LeetCode in Kotlin
1856. Maximum Subarray Min-Product
Medium
The min-product of an array is equal to the minimum value in the array multiplied by the array’s sum.
- For example, the array
[3,2,5](minimum value is2) has a min-product of2 * (3+2+5) = 2 * 10 = 20.
Given an array of integers nums, return the maximum min-product of any non-empty subarray of nums. Since the answer may be large, return it modulo 109 + 7.
Note that the min-product should be maximized before performing the modulo operation. Testcases are generated such that the maximum min-product without modulo will fit in a 64-bit signed integer.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [1,2,3,2]
Output: 14
Explanation: The maximum min-product is achieved with the subarray [2,3,2] (minimum value is 2). 2 * (2+3+2) = 2 * 7 = 14.
Example 2:
Input: nums = [2,3,3,1,2]
Output: 18
Explanation: The maximum min-product is achieved with the subarray [3,3] (minimum value is 3). 3 * (3+3) = 3 * 6 = 18.
Example 3:
Input: nums = [3,1,5,6,4,2]
Output: 60
Explanation: The maximum min-product is achieved with the subarray [5,6,4] (minimum value is 4). 4 * (5+6+4) = 4 * 15 = 60.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 107
Solution
class Solution {
fun maxSumMinProduct(nums: IntArray): Int {
val n = nums.size
val mod = (1e9 + 7).toInt()
if (n == 1) {
return (nums[0].toLong() * nums[0].toLong() % mod).toInt()
}
val left = IntArray(n)
left[0] = -1
for (i in 1 until n) {
var p = i - 1
while (p >= 0 && nums[p] >= nums[i]) {
p = left[p]
}
left[i] = p
}
val right = IntArray(n)
right[n - 1] = n
for (i in n - 2 downTo 0) {
var p = i + 1
while (p < n && nums[p] >= nums[i]) {
p = right[p]
}
right[i] = p
}
var res = 0L
val preSum = LongArray(n)
preSum[0] = nums[0].toLong()
for (i in 1 until n) {
preSum[i] = preSum[i - 1] + nums[i]
}
for (i in 0 until n) {
val sum = if (left[i] == -1) preSum[right[i] - 1] else preSum[right[i] - 1] - preSum[left[i]]
val cur = nums[i] * sum
res = Math.max(cur, res)
}
return (res % mod).toInt()
}
}