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Formatted question description: https://leetcode.ca/all/2283.html

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

  • Difficulty: Easy.
  • Related Topics: Hash Table, String, Counting.
  • Similar Questions: Self Dividing Numbers.

Problem

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.

Solution (Java, C++, Python)

  • class Solution {
    public boolean digitCount(String num) {
        int[] cnt = new int[11];
        char[] arr = num.toCharArray();
        for (char d : arr) {
            ++cnt[d - '0'];
        }
        for (int i = 0; i < arr.length; i++) {
            if (cnt[i] != arr[i] - '0') {
                return false;
            }
        }
        return true;
    }
}

############

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 (cnt[i] != num.charAt(i) - '0') {
                return false;
            }
        }
        return true;
    }
}

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

############

# 2283. Check if Number Has Equal Digit Count and Digit Value
# https://leetcode.com/problems/check-if-number-has-equal-digit-count-and-digit-value/

class Solution:
    def digitCount(self, num: str) -> bool:
        count = Counter(num)
        
        for i, x in enumerate(num):
            if count[str(i)] != int(x):
                return False
        
        return True


  • 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;
    }
};

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

  • function digitCount(num: string): boolean {
    const n = num.length;
    const count = new Array(10).fill(0);
    for (let i = 0; i < n; i++) {
        count[i] = Number(num[i]);
    }
    for (const c of num) {
        count[c]--;
    }
    return count.every(v => v === 0);
}


  • impl Solution {
    pub fn digit_count(num: String) -> bool {
        let s = num.as_bytes();
        let n = num.len();
        let mut count = [0; 10];
        for i in 0..n {
            count[i] = s[i] - b'0';
        }
        for c in s {
            count[(c - b'0') as usize] -= 1;
        }
        count.iter().all(|v| *v == 0)
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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