Problem of The Day: Check if Number Has Equal Digit Count and Digit Value

2 min read 543 words

Problem Statement

leetcode problem link

Solution [Accepted]

class Solution:
    def digitCount(self, num: str) -> bool:
        freq = {x: 0 for x in range(10)}
        for x in num:
            freq[int(x)] += 1
        for i in range(len(num)):
            if int(num[i]) != freq[i]:
                return False

        return True

time: O(n) where n is the number of digits in num space: O(1)

public class Solution {
    public bool DigitCount(string num) {
        Dictionary<int, int> frequency = new Dictionary<int, int>();
        for (int i = 0; i < 10; i++) {
            frequency[i] = 0;
        }
        foreach (var c in num) {
            if (int.TryParse(c.ToString(), out int key)) {
                frequency[key] = frequency.GetValueOrDefault(key, 0) + 1;
            }
        }
        for (int i = 0; i < num.Length; i++) {
            var ch = num[i];
            if (int.TryParse(ch.ToString(), out int currDigit)) {
                if (currDigit != frequency[i])
                    return false;
            }
        }
        return true;
    }
}

Improve CSharp code

public class Solution
{
    public bool DigitCount(string num)
    {
        // Count frequency of each digit (0-9)
        int[] freq = new int[10];
        foreach (char c in num)
        {
            int d = c - '0';     // Convert char to digit
            freq[d]++;
        }

        // Validate: at index i, num[i] should equal frequency of digit i
        for (int i = 0; i < num.Length; i++)
        {
            int expected = num[i] - '0';
            if (freq[i] != expected)
                return false;
        }

        return true;
    }
}

time: O(n) space: O(1)