LeetCode in Kotlin

1018. Binary Prefix Divisible By 5

Easy

You are given a binary array nums (0-indexed).

We define xi as the number whose binary representation is the subarray nums[0..i] (from most-significant-bit to least-significant-bit).

• For example, if nums = [1,0,1], then x0 = 1, x1 = 2, and x2 = 5.

Return an array of booleans answer where answer[i] is true if xi is divisible by 5.

Example 1:

Input: nums = [0,1,1]

Output: [true,false,false]

Explanation: The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10. Only the first number is divisible by 5, so answer[0] is true.

Example 2:

Input: nums = [1,1,1]

Output: [false,false,false]

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Solution

class Solution {
    fun prefixesDivBy5(nums: IntArray): List<Boolean> {
        val result: MutableList<Boolean> = ArrayList(nums.size)
        var remainder = 0
        for (j in nums) {
            remainder = (j + (remainder shl 1)) % 5
            result.add(remainder == 0)
        }
        return result
    }
}

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