LeetCode 3448: Count Substrings Divisible By Last Digit Solution

Master LeetCode problem 3448 (Count Substrings Divisible By Last Digit), 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.

3448. Count Substrings Divisible By Last Digit

Hard

String Dynamic Programming

Problem Explanation

Explanation:

To solve this problem, we can iterate through all substrings of the given string and check if the substring is divisible by its non-zero last digit. We can optimize this approach by calculating the count of substrings that are divisible by each digit from 1 to 9.

Steps:

Initialize an array divisibleCount of size 10 to store the count of substrings divisible by each digit from 0 to 9.
Initialize a variable result to store the final count of substrings that are divisible by their non-zero last digit.
Iterate through each digit in the input string s.
For each digit, iterate through all possible substrings ending at that digit and update the count in the divisibleCount array.
Finally, calculate the total count of substrings that are divisible by their non-zero last digit using the values in the divisibleCount array.

Time Complexity: O(n), where n is the length of the input string s. Space Complexity: O(1) as we are using a fixed-size array of size 10.

Solution Code

public int countSubstringsDivisibleByLastDigit(String s) {
    int[] divisibleCount = new int[10];
    int result = 0;
    
    for (char c : s.toCharArray()) {
        int digit = c - '0';
        int[] newDivisibleCount = new int[10];
        
        for (int i = 0; i < 10; i++) {
            newDivisibleCount[(i * 10 + digit) % 10] += divisibleCount[i];
        }
        
        newDivisibleCount[digit % 10]++;
        divisibleCount = newDivisibleCount;
        
        result += divisibleCount[0]; // Count of substrings ending in zero
    }
    
    return result;
}

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