LeetCode 1770. Maximum Score from Performing Multiplication Operations Solution in Java, C++, Python & More | Explanation + Code

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

Solution 1

PythonJavaC++GoTypeScript

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)(code-box)

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]; } }(code-box)

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); } };(code-box)

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) }(code-box)

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; }(code-box)

Solution 2

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)