LeetCode in Kotlin

1770. Maximum Score from Performing Multiplication Operations

Medium

You are given two integer arrays nums and multipliers of size n and m respectively, where n >= m. The arrays are 1-indexed.

You begin with a score of 0. You want to perform exactly m operations. On the ith operation (1-indexed), you will:

• Choose one integer x from either the start or the end of the array nums.
• Add multipliers[i] * x to your score.
• Remove x from the array nums.

Return the maximum score after performing m operations.

Example 1:

Input: nums = [1,2,3], multipliers = [3,2,1]

Output: 14

Explanation: An optimal solution is as follows:

• Choose from the end, [1,2,3], adding 3 * 3 = 9 to the score.

• Choose from the end, [1,2], adding 2 * 2 = 4 to the score.

• Choose from the end, [1], adding 1 * 1 = 1 to the score.

The total score is 9 + 4 + 1 = 14.

Example 2:

Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]

Output: 102

Explanation: An optimal solution is as follows:

• Choose from the start, [-5,-3,-3,-2,7,1], adding -5 * -10 = 50 to the score.

• Choose from the start, [-3,-3,-2,7,1], adding -3 * -5 = 15 to the score.

• Choose from the start, [-3,-2,7,1], adding -3 * 3 = -9 to the score.

• Choose from the end, [-2,7,1], adding 1 * 4 = 4 to the score.

• Choose from the end, [-2,7], adding 7 * 6 = 42 to the score.

The total score is 50 + 15 - 9 + 4 + 42 = 102.

Constraints:

• n == nums.length
• m == multipliers.length
• 1 <= m <= 103
• m <= n <= 105
• -1000 <= nums[i], multipliers[i] <= 1000

Solution

class Solution {
    fun maximumScore(nums: IntArray, multipliers: IntArray): Int {
        val n = nums.size
        val m = multipliers.size
        var row = m
        val dp = IntArray(m)
        var prev = IntArray(m + 1)
        while (--row >= 0) {
            for (i in 0..row) {
                dp[i] = Math.max(
                    prev[i] + multipliers[row] * nums[n - row + i - 1],
                    prev[i + 1] + multipliers[row] * nums[i]
                )
            }
            prev = dp
        }
        return dp[0]
    }
}

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