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

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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.

Solutions

Solution 1: Counting + Enumeration

We can use an array $\textit{cnt}$ of length $10$ to count the occurrences of each digit in the string $\textit{num}$. Then, we enumerate each digit in the string $\textit{num}$ and check if its occurrence count equals the digit itself. If this condition is satisfied for all digits, we return $\text{true}$; otherwise, we return $\text{false}$.

The time complexity is $O(n)$, and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string $\textit{num}$, and $|\Sigma|$ is the range of possible digit values, which is $10$.

Python3

class Solution:
    def digitCount(self, num: str) -> bool:
        cnt = Counter(int(x) for x in num)
        return all(cnt[i] == int(x) for i, x in enumerate(num))

Java

class Solution {
    public boolean digitCount(String num) {
        int[] cnt = new int[10];
        int n = num.length();
        for (int i = 0; i < n; ++i) {
            ++cnt[num.charAt(i) - '0'];
        }
        for (int i = 0; i < n; ++i) {
            if (num.charAt(i) - '0' != cnt[i]) {
                return false;
            }
        }
        return true;
    }
}

C++

class Solution {
public:
    bool digitCount(string num) {
        int cnt[10]{};
        for (char& c : num) {
            ++cnt[c - '0'];
        }
        for (int i = 0; i < num.size(); ++i) {
            if (cnt[i] != num[i] - '0') {
                return false;
            }
        }
        return true;
    }
};

Go

func digitCount(num string) bool {
	cnt := [10]int{}
	for _, c := range num {
		cnt[c-'0']++
	}
	for i, c := range num {
		if int(c-'0') != cnt[i] {
			return false
		}
	}
	return true
}

TypeScript

function digitCount(num: string): boolean {
    const cnt: number[] = Array(10).fill(0);
    for (const c of num) {
        ++cnt[+c];
    }
    for (let i = 0; i < num.length; ++i) {
        if (cnt[i] !== +num[i]) {
            return false;
        }
    }
    return true;
}

Rust

impl Solution {
    pub fn digit_count(num: String) -> bool {
        let mut cnt = vec![0; 10];
        for c in num.chars() {
            let x = c.to_digit(10).unwrap() as usize;
            cnt[x] += 1;
        }
        for (i, c) in num.chars().enumerate() {
            let x = c.to_digit(10).unwrap() as usize;
            if cnt[i] != x {
                return false;
            }
        }
        true
    }
}

C

bool digitCount(char* num) {
    int cnt[10] = {0};
    for (int i = 0; num[i] != '\0'; ++i) {
        ++cnt[num[i] - '0'];
    }
    for (int i = 0; num[i] != '\0'; ++i) {
        if (cnt[i] != num[i] - '0') {
            return false;
        }
    }
    return true;
}