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 i^th 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

Solution 1: Memoization

We design a function $dfs(i, j)$, which represents the maximum score that can be obtained starting from the $i$ element at the head of the nums array and starting from the $j$ element at the tail of the nums array. Then the answer is $dfs(0, 0)$.

The calculation process of function $dfs(i, j)$ is as follows:

If $i \geq m$ or $j \geq m$, or $i + j \geq m$, it means there are no elements to select, and $0$ is returned.
Otherwise, we can select the $i$ element at the head of the nums array, and the maximum score that can be obtained is $nums[i] \times multipliers[i + j] + dfs(i + 1, j)$; or we can select the $j$ element at the tail of the nums array, and the maximum score that can be obtained is $nums[n - j - 1] \times multipliers[i + j] + dfs(i, j + 1)$. We take the maximum value of the two as the return value of $dfs(i, j)$.

We can use memoized search to implement the above recursive process, where the f array is used to store the return value of the function $dfs(i, j)$ to prevent repeated calculations.

Time complexity $O(m^2)$, space complexity $O(m^2)$. Where $m$ is the length of the multipliers array.

Solution 2

We can rewrite the memorized search in method 1 into the form of dynamic programming.

We use $f[i][j]$ to represent the maximum score that can be obtained by taking the first $i$ elements of the array $nums$ and taking the last $j$ elements of the array $nums$. Initially $f[0][0] = 0$, the remaining elements are $-\infty$. The answer is $\max_{0 \leq i \leq m} f[i][m-i]$.

Considering $f[i][j]$, then currently we can select the $i$ element at the head of the nums array, or the $j$ element at the tail of the nums array. If the $i$ element at the head of the nums array is selected, the maximum score that can be obtained is $f[i-1][j] + nums[i-1] \times multipliers[i+j-1]$; if the $j$ element at the tail of the nums array is selected, the maximum score that can be obtained is $f[i][j-1] + nums[n-j] \times multipliers[i+j-1]$. We take the maximum of the two as the value of $f[i][j]$. If $i + j = m$, we update the answer $ans = \max(ans, f[i][j])$.

Finally, return the answer $ans$.

Time complexity $O(m^2)$, space complexity $O(m^2)$. Where $m$ is the length of the multipliers array.

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];
    }
}


// Solution 2
class Solution {
    public int maximumScore(int[] nums, int[] multipliers) {
        final int inf = 1 << 30;
        int n = nums.length, m = multipliers.length;
        int[][] f = new int[m + 1][m + 1];
        for (int i = 0; i <= m; i++) {
            Arrays.fill(f[i], -inf);
        }
        f[0][0] = 0;
        int ans = -inf;
        for (int i = 0; i <= m; ++i) {
            for (int j = 0; j <= m - i; ++j) {
                int k = i + j - 1;
                if (i > 0) {
                    f[i][j] = Math.max(f[i][j], f[i - 1][j] + multipliers[k] * nums[i - 1]);
                }
                if (j > 0) {
                    f[i][j] = Math.max(f[i][j], f[i][j - 1] + multipliers[k] * nums[n - j]);
                }
                if (i + j == m) {
                    ans = Math.max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
}

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);
    }
};


// Solution 2
class Solution {
public:
    int maximumScore(vector<int>& nums, vector<int>& multipliers) {
        const int inf = 1 << 30;
        int n = nums.size(), m = multipliers.size();
        vector<vector<int>> f(m + 1, vector<int>(m + 1, -inf));
        f[0][0] = 0;
        int ans = -inf;
        for (int i = 0; i <= m; ++i) {
            for (int j = 0; j <= m - i; ++j) {
                int k = i + j - 1;
                if (i > 0) {
                    f[i][j] = max(f[i][j], f[i - 1][j] + multipliers[k] * nums[i - 1]);
                }
                if (j > 0) {
                    f[i][j] = max(f[i][j], f[i][j - 1] + multipliers[k] * nums[n - j]);
                }
                if (i + j == m) {
                    ans = max(ans, f[i][j]);
                }
            }
        }
        return ans;
    }
};

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)


# Solution 2
class Solution:
    def maximumScore(self, nums: List[int], multipliers: List[int]) -> int:
        n, m = len(nums), len(multipliers)
        f = [[-inf] * (m + 1) for _ in range(m + 1)]
        f[0][0] = 0
        ans = -inf
        for i in range(m + 1):
            for j in range(m - i + 1):
                k = i + j - 1
                if i > 0:
                    f[i][j] = max(f[i][j], f[i - 1][j] + multipliers[k] * nums[i - 1])
                if j > 0:
                    f[i][j] = max(f[i][j], f[i][j - 1] + multipliers[k] * nums[n - j])
                if i + j == m:
                    ans = max(ans, f[i][j])
        return ans

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)
}


// Solution 2
func maximumScore(nums []int, multipliers []int) int {
	const inf int = 1 << 30
	n, m := len(nums), len(multipliers)
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, m+1)
		for j := range f {
			f[i][j] = -inf
		}
	}
	f[0][0] = 0
	ans := -inf
	for i := 0; i <= m; i++ {
		for j := 0; j <= m-i; j++ {
			k := i + j - 1
			if i > 0 {
				f[i][j] = max(f[i][j], f[i-1][j]+multipliers[k]*nums[i-1])
			}
			if j > 0 {
				f[i][j] = max(f[i][j], f[i][j-1]+multipliers[k]*nums[n-j])
			}
			if i+j == m {
				ans = max(ans, f[i][j])
			}
		}
	}
	return ans
}

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;
}

1770 - Maximum Score from Performing Multiplication Operations

1770. Maximum Score from Performing Multiplication Operations

Description

Solutions

Solution 1: Memoization

Solution 2

All Problems

All Solutions

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1770. Maximum Score from Performing Multiplication Operations

Description

Solutions

Solution 1: Memoization

Solution 2

All Problems

All Solutions