Step 6: Convert the operand substring to an integer `operand_val`.
Step 7: If it's the first operand (index == 0), directly pass `operand_val` as the `current_val` and `last_val` to the next `backtrack` call. It sets the starting number.
Step 8: Otherwise, recursively call `backtrack` for three cases: addition, subtraction, and multiplication.
Step 9: For addition, update `path` by appending "+" and `operand_str`, update `current_val` by adding `operand_val`, and update `last_val` to `operand_val`.
Step 10: For subtraction, update `path` by appending "-" and `operand_str`, update `current_val` by subtracting `operand_val`, and update `last_val` to `-operand_val`.
Step 11: For multiplication, update `path` by appending "*" and `operand_str`, update `current_val` by subtracting the previous `last_val` and adding `last_val * operand_val`. Update `last_val` by multiplying it with `operand_val`.
Step 12: Call `backtrack` initially with index 0, empty string for path, and 0 for current_val and last_val.
Step 13: Return `ans`.

Key Insights

Insight 1: Backtracking is essential because we need to explore all possible combinations of operators and operands.
Insight 2: The current expression value and the last operand value need to be tracked during backtracking to handle multiplication correctly due to operator precedence.
Insight 3: Leading zeros in operands should be avoided.
Insight 4: Use `long` data type to avoid overflow issues as intermediate calculations can exceed the range of `int`.

Complexity Analysis

Time Complexity: O(4^n)

Space Complexity: O(n)

Topics

This problem involves: Math, String, Backtracking.

Companies

Asked at: Pinterest.

Expression Add Operators | LeetCode 282

Hard

Math String Backtracking

Time: O(4^n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> addOperators(String num, int target) {
        int n = num.length();
        List<String> ans = new ArrayList<>();

        function backtrack(int index, String path, long current_val, long last_val) {
            if (index == n) {
                if (current_val == target) {
                    ans.add(path);
                }
                return;
            }

            for (int j = index; j < n; j++) {
                String operand_str = num.substring(index, j + 1);

                if (operand_str.length() > 1 && operand_str.charAt(0) == '0') {
                    break;
                }

                long operand_val = Long.parseLong(operand_str);

                if (index == 0) {
                    backtrack(j + 1, operand_str, operand_val, operand_val);
                } else {
                    // Addition
                    backtrack(j + 1, path + "+" + operand_str, current_val + operand_val, operand_val);

                    // Subtraction
                    backtrack(j + 1, path + "-" + operand_str, current_val - operand_val, -operand_val);

                    // Multiplication
                    long new_current_val = current_val - last_val + (last_val * operand_val);
                    backtrack(j + 1, path + "*" + operand_str, new_current_val, last_val * operand_val);
                }
            }
        }

        if (num.isEmpty()) {
            return new ArrayList<>();
        }

        backtrack(0, "", 0, 0);
        return ans;
    }

    private void backtrack(int index, String path, long current_val, long last_val) {
         int n = path.length();
         
    }
}

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Problem Info

Description

Problem 282: Expression Add Operators - Detailed explanation and solution approaches.

Topics

Math String Backtracking

Companies

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:282

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

13 steps

1
Step 1: Initialize an empty list `ans` to store the valid expressions.
2
Step 2: Define a recursive `backtrack` function with the following parameters: `index` (current index in `num`), `path` (current expression string), `current_val` (current value of the expression), and `last_val` (the last operand added/subtracted to handle multiplication).
3
Step 3: Base case: If `index` reaches the end of the `num` string, check if `current_val` is equal to `target`. If it is, add the `path` to `ans`.
4
Step 4: Iterate from `index` to the end of `num` to extract substrings as operands.
5
Step 5: Check for leading zeros in the operand substring. If leading zeros exist (e.g., "05"), skip this operand.
6
Step 6: Convert the operand substring to an integer `operand_val`.
7
Step 7: If it's the first operand (index == 0), directly pass `operand_val` as the `current_val` and `last_val` to the next `backtrack` call. It sets the starting number.
8
Step 8: Otherwise, recursively call `backtrack` for three cases: addition, subtraction, and multiplication.
9
Step 9: For addition, update `path` by appending "+" and `operand_str`, update `current_val` by adding `operand_val`, and update `last_val` to `operand_val`.
10
Step 10: For subtraction, update `path` by appending "-" and `operand_str`, update `current_val` by subtracting `operand_val`, and update `last_val` to `-operand_val`.
11
Step 11: For multiplication, update `path` by appending "*" and `operand_str`, update `current_val` by subtracting the previous `last_val` and adding `last_val * operand_val`. Update `last_val` by multiplying it with `operand_val`.
12
Step 12: Call `backtrack` initially with index 0, empty string for path, and 0 for current_val and last_val.
13
Step 13: Return `ans`.

💡

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)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity arises from exploring all possible combinations of operators between the digits. The space complexity is mainly determined by the recursion depth and the length of the `path` string during recursion.

📖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

Dynamic programming could potentially be used, but it would be significantly more complex to implement due to the need to track the intermediate expression values and the last operand. It might involve creating a table to store intermediate results for different substrings and target values.

Time: Potentially exponential, similar to backtrackingSpace: Potentially exponential, similar to backtracking

Trade-offs: Backtracking is generally preferred for its simplicity and directness in exploring the solution space, given the constraints of the problem.

✨

Key Insights

1Insight 1: Backtracking is essential because we need to explore all possible combinations of operators and operands.
2Insight 2: The current expression value and the last operand value need to be tracked during backtracking to handle multiplication correctly due to operator precedence.
3Insight 3: Leading zeros in operands should be avoided.
4Insight 4: Use `long` data type to avoid overflow issues as intermediate calculations can exceed the range of `int`.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Leading zeros in operands. The code explicitly checks for this and skips such operands. This prevents incorrect expressions like "1+05" which should not be generated from "105".
Edge case 2: Empty input string. The code handles this by returning an empty list.
Edge case 3: Integer overflow. The code uses `long` to store intermediate values, reducing the chance of integer overflow, which could lead to incorrect results. It handles cases where values might go out of integer range during intermediate calculation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling operator precedence correctly, especially multiplication. Forgetting to update the current value and the last operand can lead to incorrect results.

✗Common mistake 2: Integer overflow due to intermediate calculations exceeding the `int` range. Use `long` to avoid this.

✗Common mistake 3: Not checking for leading zeros in operands, which violates the problem constraints.

✗Common mistake 4: Incorrectly implementing backtracking leading to missed cases, especially forgetting base cases.

✅

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: Expression parsing in compilers and interpreters. This problem is a simplified version of the general problem of parsing and evaluating mathematical expressions.

🏦Real-world scenario 2: Generating possible solutions in AI planning. If you need to find a sequence of operations that results in a specific target state, this type of approach could be useful.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: September 11, 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 13

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

@CodeNinjaJul 6

❓

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.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Expression Add Operators"?

For this hard-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?