LeetCode in Kotlin
927. Three Equal Parts
Hard
You are given an array arr which consists of only zeros and ones, divide the array into three non-empty parts such that all of these parts represent the same binary value.
If it is possible, return any [i, j] with i + 1 < j, such that:
arr[0], arr[1], ..., arr[i]is the first part,arr[i + 1], arr[i + 2], ..., arr[j - 1]is the second part, andarr[j], arr[j + 1], ..., arr[arr.length - 1]is the third part.- All three parts have equal binary values.
If it is not possible, return [-1, -1].
Note that the entire part is used when considering what binary value it represents. For example, [1,1,0] represents 6 in decimal, not 3. Also, leading zeros are allowed, so [0,1,1] and [1,1] represent the same value.
Example 1:
Input: arr = [1,0,1,0,1]
Output: [0,3]
Example 2:
Input: arr = [1,1,0,1,1]
Output: [-1,-1]
Example 3:
Input: arr = [1,1,0,0,1]
Output: [0,2]
Constraints:
3 <= arr.length <= 3 * 104arr[i]is0or1
Solution
class Solution {
fun threeEqualParts(arr: IntArray): IntArray {
var ones = 0
for (num in arr) {
ones += num
}
if (ones == 0) {
return intArrayOf(0, 2)
} else if (ones % 3 != 0) {
return intArrayOf(-1, -1)
}
ones /= 3
var index1 = -1
var index2 = -1
var index3 = -1
var totalOnes = 0
for (i in arr.indices) {
if (arr[i] == 0) {
continue
}
totalOnes += arr[i]
when (totalOnes) {
1 -> index1 = i
ones + 1 -> index2 = i
2 * ones + 1 -> index3 = i
}
}
while (index3 < arr.size) {
if (arr[index1] == arr[index3] && arr[index2] == arr[index3]) {
++index1
++index2
++index3
} else {
return intArrayOf(-1, -1)
}
}
return intArrayOf(index1 - 1, index2)
}
}