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# 43. Multiply Strings

\# Medium

{% hint style="info" %}
Key idea: 

1. `num1[x]*num2[y]` will occupy `result[x+y]`(high digit) and `result[x+y+1]`(low digit)
2. `num1*num2` have at most `len(num1)+len(num2)` digits.
   {% endhint %}

![](https://3288217904-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LxJcc9A1TOyn5a5HJQ4%2F-MHnUHs_IIJTBEbsCCPV%2F-MHnWzRTWcNbpjQBkkxY%2F1600739843184.jpg?alt=media\&token=6c41a897-5b5c-46bf-980e-48328073a49f)

### Solution:

1. define result list
2. multiply each pair
3. convert into string, remove high digit which is 0

```python
class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        # edge case
        if num1 == 1 or num2 == 1:
            return str(num1*num2)
        
        result = [0 for i in range(len(num1)+len(num2))]
        tmp = 0
        for y in reversed(range(len(num1))):
            for x in reversed(range(len(num2))):
                multi = int(num1[y])*int(num2[x])
                multi += result[x+y+1]
                result[x+y+1] = multi%10
                result[x+y] += multi//10                    
                print(result)

        for i in range(len(result)):
            if result[i] != 0:
                break
        s = ""

        for j in range(i,len(result)):
            s += str(result[j])
        return s
```

{% hint style="danger" %}
`result` list = `[high digit, ..., low digit]`

思路不是特别难，但是要一次性写对太不容易了
{% endhint %}

Time complexity = Space complexity = $$O(n\*m)$$