How do you solve Count Substrings Divisible By Last Digit on LeetCode?

Count Substrings Divisible By Last Digit (LeetCode #3448) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Substrings that end with a non-zero digit can be checked for divisibility by that digit.

What is the brute force solution for Count Substrings Divisible By Last Digit?

The brute force approach for Count Substrings Divisible By Last Digit is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible substrings of the input string and checking if each substring is divisible by its last digit. This is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Substrings Divisible By Last Digit?

Count Substrings Divisible By Last Digit uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, String Manipulation.

What are the interview tips for Count Substrings Divisible By Last Digit?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as strings with leading zeros or all zeros. Discuss your thought process and approach with the interviewer before coding.

What are common mistakes when solving Count Substrings Divisible By Last Digit?

Not checking if the last digit is zero before performing the modulus operation. Failing to consider leading zeros in substrings.

#3448

Count Substrings Divisible By Last Digit

Hard

StringDynamic ProgrammingDynamic ProgrammingString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses dynamic programming to keep track of the counts of valid substrings ending at each index, leveraging the properties of modular arithmetic to avoid recalculating values.

⚙️

Algorithm

4 steps

1Step 1: Initialize a count variable to store the total number of valid substrings.
2Step 2: Use a loop to iterate through each character in the string as the end of a substring.
3Step 3: For each character, check if it is a non-zero digit and calculate the current number formed by the substring ending at this character.
4Step 4: Use modular arithmetic to check if the number is divisible by its last digit and update the count accordingly.

solution.py16 lines

1# Full working Python code
2
3def count_divisible_substrings(s):
4    count = 0
5    n = len(s)
6    for i in range(n):
7        current_number = 0
8        for j in range(i, n):
9            current_number = current_number * 10 + int(s[j])
10            last_digit = int(s[j])
11            if last_digit != 0 and current_number % last_digit == 0:
12                count += 1
13    return count
14
15# Example usage
16print(count_divisible_substrings('12936'))

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once while accumulating the current number. The space complexity is O(1) since we are using a constant amount of extra space.

1Substrings that end with a non-zero digit can be checked for divisibility by that digit.
2Dynamic programming can help optimize the counting of valid substrings without recalculating values.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.