Question

2575. Find the Divisibility Array of a String

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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 $O(1)$.

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

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;
  }
}

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;
  }
};

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
}

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;
}

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()
  }
}

/**
 * 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;
}