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This approach involves converting each number to a string and sorting them based on a custom comparator. The key idea is to determine the order of two numbers a and b by comparing the concatenated results a+b and b+a as strings. Sorting the numbers in such a manner ensures that the concatenated string is the largest possible.
Time Complexity: O(n log n), where n is the number of elements, due to sorting. Comparison operation within the sort is O(1) as it's based on string concatenation and comparison.
Space Complexity: O(n), where n is the number of integers, due to the storage of string representations of integers.
1using System;
2using System.Collections.Generic;
3using System.Linq;
4
5public class Solution {
6    public string LargestNumber(int[] nums) {
7        var numStrs = nums.Select(num => num.ToString()).ToArray();
8        Array.Sort(numStrs, (a, b) => string.Compare(b + a, a + b));
9        if (numStrs[0] == "0") {
10            return "0";
11        }
12        return string.Concat(numStrs);
13    }
14
15    public static void Main(string[] args) {
16        Solution sol = new Solution();
17        Console.WriteLine(sol.LargestNumber(new int[] {3, 30, 34, 5, 9}));
18    }
19}The C# solution makes use of LINQ to convert the integers to strings and uses a custom comparator in the Array.Sort method. The comparator compares concatenated string results to ensure proper sorting.
This approach involves using a custom comparator with a min-heap to continuously extract the largest possible number combination. This can be more efficient in some cases where we require efficient access to the largest current element for concatenation.
Time Complexity: O(n log n), dominated by the heap operations.
Space Complexity: O(n) due to the heap storage.
1import heapq
2
3def largestNumber(nums):
4    if not
The solution uses a max-heap implemented with negative values in Python’s built-in heapq, to facilitate priority retrieval of the ‘largest’ lexical integer string. The heap ensures the largest value is popped and concatenated first.