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).