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

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.

Companies: Google, JPMorgan

Related Topics:
Hash Table, String, Counting

Similar Questions:

• Self Dividing Numbers (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/check-if-number-has-equal-digit-count-and-digit-value
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    bool digitCount(string s) {
        int cnt[10] = {};
        for (int i = 0; i < s.size(); ++i) cnt[s[i] - '0']++;
        for (int i = 0; i < s.size(); ++i) {
            if (cnt[i] != s[i] - '0') return false;
        }
        return true;
    }
};