Formatted question description: https://leetcode.ca/all/1770.html
1770. Maximum Score from Performing Multiplication Operations
Level
Medium
Description
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 i-th
operation (1-indexed), you will:
- Choose one integer
x
from either the start or the end of the arraynums
. - 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 <= 10^3
m <= n <= 10^5
-1000 <= nums[i], multipliers[i] <= 1000
Solution
Use dynamic programming. Create a 2D array dp
of m + 1
rows and n
columns, where dp[i][j]
represents the maximum score using the last i
multiplers for the subarray starting from index j
. The base case is when i = 1
. The final result is dp[m][0]
.
To optimize memory, use a 1D array instead, which stores only the last row of the original dp
array.
class Solution {
public int maximumScore(int[] nums, int[] multipliers) {
int n = nums.length, m = multipliers.length;
int[] dp = new int[n];
int minWindow = n - m + 1;
for (int j = minWindow - 1; j < n; j++) {
int start = j - minWindow + 1;
dp[start] = Math.max(nums[start] * multipliers[m - 1], nums[j] * multipliers[m - 1]);
}
for (int i = 2; i <= m; i++) {
int[] dpNew = new int[n];
int window = n - m + i;
for (int j = window - 1; j < n; j++) {
int start = j - window + 1;
dpNew[start] = Math.max(dp[start + 1] + nums[start] * multipliers[m - i], dp[start] + nums[j] * multipliers[m - i]);
}
dp = dpNew;
}
return dp[0];
}
}