# 1770. Maximum Score from Performing Multiplication Operations

## Description

You are given two 0-indexed integer arrays nums and multipliers of size n and m respectively, where n >= m.

You begin with a score of 0. You want to perform exactly m operations. On the ith operation (0-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.
• Note that multipliers[0] corresponds to the first operation, multipliers[1] to the second operation, and so on.
• Remove x from 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 <= 300
• m <= n <= 105
• -1000 <= nums[i], multipliers[i] <= 1000

## Solutions

• class Solution {
private Integer[][] f;
private int[] multipliers;
private int[] nums;
private int n;
private int m;

public int maximumScore(int[] nums, int[] multipliers) {
n = nums.length;
m = multipliers.length;
f = new Integer[m][m];
this.nums = nums;
this.multipliers = multipliers;
return dfs(0, 0);
}

private int dfs(int i, int j) {
if (i >= m || j >= m || (i + j) >= m) {
return 0;
}
if (f[i][j] != null) {
return f[i][j];
}
int k = i + j;
int a = dfs(i + 1, j) + nums[i] * multipliers[k];
int b = dfs(i, j + 1) + nums[n - 1 - j] * multipliers[k];
f[i][j] = Math.max(a, b);
return f[i][j];
}
}

• class Solution {
public:
int maximumScore(vector<int>& nums, vector<int>& multipliers) {
int n = nums.size(), m = multipliers.size();
int f[m][m];
memset(f, 0x3f, sizeof f);
function<int(int, int)> dfs = [&](int i, int j) -> int {
if (i >= m || j >= m || (i + j) >= m) return 0;
if (f[i][j] != 0x3f3f3f3f) return f[i][j];
int k = i + j;
int a = dfs(i + 1, j) + nums[i] * multipliers[k];
int b = dfs(i, j + 1) + nums[n - j - 1] * multipliers[k];
return f[i][j] = max(a, b);
};
return dfs(0, 0);
}
};

• class Solution:
def maximumScore(self, nums: List[int], multipliers: List[int]) -> int:
@cache
def f(i, j, k):
if k >= m or i >= n or j < 0:
return 0
a = f(i + 1, j, k + 1) + nums[i] * multipliers[k]
b = f(i, j - 1, k + 1) + nums[j] * multipliers[k]
return max(a, b)

n = len(nums)
m = len(multipliers)
return f(0, n - 1, 0)


• func maximumScore(nums []int, multipliers []int) int {
n, m := len(nums), len(multipliers)
f := make([][]int, m)
for i := range f {
f[i] = make([]int, m)
for j := range f[i] {
f[i][j] = 1 << 30
}
}
var dfs func(i, j int) int
dfs = func(i, j int) int {
if i >= m || j >= m || i+j >= m {
return 0
}
if f[i][j] != 1<<30 {
return f[i][j]
}
k := i + j
a := dfs(i+1, j) + nums[i]*multipliers[k]
b := dfs(i, j+1) + nums[n-j-1]*multipliers[k]
f[i][j] = max(a, b)
return f[i][j]
}
return dfs(0, 0)
}

• function maximumScore(nums: number[], multipliers: number[]): number {
const inf = 1 << 30;
const n = nums.length;
const m = multipliers.length;
const f = new Array(m + 1).fill(0).map(() => new Array(m + 1).fill(-inf));
f[0][0] = 0;
let ans = -inf;
for (let i = 0; i <= m; ++i) {
for (let j = 0; j <= m - i; ++j) {
const k = i + j - 1;
if (i > 0) {
f[i][j] = Math.max(f[i][j], f[i - 1][j] + nums[i - 1] * multipliers[k]);
}
if (j > 0) {
f[i][j] = Math.max(f[i][j], f[i][j - 1] + nums[n - j] * multipliers[k]);
}
if (i + j === m) {
ans = Math.max(ans, f[i][j]);
}
}
}
return ans;
}