LeetCode 2283. Check if Number Has Equal Digit Count and Digit Value Solution in Java, C++, Python & More | Explanation + Code

Description

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.

Solutions

Solution 1: Counting + Enumeration

We can use an array cnt of length 10 to count the occurrences of each digit in the string num. Then, we enumerate each digit in the string num and check if its occurrence count equals the digit itself. If this condition is satisfied for all digits, we return true; otherwise, we return false.

The time complexity is O(n), and the space complexity is O(|Σ|). Here, n is the length of the string num, and |Σ| is the range of possible digit values, which is 10.

PythonJavaC++GoTypeScriptRustC

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

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

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; } };(code-box)

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

function digitCount(num: string): boolean { const cnt: number[] = Array(10).fill(0); for (const c of num) { ++cnt[+c]; } for (let i = 0; i < num.length; ++i) { if (cnt[i] !== +num[i]) { return false; } } return true; }(code-box)

impl Solution { pub fn digit_count(num: String) -> bool { let mut cnt = vec![0; 10]; for c in num.chars() { let x = c.to_digit(10).unwrap() as usize; cnt[x] += 1; } for (i, c) in num.chars().enumerate() { let x = c.to_digit(10).unwrap() as usize; if cnt[i] != x { return false; } } true } }(code-box)

bool digitCount(char* num) { int cnt[10] = {0}; for (int i = 0; num[i] != '\0'; ++i) { ++cnt[num[i] - '0']; } for (int i = 0; num[i] != '\0'; ++i) { if (cnt[i] != num[i] - '0') { return false; } } return true; }(code-box)