Last updated on January 20th, 2025 at 04:06 am

Here, we see the Multiply Strings LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Math, String

Companies

Facebook, Twitter

Level of Question

Medium

Multiply Strings LeetCode Solution

1. 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.

Example 1:
Input: num1 = "2", num2 = "3"
Output: "6"

Example 2:
Input: num1 = "123", num2 = "456"
Output: "56088"

2. Coding Pattern Used in Solution

The coding pattern used in the provided code is “String Multiplication Simulation”. The code simulates the process of multiplying two numbers represented as strings, similar to how multiplication is done manually on paper.

3. Code Implementation in Different Languages

3.1 Multiply Strings C++

class Solution {
public:
    string multiply(string num1, string num2) {
        if(num1=="0"||num2=="0"){return "0";}
        int n = num1.length(),m = num2.length();
        string ans(n+m,'0');

        for(int i=n-1;i>=0;i--){
            for(int j=m-1;j>=0;j--){
                int num = (num1[i] - '0') * (num2[j] - '0') + ans[i+j+1] - '0';
                ans[i+j+1] = num%10 + '0';
                ans[i+j] += num/10;
            }
        }
        for(int i=0;i<ans.length();i++){
            if(ans[i] != '0'){ return ans.substr(i);}
        } 
        return "0";      
    }
};

3.2 Multiply Strings Java

class Solution {
    public String multiply(String num1, String num2) {
    int m = num1.length(), n = num2.length();
    int[] pos = 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, p2 = i + j + 1;
            int sum = mul + pos[p2];

            pos[p1] += sum / 10;
            pos[p2] = (sum) % 10;
        }
    }  
    
    StringBuilder sb = new StringBuilder();
    for(int p : pos) if(!(sb.length() == 0 && p == 0)) sb.append(p);
    return sb.length() == 0 ? "0" : sb.toString();      
    }
}

3.3 Multiply Strings Javascript

var multiply = function(num1, num2) {
      let b = BigInt(num1) * BigInt(num2)
      return String(b)
};

3.4 Multiply Strings Python

class Solution(object):
    def multiply(self, num1, num2):
        return str(self.strint(num1)*self.strint(num2))
    def strint(self,n):
        result=0
        for i in range(len(n)):
            result = result*10 + ord(n[i])-ord('0')
        return result

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n * m)	O(n + m)
Java	O(n * m)	O(n + m)
JavaScript	O(max(n, m))	O(max(n, m))
Python	O(n + m)	O(n + m)

• The C++ and Java implementations are more algorithmic and simulate manual multiplication.
• The JavaScript and Python implementations rely on built-in features (BigInt and arbitrary-precision integers) for simplicity.
• The time complexity for C++ and Java is O(n * m), while for JavaScript and Python, it is O(max(n, m)) due to their reliance on built-in operations.