How do you solve Multiply Strings on LeetCode?

Multiply Strings (LeetCode #43) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding how to simulate multiplication manually helps in breaking down the problem.

What is the brute force solution for Multiply Strings?

The brute force approach for Multiply Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simulate the multiplication process just like how we do it manually. For each digit in num2, we multiply it with num1 and shift it accordingly, then sum all the results. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Multiply Strings?

Multiply Strings uses the following concepts: Math, String, Simulation. The recommended patterns to study are: Array, String Manipulation.

What are the interview tips for Multiply Strings?

Explain your thought process clearly as you work through the problem. Don’t hesitate to write down intermediate results to avoid confusion. Practice similar problems to get comfortable with string manipulations.

What are common mistakes when solving Multiply Strings?

Not handling leading zeros in the final result. Confusing the indices when accessing the result array.

#43

Multiply Strings

Medium

MathStringSimulationArrayString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

We can optimize the multiplication process by directly managing the carry and using a single result array to store intermediate results, which reduces the need for multiple passes.

⚙️

Algorithm

5 steps

1Step 1: Initialize a result array to hold the maximum possible digits of the product.
2Step 2: Loop through each digit of num2 from right to left.
3Step 3: For each digit in num2, loop through each digit of num1 from right to left, multiply them, and add the result to the appropriate position in the result array.
4Step 4: Handle carry for each multiplication and ensure to update the result array in place.
5Step 5: Convert the result array back to a string, skipping leading zeros.

solution.py12 lines

1# Full working Python code
2
3def multiply(num1: str, num2: str) -> str:
4    if num1 == '0' or num2 == '0': return '0'
5    result = [0] * (len(num1) + len(num2))
6    for i in range(len(num2) - 1, -1, -1):
7        for j in range(len(num1) - 1, -1, -1):
8            mul = (ord(num2[i]) - ord('0')) * (ord(num1[j]) - ord('0'))
9            sum_ = mul + result[i + j + 1]
10            result[i + j + 1] = sum_ % 10
11            result[i + j] += sum_ // 10
12    return ''.join(map(str, result)).lstrip('0')

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we efficiently manage the carries and results in a single pass. The space complexity is O(n) because we use an array to store the result.

1Understanding how to simulate multiplication manually helps in breaking down the problem.
2Managing carries properly is crucial for getting the correct result.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.