README_EN.md

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

中文文档

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

Python3

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

Java

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

C++

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

Go

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

TypeScript

...