C code works on local device but doesn't work on leetcode

Question

I'm trying to solve this leetcode problem

My solution works well on my own computer, but gives an error when I try to run it on leetcode

The error I get is this:

==31==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x602000000033 at pc 0x55679c033f3b bp 0x7ffe5ea0aca0 sp 0x7ffe5ea0ac90

When I allocate a fixed value to the string variable in the convert function instead of string_len, the error is resolved. how can I solve this problem?

excuse me for my bad English.

this is my code:

#include <stdio.h>
#include <stdlib.h>

int find(int item, int* array) {
    int index;
    for (index = 0; array[index] != item; index++);
    return index;
}

int pow2ten(int number, int time) {
    for (int count = 0; count < time; count++)
        number = number * 10;

    return number;
}

int count_len(int number) {
    int count = 1;

    for (number = number/10; number; number = number/10)
        count++;
    return count;
}

int compare(int number_1, int number_2)
{
    int len_1 = count_len(number_1);
    int len_2 = count_len(number_2);

    number_1 = pow2ten(number_1, len_2 - len_1);
    number_2 = pow2ten(number_2, len_1 - len_2);

    return number_2 > number_1;
}

char* convert(int number) {
    int string_len = count_len(number);
    char* string = malloc(string_len);
    for (int index = string_len; number; number = number / 10)
        string[--index] = (number%10) + '0';
    return string;
}

char* largestNumber(int nums[], int numsSize) {
       int number, number_len;
       number = number_len = 0;

       for (int count = 0; count < numsSize; count++)
           number_len = number_len + count_len(nums[count]);

       int part_len = 0;
       for (int time = 0; time < number_len; time = time + part_len) {
           int part = 0;

           for (int count = 0; count < numsSize; count++)
               if (compare(part, nums[count]))
                   part = nums[count];
           part_len = count_len(part);
           nums[find(part, nums)] = 0;
           part = pow2ten(part, number_len - time - part_len);

           number = number + part;
       }

       return convert(number);
}

Answer 1

In 'convert' you allocate the memory for a string for the exact length of the string. A C String is terminated by a 0-byte so you have to allocate this extra bayte and initialize it to 0.

char* string = malloc(string_len+1);
string[string_len] = 0;

This way the string is now a proper C string after the function exits and find()will not exceed the buffer.