import java.util.*; public class Solution { p...

You can eliminate the $O(n^3)$ brute-force over $(L,R)$ in your “op1” case by observing that

$$\Delta(L,R) \;=\; \sum_{i=L}^{R}\bigl((\text{cur}[i]\oplus x)-\text{cur}[i]\bigr) \;+\; \sum_{i=R+1}^{n-1}\bigl((\text{cur}[i]\oplus 2x)-\text{cur}[i]\bigr)$$

Let

java
// the gain if you XOR x at position i
long[] diff1 = new long[n];
for(int i=0;i<n;i++)
  diff1[i] = (cur[i]^x) - cur[i];

// the gain if you XOR 2*x at position i
long[] diff2 = new long[n];
for(int i=0;i<n;i++)
  diff2[i] = (cur[i]^(2*x)) - cur[i];

Then

$$\Delta(L,R) \;=\; \sum_{i=L}^R diff1[i]\;+\;\sum_{i=R+1}^{n-1}diff2[i] \;=\; (\text{prefix1}[R]-\text{prefix1}[L-1]) + \text{suffix2}[R+1].$$

You can find

$$\max_{0\le L\le R<n}\Delta(L,R)$$

in a single $O(n)$ scan by maintaining:

• a running prefix-sum of diff1,
• the minimum prefix-sum seen so far (to pick the best $L$),
• a precomputed suffix-sum array of diff2.

Likewise your “op2” (case 1) is just “pick the best suffix” of $\bigl((\text{cur}[i]\oplus 2x) - \text{cur}[i]\bigr)$, also doable in $O(n)$.

Putting it all together gives you an $O(3^m\cdot n)$ (up to a factor $m$) solution:

java
public static long maxSumAfterOps(int[] a, int[] b) {
    int n = a.length, m = b.length;
    long best = Long.MIN_VALUE;
    int totalMasks = (int) Math.pow(3, m);

    for (int mask = 0; mask < totalMasks; mask++) {
        // decode ops
        int[] ops = new int[m];
        for (int j = 0, t = mask; j < m; j++, t /= 3) {
            ops[j] = t % 3;
        }

        // working copy of a[]
        long[] cur = new long[n];
        for (int i = 0; i < n; i++) cur[i] = a[i];

        // apply each operation in O(n)
        for (int j = 0; j < m; j++) {
            long x = b[j];
            if (ops[j] == 0) {
                // -- op1: choose best (L,R)
                long[] diff1 = new long[n], suffix2 = new long[n+1];
                for (int i = 0; i < n; i++) {
                    diff1[i] = (cur[i] ^ x) - cur[i];
                }
                suffix2[n] = 0;
                for (int i = n - 1; i >= 0; i--) {
                    suffix2[i] = suffix2[i+1] + ((cur[i] ^ (2*x)) - cur[i]);
                }
                long maxGain = Long.MIN_VALUE;
                long prefixSum = 0, minPrefix = 0;
                int minPos = -1, bestL = 0, bestR = 0;
                for (int r = 0; r < n; r++) {
                    prefixSum += diff1[r];
                    long gain = prefixSum - minPrefix + suffix2[r+1];
                    if (gain > maxGain) {
                        maxGain = gain;
                        bestL = minPos + 1;
                        bestR = r;
                    }
                    if (prefixSum < minPrefix) {
                        minPrefix = prefixSum;
                        minPos = r;
                    }
                }
                // apply the XORs in one more O(n)
                for (int i = 0; i < n; i++) {
                    if (i >= bestL && i <= bestR)      cur[i] ^= x;
                    else if (i >  bestR)               cur[i] ^= 2*x;
                }

            } else if (ops[j] == 1) {
                // -- op2: choose best suffix
                long[] diff2 = new long[n];
                for (int i = 0; i < n; i++) {
                    diff2[i] = (cur[i] ^ (2*x)) - cur[i];
                }
                long maxGain = Long.MIN_VALUE;
                long suffixSum = 0;
                int bestStart = 0;
                for (int i = n - 1; i >= 0; i--) {
                    suffixSum += diff2[i];
                    if (suffixSum > maxGain) {
                        maxGain = suffixSum;
                        bestStart = i;
                    }
                }
                for (int i = bestStart; i < n; i++) {
                    cur[i] ^= 2*x;
                }
            }
            // else ops[j] == 2 → do nothing
        }

        // compute final total
        long total = 0;
        for (long v : cur) total += v;
        best = Math.max(best, total);
    }

    return best;
}

Complexity breakdown

• There are $\;3^m\;$ masks.
• For each mask, each of the $m$ ops takes $O(n)$ to
  1. build the diff arrays,
  2. scan once to pick optimal $(L,R)$ or suffix,
  3. apply the XOR in $O(n)$.
• Overall $O\bigl(3^m \times m \times n\bigr)$, which for small constant $m$ is effectively $O(3^m\cdot n)$.