Find the Divisibility Array of a String

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: [0,1,0,1]

💡 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

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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

✓ Brute Force - Convert Each Prefix

⏱️ Time: O(n²) Space: O(n)

For each position i, convert word[0:i+1] to an integer and check if it's divisible by m. This works for small numbers but fails for large inputs due to integer overflow.

Modular Arithmetic (Optimal)

⏱️ Time: O(n) Space: O(n)

Instead of converting each prefix to a full integer, we maintain the remainder when divided by m. For each new digit, we update the remainder using the formula: remainder = (remainder * 10 + digit) % m.

Brute Force - Convert Each Prefix — Algorithm Steps

For each index i from 0 to n-1
Convert substring word[0:i+1] to integer
Check if integer % m == 0
Set div[i] = 1 if divisible, 0 otherwise

Visualization

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Step-by-Step Walkthrough

Build Prefix

Extract substring from start to current position

Convert & Check

Convert to integer and check if divisible by m

Set Result

Set div[i] = 1 if divisible, 0 otherwise

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;
    
    for (int i = 0; i < n; i++) {
        // Convert prefix word[0:i+1] to integer
        char prefix[1001];
        strncpy(prefix, word, i + 1);
        prefix[i + 1] = '\0';
        long long prefixNum = atoll(prefix);
        if (prefixNum % m == 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(n²)

For each position, we convert O(n) digits to integer

⚠ Quadratic Growth

Space Complexity

O(n)

Result array of size n

⚡ Linearithmic Space

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Medium Frequency

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Related Problems

Common Approaches

Brute Force - Convert Each Prefix — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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