LeetCode-in-Java
1770. Maximum Score from Performing Multiplication Operations
Medium
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 ith operation (1-indexed), you will:
- Choose one integer
xfrom either the start or the end of the arraynums. - Add
multipliers[i] * xto your score. - Remove
xfrom the arraynums.
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.lengthm == multipliers.length1 <= m <= 103m <= n <= 105-1000 <= nums[i], multipliers[i] <= 1000
Solution
public class Solution {
public int maximumScore(int[] nums, int[] mult) {
int n = nums.length;
int m = mult.length;
int row = m;
int[] dp = new int[m];
int[] prev = new int[m + 1];
while (--row >= 0) {
for (int i = 0; i <= row; ++i) {
dp[i] =
Math.max(
prev[i] + mult[row] * nums[n - row + i - 1],
prev[i + 1] + mult[row] * nums[i]);
}
prev = dp;
}
return dp[0];
}
}