leetcode 1770 - Maximum Score from Performing Multiplication Operations

https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/

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 x from either the start or the end of the array nums.
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

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class Solution {
public:
int maximumScore(vector<int>& nums, vector<int>& multipliers) {
int size = nums.size();
int size_m = multipliers.size();
int dp[size_m+1][size_m+1];

int ans = INT_MIN;

for (int len = 1; len <= (int)multipliers.size(); len++) {
for (int left = 0; left <= len; left++) {
int right = len - left;

int l = left - 1;
int r = size - right;

if (left == 0)
dp[left][right] = (right - 1 == 0 ? 0 : dp[left][right-1]) + (multipliers[len-1] * nums[r]);
else if (right == 0)
dp[left][right] = (left - 1 == 0 ? 0 : dp[left-1][right]) + (multipliers[len-1] * nums[l]);
else {
dp[left][right] = max(dp[left][right-1] + (multipliers[len-1] * nums[r]),
dp[left-1][right] + (multipliers[len-1] * nums[l]));
}

if (len == (int)multipliers.size())
ans = max(ans, dp[left][right]);
}
}

return ans;
}
};