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Formatted question description: https://leetcode.ca/all/2575.html

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 $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  };
  

• 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
  
  

• 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()
      }
  }
  
  

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