LeetCode in Kotlin

2575. Find the Divisibility Array of a String

Medium

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

Solution

class Solution {
    fun divisibilityArray(word: String, m: Int): IntArray {
        val ans = IntArray(word.length)
        val prevRemainder = StringBuilder()

        for (i in word.indices) {
            val number = prevRemainder.append(word[i]).toString().toLong()
            if (number != 0L && number < m) {
                continue
            }

            val remainder = number % m
            prevRemainder.clear()

            if (remainder == 0L) {
                ans[i] = 1
            } else {
                prevRemainder.append(remainder)
            }
        }

        return ans
    }
}

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