# Largest Number

## Problem 

Given a list of non-negative integers `nums`, arrange them such that they form the largest number.

**Note:** The result may be very large, so you need to return a string instead of an integer.

{% hint style="info" %}
For example:

```
Input: nums = [10,2]
Output: "210"
```

```
Input: nums = [3,30,34,5,9]
Output: "9534330"
```

```
Input: nums = [1]
Output: "1"
```

```
Input: nums = [10]
Output: "10"
```

{% endhint %}

### Though Process

![](/files/-MN9OG6ZMmI81nKAHgbp)

* The idea here is to use a "comparison based" sorting algorithm 

* Given two numbers **A** and **B**, we compare two numbers AB (B appended at the end of A) and BA (A appended at the end of B). If AB is larger, then this means in the output, A should come before B, else if AB isn't larger, then B should come before A. 
  * **Example**: let A and B be 542 and 60. To compare A and B, we compare 54260 and 60542. Since 60542 is greater than 54260, we put B first.

## Solution

```
class Solution:
    def largestNumber(self, nums: List[int]) -> str:
        for i in range(len(nums)):
            for j in range(i+1,len(nums)):
                if str(nums[j])+ str(nums[i]) > str(nums[i]) + str(nums[j]):
                    nums[j], nums[i] = nums[i], nums[j]
        string = [str(i) for i in nums]
        a_string = "".join(string)
        if nums[0]==0:
            return '0'
        return(a_string)
```

## Key Facts

* We are comparing two concatenated elements to each other to see if $$AB > BA$$ or $$BA > AB$$ 

* We are arranging the elements based on comparing their different concatenations

## Time Complexity

* **Time:**$$O(n^2)$$ 
* **Space:** $$O(1)$$ 


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