u/Rocking_man24

Question: 3976. Maximum Subarray Sum After Multiplier
You are given an integer array nums and a positive integer k. You must choose exactly one subarray of nums and perform exactly one of the following operations: Return the maximum possible sum of a non-empty subarray in the resulting array. Note that the subarray chosen for the operation and the subarray chosen for the sum may be different.

Input: nums = [1,-2,3,4,-5], k = 2 Output: 14
  
class Solution {
    public long maxSubarraySum(int[] nums, int k) {
        int n = nums.length;
        long best = Long.MIN_VALUE;


        // mult
        long A = nums[0];
        long B = (long) k * nums[0];
        long C = Long.MIN_VALUE / 2;


        best = Math.max(A, B);


        for (int i = 1; i < n; i++) {
            long x = nums[i];
            long kx = k * x;


            long newA = Math.max(x, A + x);
            long newB = Math.max(kx, Math.max(B + kx, A + kx));
            long newC = Math.max(x, Math.max(C + x, B + x));


            A = newA; B = newB; C = newC;
            best = Math.max(best, Math.max(A, Math.max(B, C)));
        }


        // div
        A = nums[0];
        B = div(nums[0], k);
        C = Long.MIN_VALUE / 2;


        best = Math.max(best, Math.max(A, B));


        for (int i = 1; i < n; i++) {
            long x = nums[i];
            long dx = div(x, k);


            long newA = Math.max(x, A + x);
            long newB = Math.max(dx, Math.max(B + dx, A + dx));
            long newC = Math.max(x, Math.max(C + x, B + x));


            A = newA; B = newB; C = newC;
            best = Math.max(best, Math.max(A, Math.max(B, C)));
        }


        return best;
    }


    private long div(long x, int k) {
        // if (x >= 0) return x / k;
        // return (x + k - 1) / k;
        if (x >= 0) return x / k;
        return -((-x) / k); 
    }
}Multiply each number in the chosen subarray by k. Divide each number in the chosen subarray by k. When dividing a positive number by k, use the floor value of the division result. When dividing a negative number by k, use the ceiling value of the division result.1 <= nums.length <= 105 -105 <= nums[i] <= 105 1 <= k <= 105

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