Find the Maximum Sequence Value of Array

• The core idea is to build the maximum possible result bit by bit, from the most significant bit (MSB) to the least significant bit (LSB).
• Initialize the result ans = 0.
• Iterate from bit b = 6 down to 0 (since nums[i] < 2^7).
• In each iteration, try to set the b-th bit of the answer to 1. Let target = ans | (1 << b).
• Check if it's possible to find two disjoint k-element subsequences A and B such that their OR-sums orA and orB satisfy (orA XOR orB) having target as a prefix. This means for all bits j >= b, the j-th bit of (orA XOR orB) must match the j-th bit of target.
• This check is performed by a helper function, isPossible(target, b).
• If isPossible returns true, it means we can achieve this target prefix, so we update ans = target.
• If it returns false, we cannot set the b-th bit to 1 (while satisfying higher bits), so we leave ans as is and proceed to the next smaller bit.
• The isPossible function uses its own DP. Let dp[j1][pA] be the minimum size j2 of the second set (B) required to satisfy the prefix condition, given that the first set (A) has j1 elements and its OR-sum prefix is pA. The prefix is calculated by val & mask, where mask includes all bits from MSB down to b.
• The DP transition for isPossible considers adding the current number to set A, set B, or neither, and updates the minimum j2 values accordingly.
• After iterating through all bits, ans holds the maximum possible sequence value.

We iterate from bit 6 down to 0. Let's say our current best-achievable value is ans. For the current bit b, we test if we can achieve a value of ans | (1 << b). This test involves checking if there exist two disjoint k-element subsequences, A and B, such that (OR(A) XOR OR(B)) matches ans | (1 << b) on all bits from 6 down to b.

To perform this check, we use a DP helper function, isPossible(target). This function determines if a given target prefix is achievable. The state for this DP is dp[j1][pA], which stores the minimum number of elements (j2) needed for set B to satisfy the prefix condition, given that set A has j1 elements and its OR-sum, when masked with the current bitmask, results in pA.

The DP table dp has dimensions (k+1) x 128. dp[j1][pA] is initialized to a value larger than k (e.g., k+1), with dp[0][0] = 0.

We iterate through each num in nums. For each num, we create a new DP table new_dp based on the old dp table to represent the states after considering num. For each state (j1, pA) with j2 = dp[j1][pA], we have three choices for num:

1. Don't use num: The state is carried over, which is implicitly handled by copying dp to new_dp.
2. Add num to set A: If j1 < k, we can transition to a state with j1+1 elements in A. The new OR-sum prefix for A will be pA | (num & mask). We update new_dp[j1+1][pA | (num & mask)] with min(current_value, j2).
3. Add num to set B: If j2 < k, we can transition to a state with j2+1 elements in B. The OR-sum prefix for B changes, which in turn changes the required prefix for A to maintain the target XOR relationship. The new required pA is calculated, and we update new_dp accordingly.

After processing all numbers, if there is any pA for which dp[k][pA] <= k, it means we can form two valid sets of size k, and the function returns true.

1class Solution {2    public int maxSequenceValue(int[] nums, int k) {3        int ans = 0;4        for (int b = 6; b >= 0; b--) {5            int target = ans | (1 << b);6            if (isPossible(nums, k, target, b)) {7                ans = target;8            }9        }10        return ans;11    }12 13    private boolean isPossible(int[] nums, int k, int target, int bit) {14        int mask = 0;15        for (int i = 6; i >= bit; i--) {16            mask |= (1 << i);17        }18 19        int[][] dp = new int[k + 1][128];20        for (int i = 0; i <= k; i++) {21            for (int j = 0; j < 128; j++) {22                dp[i][j] = k + 1; // Use k+1 as infinity23            }24        }25        dp[0][0] = 0;26 27        for (int num : nums) {28            int numPrefix = num & mask;29            int[][] newDp = new int[k + 1][128];30            for(int i=0; i<=k; i++) {31                System.arraycopy(dp[i], 0, newDp[i], 0, 128);32            }33 34            for (int j1 = 0; j1 <= k; j1++) {35                for (int pA = 0; pA < 128; pA++) {36                    if (dp[j1][pA] > k) continue;37                    int j2 = dp[j1][pA];38 39                    // Option 1: Add num to set A40                    if (j1 < k) {41                        newDp[j1 + 1][pA | numPrefix] = Math.min(newDp[j1 + 1][pA | numPrefix], j2);42                    }43 44                    // Option 2: Add num to set B45                    if (j2 < k) {46                        int pB = pA ^ target;47                        int newPB = pB | numPrefix;48                        int newPA = newPB ^ target;49                        newDp[j1][newPA] = Math.min(newDp[j1][newPA], j2 + 1);50                    }51                }52            }53            dp = newDp;54        }55 56        for (int pA = 0; pA < 128; pA++) {57            if (dp[k][pA] <= k) {58                return true;59            }60        }61        return false;62    }63}