Solutions

1. Brute Force

class Solution:
    def largestNumber(self, nums: List[int]) -> str:
        arr = [str(num) for num in nums]

        res = []
        while arr:
            maxi = 0
            for i in range(1, len(arr)):
                if arr[i] + arr[maxi] > arr[maxi] + arr[i]:
                    maxi = i
            res.append(arr[maxi])
            arr.pop(maxi)

        result = "".join(res)
        return result if result[0] != '0' else '0'

Time & Space Complexity

Time complexity: $O(n * N)$
Space complexity: $O(N)$

Where $n$ is the size of the array $nums$ and $N$ is the total number of digits in the array $nums$ .

2. Sorting

class Solution:
    def largestNumber(self, nums: List[int]) -> str:
        for i, num in enumerate(nums):
            nums[i] = str(num)

        def compare(n1, n2):
            if n1 + n2 > n2 + n1:
                return -1
            else:
                return 1
        nums = sorted(nums, key=cmp_to_key(compare))
        return str(int("".join(nums)))

Time & Space Complexity

Time complexity: $O(N \log N)$
Space complexity: $O(N)$

Where $N$ is the total number of digits in the array $nums$ .