LeetCode #3448 — HARD

Count Substrings Divisible By Last Digit

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given a string s consisting of digits.

Return the number of substrings of s divisible by their non-zero last digit.

Note: A substring may contain leading zeros.

Example 1:

Input: s = "12936"

Output: 11

Explanation:

Substrings "29", "129", "293" and "2936" are not divisible by their last digit. There are 15 substrings in total, so the answer is 15 - 4 = 11.

Example 2:

Input: s = "5701283"

Output: 18

Explanation:

Substrings "01", "12", "701", "012", "128", "5701", "7012", "0128", "57012", "70128", "570128", and "701283" are all divisible by their last digit. Additionally, all substrings that are just 1 non-zero digit are divisible by themselves. Since there are 6 such digits, the answer is 12 + 6 = 18.

Example 3:

Input: s = "1010101010"

Output: 25

Explanation:

Only substrings that end with digit '1' are divisible by their last digit. There are 25 such substrings.

Constraints:

1 <= s.length <= 10⁵
s consists of digits only.

Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of digits. Return the number of substrings of s divisible by their non-zero last digit. Note: A substring may contain leading zeros.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

"12936"

Example 2

"5701283"

Example 3

"1010101010"

Core Insight

What unlocks the optimal approach

Let <code>dp[index][i][j]</code> be the number of subarrays <code>s[start...index]</code> such that <code>s[start...index] % i == j</code>.
For every pair <code>(i, j)</code>, add <code>dp[index - 1][i][j]</code> to <code>dp[index][i][(j * 10 + x)%i)]</code>.
You should optimize this solution so that it can fit into the memory limit.
In order to find <code>dp[index][i][j]</code> we use values from <code>dp[index - 1][i][j]</code>. Hence, we can keep only <code>dp[index][i][j]</code> and <code>dp[index - 1][i][j]</code> at every iteration of the loop.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3448: Count Substrings Divisible By Last Digit
class Solution {
  public long countSubstrings(String s) {
    long ans = 0;
    // dp[i][num][rem] := the number of first `i` digits of s that have a
    // remainder of `rem` when divided by `num`
    int[][][] dp = new int[s.length() + 1][10][10];

    for (int i = 1; i <= s.length(); ++i) {
      final int digit = s.charAt(i - 1) - '0';
      for (int num = 1; num < 10; ++num) {
        for (int rem = 0; rem < num; ++rem)
          dp[i][num][(rem * 10 + digit) % num] += dp[i - 1][num][rem];
        ++dp[i][num][digit % num];
      }
      ans += dp[i][digit][0];
    }

    return ans;
  }
}

// Accepted solution for LeetCode #3448: Count Substrings Divisible By Last Digit
package main

// https://space.bilibili.com/206214
func countSubstrings(s string) (ans int64) {
	f := [10][9]int{}
	for _, c := range s {
		d := int(c - '0')
		for m := 1; m < 10; m++ { // 枚举模数 m
			// 滚动数组计算 f
			nf := [9]int{}
			nf[d%m] = 1
			for rem, fv := range f[m][:m] { // 枚举模 m 的余数 rem
				nf[(rem*10+d)%m] += fv // 刷表法
			}
			f[m] = nf
		}
		// 以 s[i] 结尾的，模 s[i] 余数为 0 的子串个数
		ans += int64(f[d][0])
	}
	return
}

# Accepted solution for LeetCode #3448: Count Substrings Divisible By Last Digit
class Solution:
  def countSubstrings(self, s: str) -> int:
    ans = 0
    # dp[i][num][rem] := the number of first `i` digits of s that have a
    # remainder of `rem` when divided by `num`
    dp = [[[0] * 10 for _ in range(10)] for _ in range(len(s) + 1)]

    for i in range(1, len(s) + 1):
      digit = int(s[i - 1])
      for num in range(1, 10):
        for rem in range(num):
          dp[i][num][(rem * 10 + digit) % num] += dp[i - 1][num][rem]
        dp[i][num][digit % num] += 1
      ans += dp[i][digit][0]

    return ans

// Accepted solution for LeetCode #3448: Count Substrings Divisible By Last Digit
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3448: Count Substrings Divisible By Last Digit
// class Solution {
//   public long countSubstrings(String s) {
//     long ans = 0;
//     // dp[i][num][rem] := the number of first `i` digits of s that have a
//     // remainder of `rem` when divided by `num`
//     int[][][] dp = new int[s.length() + 1][10][10];
// 
//     for (int i = 1; i <= s.length(); ++i) {
//       final int digit = s.charAt(i - 1) - '0';
//       for (int num = 1; num < 10; ++num) {
//         for (int rem = 0; rem < num; ++rem)
//           dp[i][num][(rem * 10 + digit) % num] += dp[i - 1][num][rem];
//         ++dp[i][num][digit % num];
//       }
//       ans += dp[i][digit][0];
//     }
// 
//     return ans;
//   }
// }

// Accepted solution for LeetCode #3448: Count Substrings Divisible By Last Digit
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3448: Count Substrings Divisible By Last Digit
// class Solution {
//   public long countSubstrings(String s) {
//     long ans = 0;
//     // dp[i][num][rem] := the number of first `i` digits of s that have a
//     // remainder of `rem` when divided by `num`
//     int[][][] dp = new int[s.length() + 1][10][10];
// 
//     for (int i = 1; i <= s.length(); ++i) {
//       final int digit = s.charAt(i - 1) - '0';
//       for (int num = 1; num < 10; ++num) {
//         for (int rem = 0; rem < num; ++rem)
//           dp[i][num][(rem * 10 + digit) % num] += dp[i - 1][num][rem];
//         ++dp[i][num][digit % num];
//       }
//       ans += dp[i][digit][0];
//     }
// 
//     return ans;
//   }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Count Substrings Divisible By Last Digit

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

State misses one required dimension