What is the time complexity of Multiply Strings?

The optimal solution has a time complexity of O(m*n).

What is the space complexity of Multiply Strings?

The space complexity is O(m+n).

What data structures are used to solve Multiply Strings?

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

What are the key insights for solving Multiply Strings?

Insight 1: Multiplying strings requires simulating manual multiplication, keeping track of carries. Insight 2: The maximum length of the resulting string is the sum of the lengths of the input strings (m + n).

Which companies ask LeetCode 43. Multiply Strings in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Meta and 7 more companies.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Multiply Strings - LeetCode 43 Solution

Q: How to solve Multiply Strings?

The provided solution simulates manual multiplication. It iterates through the digits of `num1` and `num2` from right to left. For each pair of digits, it calculates their product and adds it to the appropriate position in the `res` array. The `res` array is initialized with zeros and has a size equal to the sum of the lengths of `num1` and `num2`. Carry-over values are handled by propagating them to the next higher digit position in the `res` array. Finally, the `res` array is converted to a string, removing any leading zeros.

Multiply Strings - Complete Solution Guide

Multiply Strings is LeetCode problem 43, 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 two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2 , also represented as a string. Note: You must not use any built-in BigInteger library or convert the inputs to integer directly. Example 1: Input: num1 = "2", num2 = "3" Output: "6" Example 2: Input: num1 = "123", num2 = "456" Output: "56088" Constraints: 1 <= num1.length, num2.length <= 200 num1 and num2 consist of digits only. Both num1 and num2 do not contain any leading zero, except the

Detailed Explanation

The problem asks us to multiply two non-negative integers, `num1` and `num2`, which are given as strings. We need to return the product of these numbers, also represented as a string. The crucial constraint is that we cannot use any built-in BigInteger library or convert the inputs to integers directly. This means we must implement the multiplication algorithm manually, digit by digit, similar to how we would do it on paper.

Solution Approach

The provided solution simulates manual multiplication. It iterates through the digits of `num1` and `num2` from right to left. For each pair of digits, it calculates their product and adds it to the appropriate position in the `res` array. The `res` array is initialized with zeros and has a size equal to the sum of the lengths of `num1` and `num2`. Carry-over values are handled by propagating them to the next higher digit position in the `res` array. Finally, the `res` array is converted to a string, removing any leading zeros.

Step-by-Step Algorithm

Step 1: Handle the edge case where either `num1` or `num2` is '0'. In this case, the product is '0', so return '0'.
Step 2: Initialize an integer array `res` of size `m + n` with all elements set to 0. Here, `m` is the length of `num1` and `n` is the length of `num2`.
Step 3: Iterate through `num1` from right to left (from index `m - 1` to 0).
Step 4: In the outer loop, iterate through `num2` from right to left (from index `n - 1` to 0).
Step 5: Calculate the product of the current digits from `num1` and `num2`. Convert the character digits to integer values.
Step 6: Determine the positions `p1` and `p2` in the `res` array where the product should be added. `p1 = i + j` and `p2 = i + j + 1`. These positions correspond to the tens and units place of the result of multiplying the digits at indices i and j.
Step 7: Add the product to the value already present at `res[p2]`. Store the ones digit of the sum at `res[p2]`, i.e., `res[p2] = total % 10`. Add the tens digit (carry) to `res[p1]`, i.e., `res[p1] += total // 10`.
Step 8: After the nested loops, iterate through the `res` array to find the first non-zero digit.
Step 9: Construct a string from the remaining digits in the `res` array, starting from the first non-zero digit. If all digits are zero, the resulting string will be empty, return '0' instead. Otherwise, return the constructed string.

Key Insights

Insight 1: Multiplying strings requires simulating manual multiplication, keeping track of carries.
Insight 2: The maximum length of the resulting string is the sum of the lengths of the input strings (m + n).
Insight 3: We can use an array to store intermediate results. Each index represents a digit in the final product.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m+n)

Topics

This problem involves: Math, String, Simulation.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Meta, Microsoft, Pinterest, Roku, TikTok, Two Sigma, Uber, X.

Multiply Strings | LeetCode 43

Medium

Math String Simulation

Time: O(m*n)Space: O(m+n)

Solution Code

Available in:

class Solution {
    public String multiply(String num1, String num2) {
        if (num1.equals("0") || num2.equals("0")) {
            return "0";
        }

        int m = num1.length();
        int n = num2.length();
        int[] res = new int[m + n];

        for (int i = m - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                int mul = (num1.charAt(i) - '0') * (num2.charAt(j) - '0');
                int p1 = i + j;
                int p2 = i + j + 1;

                int total = mul + res[p2];

                res[p2] = total % 10;
                res[p1] += total / 10;
            }
        }

        int i = 0;
        while (i < res.length && res[i] == 0) {
            i++;
        }

        StringBuilder sb = new StringBuilder();
        for (; i < res.length; i++) {
            sb.append(res[i]);
        }

        return sb.toString();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 43: Multiply Strings - Detailed explanation and solution approaches.

Topics

Math String Simulation

Companies

AdobeAmazonAppleBloombergMetaMicrosoftPinterestRoku

Stats

Difficulty:Medium

Languages:4

Companies:12

Problem #:43

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

9 steps

1
Step 1: Handle the edge case where either `num1` or `num2` is '0'. In this case, the product is '0', so return '0'.
2
Step 2: Initialize an integer array `res` of size `m + n` with all elements set to 0. Here, `m` is the length of `num1` and `n` is the length of `num2`.
3
Step 3: Iterate through `num1` from right to left (from index `m - 1` to 0).
4
Step 4: In the outer loop, iterate through `num2` from right to left (from index `n - 1` to 0).
5
Step 5: Calculate the product of the current digits from `num1` and `num2`. Convert the character digits to integer values.
6
Step 6: Determine the positions `p1` and `p2` in the `res` array where the product should be added. `p1 = i + j` and `p2 = i + j + 1`. These positions correspond to the tens and units place of the result of multiplying the digits at indices i and j.
7
Step 7: Add the product to the value already present at `res[p2]`. Store the ones digit of the sum at `res[p2]`, i.e., `res[p2] = total % 10`. Add the tens digit (carry) to `res[p1]`, i.e., `res[p1] += total // 10`.
8
Step 8: After the nested loops, iterate through the `res` array to find the first non-zero digit.
9
Step 9: Construct a string from the remaining digits in the `res` array, starting from the first non-zero digit. If all digits are zero, the resulting string will be empty, return '0' instead. Otherwise, return the constructed string.

💡

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(m*n)

💾

Space Complexity

O(m+n)

📝

Why?

The time complexity arises directly from the nested loops used to multiply each digit of the two input strings. The space complexity is determined by the size of the result array `res`, which can have at most `m + n` digits.

📖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

Using BigInteger (Not Allowed)

If the constraint of not using BigInteger was removed, one could convert the strings to BigInteger objects, multiply them, and then convert the result back to a string.

Time: Implementation dependent but likely close to O(n log n) for multiplication.Space: O(n), where n is the number of digits in the resulting number.

Trade-offs: This would be much simpler to implement, but violates the problem constraints. It's significantly faster for very large numbers.

✨

Key Insights

1Insight 1: Multiplying strings requires simulating manual multiplication, keeping track of carries.
2Insight 2: The maximum length of the resulting string is the sum of the lengths of the input strings (m + n).
3Insight 3: We can use an array to store intermediate results. Each index represents a digit in the final product.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: One or both input strings are '0'. The solution correctly handles this by returning '0'.
Edge case 2: Input strings contain leading zeros (although the problem states they won't, good to consider). The solution correctly handles leading zeros in the `res` array by skipping them when constructing the final string. Note: the initial check avoids the case where the leading zeroes become all the digits, which would otherwise result in returning an empty string when it should be '0'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling carry-over values, leading to wrong results.

✗Common mistake 2: Not properly converting characters to integers and back to characters/strings, leading to unexpected behavior.

✗Common mistake 3: Incorrectly calculating the positions p1 and p2 where the product of digits should be added in the intermediate result array.

✗Common mistake 4: Forgetting to remove leading zeros from the final result string, resulting in strings like '00123' instead of '123'.

✅

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: Cryptography. Large number multiplication is a fundamental operation in many cryptographic algorithms.

🏦Real-world scenario 2: Scientific computing. Calculations involving very large numbers often arise in scientific simulations and modeling.

💡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: July 14, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

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 "Multiply Strings"?

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?