3287. Find the Maximum Sequence Value of Array - Detailed Explanation

Java Implementation (Greedy heuristic):

Note: This greedy approach is not guaranteed to find the optimal solution for all inputs. It’s a heuristic that can be used to get a quick answer that is sometimes correct (and runs fast), but there are cases where it will miss the true maximum.

3. Optimal Dynamic Programming Approach (Prefix/Suffix DP)

Idea: Use dynamic programming to systematically explore possible OR values for subsets of size k in different parts of the array, then combine those results to find the best XOR. The insight is that because each number is at most 7 bits, the OR of any subset is limited to at most 7 bits (values 0 through 127). We can leverage this small range of OR outcomes to do DP efficiently.

Approach Overview:

1. Divide the problem at a split point: Suppose in the optimal solution, the subsequence of length 2*k is such that the first half (k elements) comes from some prefix of the array, and the second half (k elements) comes from the remaining suffix. In other words, there is some index i where the first half of the subsequence lies entirely in nums[0 .. i-1] and the second half lies in nums[i .. n-1]. We don’t know i in advance, but we can try all possible split positions i (with at least k elements before and after, so k <= i <= n-k).
2. DP on prefixes: For each prefix ending at position i-1, determine all possible bitwise OR results we can get by choosing exactly j elements from that prefix (for all j from 0 to k). We particularly need the results for j = k. Let’s call this set (or boolean table) f[i][j], where f[i][x] represents all possible OR values using x elements from the first i elements of the array.
3. DP on suffixes: Do a similar computation on suffixes. For each suffix starting at position i, determine all possible OR results by choosing j elements from that suffix (for 0 <= j <= k). We’ll call this g[i][j], where g[i][y] is the set of OR values achievable using y elements from nums[i .. n-1].
4. Combine prefix and suffix: Now, for each potential split index i (where the boundary is between i-1 and i in the array), consider taking k elements in the prefix and k in the suffix. We will have some possible OR values for the prefix group (say values x in f[i][k]) and some possible OR values for the suffix group (values y in g[i][k]). The value of that particular subsequence (with that split) could be x XOR y. We choose the maximum such XOR over all possible choices of x and y. Then among all split positions, take the global maximum.

This approach ensures we consider every valid way to pick k on the left and k on the right, but in a smart way using DP to avoid re-enumerating combinations each time.

DP State Definition:
Let f[i][j] be a boolean or set representing which OR values are achievable by picking j elements from the first i elements of the array (nums[0 ... i-1] if we use 0-indexing for array). Similarly, let g[i][j] represent OR values achievable by picking j elements from the subarray nums[i ... n-1] (i.e., starting at index i through the end). We only care about j up to k. The OR values range from 0 to 127 (inclusive) because of the 7-bit limit.

DP Transition (for f):

• Base case: f[0][0] = {0} (with 0 elements, OR is 0). For any j > 0, f[0][j] is empty (can’t pick >0 elements from an empty prefix).
• Iterating through array (prefix): For each new element nums[i] (at index i-1 in 1-indexed DP terms), we update the possible OR values:
  • We can skip this element. Any OR value that was achievable with j picks from the first i-1 elements is still achievable with j picks from first i elements. (Carry over f[i-1][j] into f[i][j].)
  • We can take this element as part of the subset. Then any OR value that was achievable with j-1 picks from first i-1 elements, combined with this element, will yield a new OR value for j picks. Specifically, for each OR value val in f[i-1][j-1], we can achieve val OR nums[i-1] in f[i][j].

In formula form, for each i (1-indexed for DP length) and for each j from 0 to k:

• All OR values in f[i-1][j] are included in f[i][j] (skip nums[i-1]).
• For each x in f[i-1][j-1], include x OR nums[i-1] in f[i][j] (take nums[i-1]).

We perform this DP for i = 1 to n. After processing the whole array, f[n][k] would represent all OR values achievable by picking k from all n elements (but we will also need intermediate results for splits).

DP Transition (for g): It’s similar but we build from the end of the array backward.

• Base: g[n][0] = {0} (0 elements from an empty suffix OR is 0).
• Iterate i = n-1 down to 0: when considering element nums[i] for the suffix:
  • Skipping it: carry over all OR values in g[i+1][j] to g[i][j].
  • Taking it: for each OR value y achievable with j-1 picks from nums[i+1 ... n-1], include y OR nums[i] in g[i][j].

By the end, g[0][k] would be all OR values achievable by picking k elements from the entire array (from index 0 to end), but more importantly g[i][k] gives OR values for picking k from the suffix starting at i.

Combining Results: Now, we try each possible split position i (where the subsequence splits just before index i):

• We require that the prefix (first part of subsequence) ends before index i and the suffix (second part) starts at i. This means we pick k elements from nums[0 .. i-1] and k from nums[i .. n-1]. We have the sets of OR outcomes: f[i][k] for the prefix and g[i][k] for the suffix.
• For each possible OR value x in f[i][k] and each possible OR value y in g[i][k], compute x XOR y. Track the maximum of these XORs across all such pairs. That is our candidate answer for that split.
• Do this for every i from k to n-k (to ensure both sides can have k picks). The overall answer is the maximum XOR found among all splits.

This might sound expensive (nested loops over OR values for each i), but recall each OR value x or y ranges only over 0..127. So at most 128 possible values on each side. Checking all pairs is at most 128128 = 16,384 operations per split. With at most ~400 possible split positions, that’s about 40016384 ≈ 6.5 million checks in worst case, which is manageable in C++/Java. The DP itself is O(n * k * 128) for building prefix and similarly for suffix, which for n=400, k=200 is 400200128 ≈ 10,240,000 – also manageable.

Memory: We can compress the DP states to use less memory (since each state f[i] only depends on f[i-1], we don’t need to store the full 3D table at once), but conceptually it’s fine. A full DP table would be size ~400200128 ≈ 10 million booleans, which is okay.

Example to Illustrate DP (Small):
Consider nums = [4, 2, 5] and k = 2. We want to pick 4 elements total, but n=3 so obviously we must pick all 3? (This example is just to illustrate DP states; for valid subsequence we’d need 2*k <= n, so not a valid input. Let’s modify for example: use nums = [4, 2, 6, 7], n=4, k=2, as in skipping one scenario.) For nums = [4, 2, 6, 7], k=2:

• Prefix DP f results:

  • f[1] (using first element 4):
    • f[1][0] contains {0} (pick nothing),
    • f[1][1] contains {4} (pick 4),
    • f[1][2] is empty (can’t pick 2 out of 1 element).
  • f[2] (first two elements 4,2):
    • f[2][0] = {0},
    • f[2][1]: we can pick one of {4,2} → {4,2},
    • f[2][2]: pick both 4 and 2 → OR = 4|2 = 6 → {6}.
  • f[3] (first three 4,2,6):
    • f[3][0] = {0},
    • f[3][1]: can be {4,2,6} (pick any one),
    • f[3][2]: possibilities: pick any two -> {4|2=6, 4|6=6 (since 6 already has bits of 4), 2|6=6} = {6}, also {4|2 (we did), 4|6, 2|6} all yield 6 in this case,
    • f[3][2] actually yields {6}, and
    • f[3][3]: pick all three -> OR = 4|2|6 = 6 (because 6 in binary 110 already covers bits of 4 and 2).
  • f[4] (all four 4,2,6,7):
    • f[4][0] = {0}
    • f[4][1] = {2,4,6,7} (any single element OR is itself)
    • f[4][2]: many combos:
      • {4|2=6, 4|6=6, 4|7=7 (0111|0100=0111), 2|6=6, 2|7=7 (0010|0111=0111), 6|7=7 (0110|0111=0111)} ⇒ {6,7}
    • f[4][2] = {6,7}
    • f[4][3]: choose 3 out of 4:
      • likely results in {7} (e.g., 4|2|6=6, 4|2|7=7, 4|6|7=7, 2|6|7=7) ⇒ {6,7}? Actually check: 4|2|6 = 6, the others have 7 in them, so {6,7}.
    • f[4][4]: OR of all 4 = 4|2|6|7 = 7. For our purpose k=2, we see f[4][2] = {6,7}: possible ORs picking 2 from full array.

• Suffix DP g results (we do similarly from end):

  • g[4] (suffix starting at index4 (past the end)): g[4][0]={0}, others empty.
  • g[3] (suffix [7]):
    • g[3][0]={0},
    • g[3][1]={7},
    • g[3][2] empty.
  • g[2] (suffix [6,7]):
    • g[2][0]={0},
    • g[2][1]={6,7},
    • g[2][2]={6|7=7},
  • g[1] (suffix [2,6,7]):
    • g[1][0]={0},
    • g[1][1]={2,6,7},
    • g[1][2]: pick 2 from [2,6,7] -> {2|6=6, 2|7=7, 6|7=7} = {6,7},
    • g[1][3]: pick all three -> {2|6|7 = 7} = {7}.
  • g[0] (suffix [4,2,6,7] whole array):
    • g[0][1]={4,2,6,7},
    • g[0][2]={6,7} (similarly to above),
    • etc.

Now to get answer for k=2, we try splits:

• Split i=2 (prefix [4,2], suffix [6,7]):
  • ORs from prefix with 2 picks: f[2][2] = {6}
  • ORs from suffix with 2 picks: g[2][2] = {7}
  • XOR = 6 ^ 7 = 1.
• Split i=3 (prefix [4,2,6], suffix [7]):
  • f[3][2] = {6},
  • g[3][2] – does suffix [7] have 2 picks? No, g[3][2] is empty (cannot pick 2 out of one element), so this split can’t yield 2 picks in suffix.
• Split i=1 (prefix [4], suffix [2,6,7]):
  • f[1][2] empty (can't pick 2 from one element), so invalid.
• Split i=4 (prefix [4,2,6,7], suffix []): suffix can’t provide 2 picks, invalid.

So the only viable split was i=2 here, giving value 1. Indeed, from earlier brute check, [4,5,6,7] was optimal for that full example with skip, but here we used [4,2,6,7] example to illustrate DP, which might not have had a different option? (In [4,2,6,7], the optimal subsequence of length4 is either skip none since n=4 and 2k=4, you must take all. That yields value (4|2) XOR (6|7) = 6 XOR 7 = 1, which matches our DP result.)

The DP correctly would enumerate possibilities and find the maximum.

Complexity Revisited: Time complexity is roughly O(n * k * 128 + n * 128 * 128), which simplifies to about O(n * k + n) with a constant factor, given 128 and 16384 are constants . For worst-case n=400, k=200, this is on the order of a few million operations – well within limits in C++/Java. Python might be slower, but with optimizations (using bit sets or Python sets to store OR values), it can pass since the state space is limited.

Python Implementation (Optimal DP):

The Java implementation above follows the described DP logic closely:

• f[i][j][x] is true if an OR value x is achievable using j picks from first i elements.
• g[i][j][y] is true if OR y is achievable using j picks from the suffix starting at i. It then checks all splits i and XORs all pairs x and y to find the maximum.

Optimization: In practice, one can optimize space by using rolling arrays (since each f[i] depends only on f[i-1], etc.) or bitsets to represent the OR values. Also, we might break early if certain XOR is already 127 (the theoretical max possible, since 127 is the highest 7-bit number) as we can’t exceed that. However, implementing these is complex; the above straightforward DP is easier to understand and is efficient enough given the input limits.

Time Complexity: As derived, roughly O(n * k * 128 + n * 128^2). For n up to 400 and k up to 200, this is on the order of millions of operations (worst-case ~16 million), which is fine in C++/Java. The solution is optimal for the given constraints.

Space Complexity: About O(n * k * 128) to store the DP tables, which is roughly 400200128 ≈ 10^8 booleans at maximum, which is around 100 million. This is borderline but since each boolean can be bit-packed or memory is just about a few tens of MB, it’s acceptable. If memory is a concern, using bitsets or compressing the DP can significantly reduce it (e.g., using a 2D array of bitset[128] for each j).

Common Mistakes

• Assuming contiguous subarrays: Some might misread the problem and think the subsequence must be contiguous (a subarray). The problem allows any subsequence (not necessarily contiguous). If you restrict to contiguous segments, you might miss the optimal solution.
• Greedy bit selection pitfalls: Trying to greedily assign highest-value or highest-bit elements to different groups without full search can fail. A common mistake is to assume taking the 2*k largest numbers yields the best XOR – as we saw, two large numbers can cancel out bits in XOR. Without careful consideration, greedy approaches might produce a suboptimal grouping.
• Ignoring the ordering constraint: When dividing into two groups of k, remember that the first group’s elements must come from earlier in the array than the second group’s elements (because they form a subsequence in that order). If you choose two sets of k elements without regard to original indices, you might end up with an invalid subsequence (one of the “first half” elements actually comes after a “second half” element in the array). The DP approach inherently respects ordering by using prefix and suffix divisions.
• Not utilizing the 7-bit limit: If one tries a DP or bitmask approach without noticing the values are <128, they might attempt a DP state for OR values up to, say, 2^31 which is impossible to iterate over. The small bit-size is crucial – iterating all OR combinations up to 127 is what makes the DP feasible.
• Off-by-one errors in DP indices: When implementing the DP, it’s easy to slip on indices for prefix vs suffix. For example, ensuring that f[i][j] corresponds to first i elements (which are indices 0 to i-1) and that g[i][j] corresponds to suffix from i (index i to n-1). Mixing these up can lead to incorrect results. It’s helpful to clearly document or draw the interpretation of i in each DP.
• Memory optimization issues: If implementing in a low-level language, one might try to compress the DP but introduce bugs. It’s safer to get the correct DP working before optimizing. Also, watch out for using bit operations: using XOR in transitions instead of OR (since the problem is about OR of halves, XOR is only for final combination). The DP transitions should use OR for adding a number to the set.

Edge Cases

• Minimum size array: If nums.length = 2, then k must be 1 (since k <= n/2). We simply take both elements; the answer will be nums[0] XOR nums[1]. For example, nums=[a,b], k=1 yields a XOR b. Our solution handles this as there’s only one possible subsequence of length 2.
• All numbers have the same bits: If every number in the array shares a certain bit, that bit will inevitably appear in both halves’ OR (as long as both halves have at least one element). For instance, if all numbers are odd (all have the lowest bit = 1), then both groups of any valid subsequence will have OR that is odd, making the XOR’s lowest bit 0. In extreme cases, if the array elements are all identical, any selection will result in both ORs being that number, and XOR will be 0. The maximum value could be 0 in such scenarios. Our algorithms correctly handle this (they will find 0 as the max if no bit can be isolated to one side).
• k equal to n/2: This means we are using the entire array as our subsequence (since 2*k = n). In this case, there is no choice of which elements to pick (we pick all), only how to split them into two groups of equal size. The split is actually fixed as the first n/2 elements and the last n/2 elements of the array (since the subsequence is the whole array). The answer is then just OR of first half XOR OR of second half of the original array. (Our DP would consider splits, but only one split is actually possible: i = n/2 in 0-indexed terms.)
• k = 1 (special simple case): This reduces to finding two elements in the array with maximum XOR (since OR of a single element is itself, so we want max of a XOR b). There are known efficient solutions (trie-based) for this specific case. Our DP approach still works for k=1 – it will compute all 1-element ORs (which are just the elements themselves) in f and g, then try all splits. Essentially it will end up considering all pairs x and y from the array, which is correct. The DP is overkill for k=1 but still fine.

Alternative Variations

• Different Group Sizes: If the problem allowed different sizes for the two halves (say one half of length p and another of length q, not necessarily equal), the XOR value would then be (OR of p elements) XOR (OR of q elements). The approach could be adapted by iterating possible values of p and q (with p+q = some total). However, here 2*k forces equal halves. Unequal halves would complicate the search, but a similar DP could be done for each possible split length combination.

• Maximize OR or other operations: A variation might ask to maximize (OR of first half) + (OR of second half) or minimize the XOR, etc. The DP approach could often be adapted, but the objective function change might require different combining logic. For sum instead of XOR, you’d try to maximize the sum of two ORs – likely you’d end up taking as many bits as possible in both halves (which tends to just combine all bits, less interesting as a problem).

• Contiguous subsequence (subarray): If the sequence of length 2k had to be contiguous in the array, we would then be looking for a subarray of length 2k that maximizes the XOR of ORs of its halves. This becomes a different problem – one could slide a window of size 2*k and compute the metric for each window, using two-pointer or deque techniques to maintain ORs, or recompute efficiently. The contiguous constraint simplifies the selection (no combinatorial picking, just fixed segments to evaluate).

• Allowing any subsequence length (not exactly 2*k): If we were allowed to pick any even number of elements (up to n) and maximize the sequence value, we’d have to consider all possible even lengths. Intuitively, if larger groups allow more bits to appear, they might not always be better because more elements could introduce overlapping bits in both ORs. That variation would require checking multiple values of group size k and would be more complex. However, the given problem fixes the length at 2*k.

• Minimizing the value: If instead we were asked for the minimum sequence value for length 2*k, the approach would still work – we’d just take the minimum XOR rather than max when combining. The minimum could often be 0 (especially if some bit must appear in both halves unavoidably).

• Other bitwise operations: We could imagine similar tasks with AND instead of OR: e.g., define sequence value = (AND of first half) XOR (AND of second half). That becomes quite different, because AND behaves differently (it tends to lose 1 bits when combining numbers). The DP strategy would change since the range of possible AND results isn’t limited in the same simple way (though if numbers are bounded, perhaps it is). OR was nice because any OR result is a combination of bits present in some picks.

Related Problems

• Count Number of Maximum Bitwise OR Subsets: In this problem, you enumerate subsets and look at their OR results. It asks how many subsets achieve the maximum OR value. While a different goal, it also uses the bitwise OR of subsets and often is approached with DP or BFS over OR states, utilizing the limited bit-size of numbers.

• Bitwise XOR of All Pairings: This problem deals with XOR and pairings in a different way. Given two arrays, it asks for the XOR of every possible pair (one from each array). While not directly related to subset OR, it involves XOR combinations and has a combinatorial flavor. The solution is simpler (it uses properties of XOR and frequency), but it’s a reminder to consider bit properties in combinations.

• Split Array Largest Sum