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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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