README.md

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

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.

Solutions (Rust)

1. Solution

impl Solution {
    pub fn digit_count(num: String) -> bool {
        let mut count = [0; 10];

        for d in num.bytes() {
            count[(d - b'0') as usize] += 1;
        }

        num.bytes()
            .enumerate()
            .all(|(i, d)| count[i] == (d - b'0') as usize)
    }
}