carloscn
commented
11 months ago Analysis
static int32_t digit_count(const char *num, bool *result)
{
int32_t ret = 0;
size_t len;
UTILS_CHECK_PTR(num);
UTILS_CHECK_PTR(result);
UTILS_CHECK_LEN(len = strlen(num));
int32_t bucket[11] = {0};
for (size_t i = 0; i < len; i ++) {
bucket[(size_t)(num[i] - '0')] ++;
}
for (size_t i = 0; i < len; i ++) {
if ((int32_t)(num[i] - '0') != bucket[i]) {
*result = false;
goto finish;
}
}
*result = true;
finish:
return ret;
}
Description
You are given a 0-indexed string num of length n consisting of digits.
Return true if for every index i in the range 0 <= i < n, the digit i occurs num[i] times in num, otherwise return false.
Example 1:
Input: num = "1210" Output: true Explanation: num[0] = '1'. The digit 0 occurs once in num. num[1] = '2'. The digit 1 occurs twice in num. num[2] = '1'. The digit 2 occurs once in num. num[3] = '0'. The digit 3 occurs zero times in num. The condition holds true for every index in "1210", so return true.
Example 2:
Input: num = "030" Output: false Explanation: num[0] = '0'. The digit 0 should occur zero times, but actually occurs twice in num. num[1] = '3'. The digit 1 should occur three times, but actually occurs zero times in num. num[2] = '0'. The digit 2 occurs zero times in num. The indices 0 and 1 both violate the condition, so return false.
Constraints:
n == num.length 1 <= n <= 10 num consists of digits.