Number of Divisible Substrings

Number of Divisible Substrings - Problem

Each character of the English alphabet has been mapped to a digit as shown below:

a → 1, b → 2, c → 3, d → 4, e → 5, f → 6, g → 7, h → 8, i → 9, j → 1, k → 2, l → 3, m → 4, n → 5, o → 6, p → 7, q → 8, r → 9, s → 1, t → 2, u → 3, v → 4, w → 5, x → 6, y → 7, z → 8

A string is divisible if the sum of the mapped values of its characters is divisible by its length.

Given a string s, return the number of divisible substrings of s.

A substring is a contiguous non-empty sequence of characters within a string.

Input & Output

Example 1 — Simple String

$ Input: s = "asdf"

› Output: 7

💡 Note: Character values: a=1, s=1, d=4, f=6. Divisible substrings: "a"(1÷1=0), "s"(1÷1=0), "d"(4÷1=0), "f"(6÷1=0), "as"(2÷2=0), "df"(10÷2=0), "asd"(6÷3=0), "asdf"(12÷4=0). Non-divisible: "sd"(5÷2≠0), "sdf"(11÷3≠0). Total divisible: 7.

Example 2 — Single Character

$ Input: s = "a"

› Output: 1

💡 Note: Only one substring "a" with value 1 and length 1. Since 1 % 1 = 0, it's divisible.

Example 3 — Two Characters

$ Input: s = "ab"

› Output: 2

💡 Note: Substrings: "a"(1÷1=0), "b"(2÷1=0), "ab"(3÷2≠0). Only "a" and "b" are divisible. Total = 2.

Constraints

1 ≤ s.length ≤ 1000
s consists of lowercase English letters only

Visualization

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Asked in

G Google 25 a Amazon 18 M Microsoft 12

The key insight is using character-to-digit mapping and checking if substring sum is divisible by length. Best approach uses prefix sum optimization to avoid recalculating sums. Time: O(n²), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n³) Space: O(1)

Generate all possible substrings, calculate sum of character values for each, and check if sum is divisible by substring length.

Hash Map with Remainder Tracking

⏱️ Time: O(n²) Space: O(1)

For each starting position, track remainder values (sum % length) and use mathematical properties to identify when sum is divisible by length.

Prefix Sum Optimization

⏱️ Time: O(n²) Space: O(1)

For each starting position, maintain a running sum as we extend the substring, avoiding recalculation of character values.

Brute Force — Algorithm Steps

Generate all substrings
Calculate sum of mapped values
Check if sum % length == 0

Visualization

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Step-by-Step Walkthrough

Generate Substrings

Create all possible substrings

Calculate Sum

Map characters to values and sum

Check Divisible

Test if sum % length == 0

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int charValue(char c) {
    return ((c - 'a') % 9) + 1;
}

int solution(char* s) {
    int count = 0;
    int n = strlen(s);
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int charSum = 0;
            int length = j - i + 1;
            
            for (int k = i; k <= j; k++) {
                charSum += charValue(s[k]);
            }
            
            if (charSum % length == 0) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char s[10001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    printf("%d\n", solution(s));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Two nested loops for all substrings O(n²), inner loop to calculate sum O(n)

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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