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 is 2) has a min-product of 2 * (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 <= 105
• 1 <= 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()
    }
}

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