LeetCode in Kotlin
1191. K-Concatenation Maximum Sum
Medium
Given an integer array arr and an integer k, modify the array by repeating it k times.
For example, if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].
Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0.
As the answer can be very large, return the answer modulo 109 + 7.
Example 1:
Input: arr = [1,2], k = 3
Output: 9
Example 2:
Input: arr = [1,-2,1], k = 5
Output: 2
Example 3:
Input: arr = [-1,-2], k = 7
Output: 0
Constraints:
1 <= arr.length <= 1051 <= k <= 105-104 <= arr[i] <= 104
Solution
class Solution {
private var mod = (1e9 + 7).toInt()
fun kConcatenationMaxSum(arr: IntArray, k: Int): Int {
var sum: Long = 0
for (i in arr.indices) {
sum += arr[i].toLong()
}
return if (sum <= 0 || k == 1) {
var cb: Long = 0
var ob: Long = 0
for (i in arr.indices) {
cb = if (arr[i] + cb > arr[i]) {
arr[i] + cb
} else {
arr[i].toLong()
}
if (ob < cb) {
ob = cb
}
}
if (k == 1) {
return (ob % mod).toInt()
}
for (i in arr.indices) {
cb = if (arr[i] + cb > arr[i]) {
arr[i] + cb
} else {
arr[i].toLong()
}
if (ob < cb) {
ob = cb
}
}
(ob % mod).toInt()
} else {
var max1: Long = 0
var smax: Long = 0
for (i in arr.indices.reversed()) {
smax += arr[i].toLong()
if (smax > max1) {
max1 = smax
}
}
max1 %= mod.toLong()
var max2: Long = 0
smax = 0
for (i in arr.indices) {
smax += arr[i].toLong()
if (smax > max2) {
max2 = smax
}
}
max2 %= mod.toLong()
val ans = max1 + (k - 2) * sum + max2
(ans % mod).toInt()
}
}
}