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 trueif for every indexiin the range0 <= i < n, the digitioccursnum[i]times innum, otherwise returnfalse.

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

implSolution{pubfndigit_count(num:String) -> bool{letmut count = [0;10];for d in num.bytes(){ count[(d - b'0')asusize] += 1;} num.bytes().enumerate().all(|(i, d)| count[i] == (d - b'0')asusize)}}