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2575. Find the Divisibility Array of a String

Description

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.

 

Example 1:

Input: word = "998244353", m = 3
Output: [1,1,0,0,0,1,1,0,0]
Explanation: There are only 4 prefixes that are divisible by 3: "9", "99", "998244", and "9982443".

Example 2:

Input: word = "1010", m = 10
Output: [0,1,0,1]
Explanation: There are only 2 prefixes that are divisible by 10: "10", and "1010".

 

Constraints:

  • 1 <= n <= 105
  • word.length == n
  • word consists of digits from 0 to 9
  • 1 <= m <= 109

Solutions

Solution 1: Traversal + Modulo

We iterate over the string word, using a variable $x$ to record the modulo result of the current prefix with $m$. If $x$ is $0$, then the divisible array value at the current position is $1$, otherwise it is $0$.

The time complexity is $O(n)$, where $n$ is the length of the string word. The space complexity is $O(1)$.

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class Solution:
    def divisibilityArray(self, word: str, m: int) -> List[int]:
        ans = []
        x = 0
        for c in word:
            x = (x * 10 + int(c)) % m
            ans.append(1 if x == 0 else 0)
        return ans

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class Solution {
    public int[] divisibilityArray(String word, int m) {
        int n = word.length();
        int[] ans = new int[n];
        long x = 0;
        for (int i = 0; i < n; ++i) {
            x = (x * 10 + word.charAt(i) - '0') % m;
            if (x == 0) {
                ans[i] = 1;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> divisibilityArray(string word, int m) {
        vector<int> ans;
        long long x = 0;
        for (char& c : word) {
            x = (x * 10 + c - '0') % m;
            ans.push_back(x == 0 ? 1 : 0);
        }
        return ans;
    }
};

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func divisibilityArray(word string, m int) (ans []int) {
    x := 0
    for _, c := range word {
        x = (x*10 + int(c-'0')) % m
        if x == 0 {
            ans = append(ans, 1)
        } else {
            ans = append(ans, 0)
        }
    }
    return ans
}

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function divisibilityArray(word: string, m: number): number[] {
    const ans: number[] = [];
    let x = 0;
    for (const c of word) {
        x = (x * 10 + Number(c)) % m;
        ans.push(x === 0 ? 1 : 0);
    }
    return ans;
}

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impl Solution {
    pub fn divisibility_array(word: String, m: i32) -> Vec<i32> {
        let m = m as i64;
        let mut x = 0i64;
        word.as_bytes()
            .iter()
            .map(|&c| {
                x = (x * 10 + i64::from(c - b'0')) % m;
                if x == 0 {
                    1
                } else {
                    0
                }
            })
            .collect()
    }
}

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/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* divisibilityArray(char* word, int m, int* returnSize) {
    int n = strlen(word);
    int* ans = malloc(sizeof(int) * n);
    long long x = 0;
    for (int i = 0; i < n; i++) {
        x = (x * 10 + word[i] - '0') % m;
        ans[i] = x == 0 ? 1 : 0;
    }
    *returnSize = n;
    return ans;
}

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