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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.
1#include <stdio.h>
2#include <stdlib.h>
3#include <string.h>
4
5int compare(const void *a, const void *b) {
6 char *num1 = *(char **)a;
7 char *num2 = *(char **)b;
8 char option1[20], option2[20];
9 sprintf(option1, "%s%s", num1, num2);
10 sprintf(option2, "%s%s", num2, num1);
11 return strcmp(option2, option1);
12}
13
14char* largestNumber(int* nums, int numsSize) {
15 char* numsStr[numsSize];
16 for (int i = 0; i < numsSize; i++) {
17 numsStr[i] = malloc(11);
18 sprintf(numsStr[i], "%d", nums[i]);
19 }
20 qsort(numsStr, numsSize, sizeof(char*), compare);
21 if (numsStr[0][0] == '0') {
22 return "0";
23 }
24 int totalLength = 0;
25 for (int i = 0; i < numsSize; i++) {
26 totalLength += strlen(numsStr[i]);
27 }
28 char* result = malloc(totalLength + 1);
29 result[0] = '\0';
30 for (int i = 0; i < numsSize; i++) {
31 strcat(result, numsStr[i]);
32 free(numsStr[i]);
33 }
34 return result;
35}
36
37int main() {
38 int nums[] = {3, 30, 34, 5, 9};
39 int size = sizeof(nums) / sizeof(nums[0]);
40 char* result = largestNumber(nums, size);
41 printf("%s\n", result);
42 free(result);
43 return 0;
44}This solution first converts each number to a string and stores it in an array. It then sorts the array using a custom comparator. The comparator function forms two possible concatenated strings and orders based on which string is larger. After sorting, it concatenates the sorted numbers to form the largest number and checks for leading zeros.
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.