LeetCode 2376: Count Special Integers Solution

Master LeetCode problem 2376 (Count Special Integers), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2376. Count Special Integers

Hard

Math Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we need to count the number of special integers in the range [1, n]. A special integer is defined as having all distinct digits. We can iterate through each number in the range and check if it is special by converting it to a string and checking if all its characters are unique. We then increment a counter for each special integer found.

Algorithm Steps:
1. Initialize a counter to 0 to keep track of special integers.
2. Iterate through each number from 1 to n.
3. For each number, convert it to a string and check if all characters are unique (special).
4. If the number is special, increment the counter.
5. Finally, return the counter as the result.
Time Complexity: O(n * log n) where n is the input number.
Space Complexity: O(log n) for storing the string representation of each number.

Solution Code

class Solution {
    public int countSpecialIntegers(int n) {
        int count = 0;
        for (int i = 1; i <= n; i++) {
            if (isSpecial(i)) {
                count++;
            }
        }
        return count;
    }
    
    private boolean isSpecial(int num) {
        String numStr = String.valueOf(num);
        HashSet<Character> set = new HashSet<>();
        for (char c : numStr.toCharArray()) {
            if (set.contains(c)) {
                return false;
            }
            set.add(c);
        }
        return true;
    }
}

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