LeetCode in Kotlin
1073. Adding Two Negabinary Numbers
Medium
Given two numbers arr1 and arr2 in base -2, return the result of adding them together.
Each number is given in array format: as an array of 0s and 1s, from most significant bit to least significant bit. For example, arr = [1,1,0,1] represents the number (-2)^3 + (-2)^2 + (-2)^0 = -3. A number arr in array, format is also guaranteed to have no leading zeros: either arr == [0] or arr[0] == 1.
Return the result of adding arr1 and arr2 in the same format: as an array of 0s and 1s with no leading zeros.
Example 1:
Input: arr1 = [1,1,1,1,1], arr2 = [1,0,1]
Output: [1,0,0,0,0]
Explanation: arr1 represents 11, arr2 represents 5, the output represents 16.
Example 2:
Input: arr1 = [0], arr2 = [0]
Output: [0]
Example 3:
Input: arr1 = [0], arr2 = [1]
Output: [1]
Constraints:
1 <= arr1.length, arr2.length <= 1000arr1[i]andarr2[i]are0or1arr1andarr2have no leading zeros
Solution
class Solution {
fun addNegabinary(arr1: IntArray, arr2: IntArray): IntArray {
val len1 = arr1.size
val len2 = arr2.size
val reverseArr1 = IntArray(len1)
for (i in len1 - 1 downTo 0) {
reverseArr1[len1 - i - 1] = arr1[i]
}
val reverseArr2 = IntArray(len2)
for (i in len2 - 1 downTo 0) {
reverseArr2[len2 - i - 1] = arr2[i]
}
val sumArray = IntArray(len1.coerceAtLeast(len2) + 2)
System.arraycopy(reverseArr1, 0, sumArray, 0, len1)
for (i in sumArray.indices) {
if (i < len2) {
sumArray[i] += reverseArr2[i]
}
if (sumArray[i] > 1) {
sumArray[i] -= 2
sumArray[i + 1]--
} else if (sumArray[i] == -1) {
sumArray[i] = 1
sumArray[i + 1]++
}
}
var resultLen = sumArray.size
for (i in sumArray.indices.reversed()) {
if (sumArray[i] == 0) {
resultLen--
} else {
break
}
}
if (resultLen == 0) {
return intArrayOf(0)
}
val result = IntArray(resultLen)
for (i in resultLen - 1 downTo 0) {
result[resultLen - i - 1] = sumArray[i]
}
return result
}
}