2283. Check if Number Has Equal Digit Count and Digit Value

Easy

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.

解題

Runtime: 3 ms, faster than 35.71%

Memory Usage: 1.9 MB, less than 100.00%

func digitCount(num string) bool {
    m := make(map[byte]int)
    
    for i:=0; i<len(num); i++ {
        m[num[i]]++
    }
    
    for i:=0; i<len(num); i++ {
        if m[byte(i+48)]!= int(num[i]-48) {
            return false
        }
    }
    
    return true
}