carloscn
commented
1 year ago Analysis
static int32_t remove_digit(const char *number, char digit, char *out)
{
int32_t ret = 0;
size_t len;
UTILS_CHECK_PTR(number);
UTILS_CHECK_PTR(out);
UTILS_CHECK_LEN(len = strlen(number));
char buffer[15] = {0};
int32_t max_number = INT32_MIN;
for (size_t i = 0; i < len; i ++) {
if (number[i] == digit) {
size_t n = 0;
for (size_t j = 0; j < len + 1; j ++) {
if (j != i) {
buffer[n] = number[j];
n ++;
}
}
int32_t number = atoi(buffer);
max_number = (max_number > number) ? max_number : number;
}
}
sprintf(out, "%d", max_number);
finish:
return ret;
}
Description
You are given a string number representing a positive integer and a character digit.
Return the resulting string after removing exactly one occurrence of digit from number such that the value of the resulting string in decimal form is maximized. The test cases are generated such that digit occurs at least once in number.
Example 1:
Input: number = "123", digit = "3" Output: "12" Explanation: There is only one '3' in "123". After removing '3', the result is "12". Example 2:
Input: number = "1231", digit = "1" Output: "231" Explanation: We can remove the first '1' to get "231" or remove the second '1' to get "123". Since 231 > 123, we return "231". Example 3:
Input: number = "551", digit = "5" Output: "51" Explanation: We can remove either the first or second '5' from "551". Both result in the string "51".
Constraints:
2 <= number.length <= 100 number consists of digits from '1' to '9'. digit is a digit from '1' to '9'. digit occurs at least once in number.