Find the Divisibility Array of a String - Practice Coding Problems

Find the Divisibility Array of a String - Problem

You are given a 0-indexed string word of length n consisting of digits, and a positive integer m.

The divisibility array div of word is an integer array of length n such that:

div[i] = 1 if the numeric value of word[0,...,i] is divisible by m, or
div[i] = 0 otherwise.

Return the divisibility array of word.

Input & Output

Example 1 — Basic Case

$ Input: word = "998", m = 7

› Output: [0,0,1]

💡 Note: "9" = 9, 9 % 7 = 2 ≠ 0, so div[0] = 0. "99" = 99, 99 % 7 = 1 ≠ 0, so div[1] = 0. "998" = 998, 998 % 7 = 0, so div[2] = 1.

Example 2 — Multiple Divisible Prefixes

$ Input: word = "1010", m = 10

› Output: [0,0,1,1]

💡 Note: "1" = 1 % 10 ≠ 0, "10" = 10 % 10 = 0, "101" = 101 % 10 ≠ 0, "1010" = 1010 % 10 = 0.

Example 3 — Single Digit

$ Input: word = "5", m = 5

› Output: [1]

💡 Note: "5" = 5, and 5 % 5 = 0, so div[0] = 1.

Constraints

1 ≤ word.length ≤ 10⁵
word consists of digits only
1 ≤ m ≤ 10⁹

Visualization

Tap to expand

Understanding the Visualization

Input

String of digits and divisor m

Process

Check each prefix for divisibility using modular arithmetic

Output

Array where 1 means prefix is divisible by m, 0 otherwise

Key Takeaway

🎯 Key Insight: Use modular arithmetic to track remainder without integer overflow

Asked in

G Google 35 M Microsoft 28

The key insight is using modular arithmetic to avoid integer overflow when checking divisibility of large numbers. Instead of converting each prefix to a full integer, we maintain only the remainder when divided by m. Best approach uses the formula: remainder = (remainder * 10 + digit) % m. Time: O(n), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Modular Arithmetic (Optimal)	O(n)	O(n)	Use modular arithmetic to track remainder without converting full numbers
Brute Force - Convert Each Prefix	O(n²)	O(n)	Convert each prefix substring to integer and check divisibility

Modular Arithmetic (Optimal) — Algorithm Steps

Initialize remainder = 0 and result array
For each digit in the string
Update remainder = (remainder * 10 + digit) % m
Set div[i] = 1 if remainder == 0, else 0

Visualization

Tap to expand

Step-by-Step Walkthrough

Initialize

Start with remainder = 0

Process Digit

remainder = (remainder * 10 + digit) % m

Check Divisibility

If remainder == 0, prefix is divisible by m

Code -

solution.c — C

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

int* solution(char* word, int m, int* returnSize) {
    int n = strlen(word);
    int* div = (int*)malloc(n * sizeof(int));
    *returnSize = n;
    int remainder = 0;
    
    for (int i = 0; i < n; i++) {
        int digit = word[i] - '0';
        remainder = (remainder * 10 + digit) % m;
        if (remainder == 0) {
            div[i] = 1;
        } else {
            div[i] = 0;
        }
    }
    
    return div;
}

int main() {
    char word[1001];
    int m;
    fgets(word, sizeof(word), stdin);
    word[strcspn(word, "\n")] = 0;
    scanf("%d", &m);
    
    int returnSize;
    int* result = solution(word, m, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string

✓ Linear Growth

Space Complexity

O(n)

Result array of size n

⚡ Linearithmic Space

23.4K Views

Medium Frequency

~15 min Avg. Time

892 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Modular Arithmetic (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler