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1458. Max Dot Product of Two Subsequences

Description

Given two arrays nums1 and nums2.

Return the maximum dot product between non-empty subsequences of nums1 and nums2 with the same length.

A subsequence of an array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, [2,3,5] is a subsequence of [1,2,3,4,5] while [1,5,3] is not).

 

Example 1:

Input: nums1 = [2,1,-2,5], nums2 = [3,0,-6]
Output: 18
Explanation: Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2.
Their dot product is (2*3 + (-2)*(-6)) = 18.

Example 2:

Input: nums1 = [3,-2], nums2 = [2,-6,7]
Output: 21
Explanation: Take subsequence [3] from nums1 and subsequence [7] from nums2.
Their dot product is (3*7) = 21.

Example 3:

Input: nums1 = [-1,-1], nums2 = [1,1]
Output: -1
Explanation: Take subsequence [-1] from nums1 and subsequence [1] from nums2.
Their dot product is -1.

 

Constraints:

  • 1 <= nums1.length, nums2.length <= 500
  • -1000 <= nums1[i], nums2[i] <= 1000

Solutions

Solution 1: Dynamic Programming

Thinking

Subsequences must be non-empty and values may be negative. \(m,n\le 500\). Taking only positive products like LCS would miss a single negative pair.

\(f[i][j]\) is the best dot product of the two prefixes: drop one end, or pair them and optionally discard a negative prefix via \(\max(0,f[i-1][j-1])+x\cdot y\). \(-\infty\) forces at least one pair.

We define \(f[i][j]\) to represent the maximum dot product of two subsequences formed by the first \(i\) elements of \(\textit{nums1}\) and the first \(j\) elements of \(\textit{nums2}\). Initially, \(f[i][j] = -\infty\).

For \(f[i][j]\), we have the following cases:

  1. Do not select \(\textit{nums1}[i-1]\) or do not select \(\textit{nums2}[j-1]\), i.e., \(f[i][j] = \max(f[i-1][j], f[i][j-1])\);
  2. Select \(\textit{nums1}[i-1]\) and \(\textit{nums2}[j-1]\), i.e., \(f[i][j] = \max(f[i][j], \max(0, f[i-1][j-1]) + \textit{nums1}[i-1] \times \textit{nums2}[j-1])\).

The final answer is \(f[m][n]\).

The time complexity is \(O(m \times n)\), and the space complexity is \(O(m \times n)\). Here, \(m\) and \(n\) are the lengths of the arrays \(\textit{nums1}\) and \(\textit{nums2}\), respectively.

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class Solution:
    def maxDotProduct(self, nums1: List[int], nums2: List[int]) -> int:
        m, n = len(nums1), len(nums2)
        f = [[-inf] * (n + 1) for _ in range(m + 1)]
        for i, x in enumerate(nums1, 1):
            for j, y in enumerate(nums2, 1):
                v = x * y
                f[i][j] = max(f[i - 1][j], f[i][j - 1], max(0, f[i - 1][j - 1]) + v)
        return f[m][n]

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class Solution {
    public int maxDotProduct(int[] nums1, int[] nums2) {
        int m = nums1.length, n = nums2.length;
        int[][] f = new int[m + 1][n + 1];
        for (var g : f) {
            Arrays.fill(g, Integer.MIN_VALUE);
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int v = nums1[i - 1] * nums2[j - 1];
                f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
                f[i][j] = Math.max(f[i][j], Math.max(f[i - 1][j - 1], 0) + v);
            }
        }
        return f[m][n];
    }
}

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class Solution {
public:
    int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        int f[m + 1][n + 1];
        memset(f, 0xc0, sizeof f);
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                int v = nums1[i - 1] * nums2[j - 1];
                f[i][j] = max(f[i - 1][j], f[i][j - 1]);
                f[i][j] = max(f[i][j], max(0, f[i - 1][j - 1]) + v);
            }
        }
        return f[m][n];
    }
};

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func maxDotProduct(nums1 []int, nums2 []int) int {
    m, n := len(nums1), len(nums2)
    f := make([][]int, m+1)
    for i := range f {
        f[i] = make([]int, n+1)
        for j := range f[i] {
            f[i][j] = math.MinInt32
        }
    }
    for i := 1; i <= m; i++ {
        for j := 1; j <= n; j++ {
            v := nums1[i-1] * nums2[j-1]
            f[i][j] = max(f[i-1][j], f[i][j-1])
            f[i][j] = max(f[i][j], max(0, f[i-1][j-1])+v)
        }
    }
    return f[m][n]
}

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function maxDotProduct(nums1: number[], nums2: number[]): number {
    const m = nums1.length;
    const n = nums2.length;
    const f = Array.from({ length: m + 1 }, () => Array.from({ length: n + 1 }, () => -Infinity));
    for (let i = 1; i <= m; ++i) {
        for (let j = 1; j <= n; ++j) {
            const v = nums1[i - 1] * nums2[j - 1];
            f[i][j] = Math.max(f[i - 1][j], f[i][j - 1]);
            f[i][j] = Math.max(f[i][j], Math.max(0, f[i - 1][j - 1]) + v);
        }
    }
    return f[m][n];
}

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impl Solution {
    pub fn max_dot_product(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {
        let m = nums1.len();
        let n = nums2.len();
        let mut f = vec![vec![i32::MIN; n + 1]; m + 1];

        for i in 1..=m {
            for j in 1..=n {
                let v = nums1[i - 1] * nums2[j - 1];
                f[i][j] = f[i][j].max(f[i - 1][j]).max(f[i][j - 1]);
                f[i][j] = f[i][j].max(f[i - 1][j - 1].max(0) + v);
            }
        }

        f[m][n]
    }
}

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