3133. Minimum Array End - Detailed Explanation

Explanation: The Java optimized version follows the same logic as the Python one. We use bit manipulation (<< and &) to test and set bits. We loop through bit positions: if x has a 1-bit, we leave it (result keeps it), if x has a 0-bit, we assign a bit from k to that position in the result. We stop when we've assigned all bits of k. The while loop condition here is written to avoid infinite looping, using Java’s Long.SIZE (64 bits) as an upper bound and checking we don't overshoot the significant bits of k.

Complexity Analysis:

• Brute Force Approach: The time complexity is O(n * B), where B is the bit-length of the numbers (for checking the condition (next & x) == x, which is O(B) per check, but B is at most 30 for our range, so mainly O(n)). In the worst case, it could check nearly every number until building n elements. For large n (up to 100 million), this is clearly infeasible. The space complexity is O(1) (we only store a few counters and current number). While this approach is conceptually simple, it will time out or run for too long for large inputs.

• Optimized Approach: The time complexity is O(log n + log x), which comes from iterating through the bit positions needed. In the worst case, log2(n) and log2(x) are around 27 (for n or x up to 1e8), so maybe on the order of 30-35 iterations, which is effectively constant time for practical input sizes. This is extremely fast and easily handles the maximum constraints. Space complexity is O(1), as we only use a few variables to compute the result. This approach is efficient because it uses bit operations that execute in constant time per bit and doesn’t grow with n linearly.

Why O(log n + log x)? We combine bits of n-1 and x. In the worst case, if x is 0 (which it can’t be in our constraints because x>=1, but imagine conceptually) or very small, we might have to consider bits up to log2(n) because n-1 could have that many bits. If x is large (with bits near that range), it might itself contribute up to log2(x) bits to consider. We iterate through the union of those bit-lengths. Still, even summing those, it’s on the order of at most ~60 bits for our input range, which is constant-ish. Thus, this solution is extremely scalable for the given limits.

Step-by-Step Walkthrough (Optimized Solution with Example):

To solidify the understanding, let’s do a detailed binary walkthrough for a specific example using the optimized method. We’ll use Example 1 (n=3, x=4) for clarity:

• Input: n = 3, x = 4.
  x in binary is 100. The bits of x that are 1: only bit2 is 1. The bits of x that are 0: bit0 and bit1 (and any higher bits beyond bit2 are also 0, but we may not need those).
  We need n = 3 numbers, so consider k = n - 1 = 2.
  k = 2 in binary is 10.

Now, we fill the zero bits of x (from LSB upwards) with the bits of k (10):

• Bit position 0:
  • x has a 0 in this position (since 100 has 0 at bit0).
  • The least significant bit of k (2, which is 10 in binary) is 0.
  • We set bit0 of result to 0 (leave it as 0).
  • Move to the next bit of k (now consider the next bit in 10, which is 1) and next zero position.
• Bit position 1:
  • x has 0 in this position as well (bit1 of 100 is 0).
  • The next bit of k is 1.
  • We set bit1 of result to 1 (to include this bit from k).
  • Move to the next bit of k (we have used both bits of k now: first 0, then 1) and next zero position.
• Bit position 2:
  • x has 1 in this position (bit2 of 100 is 1). This is a fixed bit that all numbers must have.
  • We ensure the result has bit2 = 1 (it already will, since we start result = x). We do not consume a k bit here because this bit was not a “free” position; it’s mandated by x.
  • (We continue scanning higher, but since k’s bits are already all used, we can stop once we’ve ensured all k bits are placed.)
• Higher bits (bit3, bit4, ...):
  • x has 0 in these higher bits, but we have no more bits in k (we placed both bits of k already). So all higher bits remain 0 in the result.

Now compile the result bits we determined:

• Bit0 = 0
• Bit1 = 1
• Bit2 = 1 (from x)
• Higher bits = 0

So the result in binary is 110, which is 6 in decimal. This matches the output we expected. The sequence constructed implicitly here is: 100 (4) -> 101 (5) -> 110 (6). And indeed [4,5,6] was our example array.

Visualizing the bit placement:

Let's illustrate this process in a compact form:

    x = 100   (binary for 4)
  n-1 =  10   (binary for 2)

Positions (from LSB) :  bit2  bit1  bit0
x bits               :   1     0     0
n-1 bits to place    :        1     0    (aligning n-1 = "10" with x's zero bits)
----------------------------------------
Result bits          :   1     1     0   => 110 (binary) = 6 (decimal)

In the above diagram, we aligned n-1’s binary (10) with the zero bits of x. Bit1 and bit0 of x were zero, so those became targets for n-1's bits (1 and 0 respectively). The result’s bits are then the combination of x’s bits (bit2) and the placed bits.

This method scales to bigger numbers the same way. We always ensure the x bits stay, and we fill in extra bits from n-1 into the gaps.

Common Mistakes & Edge Cases:

• Dropping bits of x: A common mistake is to start the array with a number smaller than x or to allow a number in the sequence that doesn’t contain all bits of x. Remember that every element must have all the 1-bits of x. The first element should essentially be x itself (if you choose anything lower, it would miss a bit that x has, causing the AND to drop that bit). Always check this condition; (num & x) == x should hold for every element num in the array.

• Assuming OR instead of AND: Sometimes confusion arises between AND and OR in bitwise problems. Here we require the AND of all numbers to equal x. This is a stricter condition than OR. For clarity: AND = x means each bit that is 1 in x is 1 in every number (and any bit that ends up 0 in x means at least one number had 0 in that bit position). If you mistakenly approach this as a bitwise OR problem, you might try to ensure each bit of x appears somewhere – that would be a different problem. Make sure to enforce the AND condition correctly.

• Performance pitfalls: Iterating n times (or more) can be a huge pitfall given n can be up to 100 million. Even if each step is a simple operation, 100 million steps in Python (or even Java) can be borderline. The optimized approach avoids iterating through each number. Be cautious about any approach that loops n times or tries to build the entire array explicitly for large n.

• Edge Case – n = 1: If there’s only one element in the array, the problem trivializes: the array AND must equal x, and there’s only one number to pick, so that number must be x. The minimum (and only) last element is just x itself. Our formula also reflects that: if n=1, then n-1=0, we fill no extra bits, and the result remains x.

• Edge Case – x with all bits 1 up to a certain position: If x is like 7 (111) or 15 (1111), it has no zero bits up to its highest bit. That means you can’t create a second number without jumping to a higher bit that was zero (because any change in lower bits would drop a 1). For example, x=7 (111) as in Example 2: the next number had to use bit3 (giving 15). This often results in a significant jump for the second number. Ensure the logic correctly handles when it needs to go to a higher bit position that x doesn’t have. The formula naturally does this by using the next available zero bit (bit3 in that case) and placing the next bit of k there.

• Checking bit positions beyond the range of x: If n is large, you might end up setting bits higher than any bit present in x. That’s normal. For instance, if x=1 and n is large, the last element could be quite big because you’ll be using many higher bits to accommodate all n combinations. Make sure your solution doesn’t assume a fixed bit-size of x and iterates enough to place all needed bits of n-1.

• Using the wrong data type: In languages like Java or C++, the result might exceed the range of a 32-bit integer if a lot of high-order bits get set. For safety, use a larger type (like 64-bit) for calculations. In our Java solution we used long for this reason. In Python, integers can grow arbitrarily large, so it’s not an issue there.

Alternative Variations:

This problem showcases a neat technique of combining bit patterns. There are a few variations or related concepts worth noting:

• Bitwise OR instead of AND: Imagine a variation where the condition was that the bitwise OR of all elements equals some target x. In that case, to minimize the last element, you would do the opposite: every bit that is 0 in x must be 0 in all numbers (instead of at least one number), and bits that are 1 in x can be distributed. In fact, if OR of all = x, the simplest way is often to take the first element as x and all others as 0 (except 0 might not be allowed as "positive", so maybe x and some numbers that don’t introduce new 1 bits). That problem is much easier – the last element could just be x itself in many cases.

• Bitwise XOR conditions: Another variation could be requiring the XOR of all elements to equal x. XOR has different properties (it’s 1 in a bit if an odd number of elements have 1 in that bit). Constructing arrays for XOR conditions would be a different challenge, often solvable by greedy or pairing approaches.

• Different update operations: The way we generated the next number as (prev + 1) OR x in the simulation is a specific operation sequence. One could imagine an interview question asking to derive that operation or use it directly. This (num + 1) | x trick effectively finds the next number that has all bits of x by adding 1 and then restoring any dropped bits (OR with x). This kind of operation is useful in bit manipulations for constructing constrained sequences.

• Interleaving bit patterns: The optimized solution interleaves the bits of n-1 into the zeros of x. This is reminiscent of bit interleaving problems (for example, merging two binary numbers by alternating their bits). In our case, one number provides positions (the pattern of x), and another provides content for the gaps (n-1). This technique of merging two bit patterns could apply to other problems, such as encoding two numbers into one, creating Gray codes, or generating combinatorial subsets by bit masking.

• Binary combinatorics: Essentially, we treated the sequence of valid numbers as a binary counting problem in a reduced bit-space (only bits where x is 0). This approach is related to combinatorics – generating combinations of “extra” bits. If a different problem asked for k-th smallest number that has certain bits fixed, a similar approach of treating it as a combinatorial index could be used.

Understanding this technique can help in any scenario where you need to enforce that certain bits remain constant across a set or when merging two sets of bits. Always consider the bitwise interpretation of constraints – it often leads to elegant solutions like this one.

Related Problems for Further Practice:

Practicing similar concepts will reinforce the understanding of bitwise operations and sequence construction:

• LeetCode 201: Bitwise AND of Numbers Range – This problem asks for the AND of all numbers in a range [m, n]. It has a similar flavor because as the range grows, bits drop out. The solution involves finding common prefix bits of m and n (which is conceptually related to keeping certain bits constant, much like we did here).

• LeetCode 390: Elimination Game – Although not obviously a bitwise problem, the elimination pattern leads to a solution using bit manipulation (power of two reasoning). It’s good practice for thinking in terms of binary when iterating through sequences.

• LeetCode 136 & 137: Single Number problems – These involve XOR properties (finding a unique number where others repeat). They’re simpler but build comfort with bitwise operations.

• LeetCode 260: Single Number III – Another XOR-based puzzle for finding two unique numbers in an array. It requires using a bit mask trick to separate numbers into two groups.

• LeetCode 784: Letter Case Permutation – Not a bitwise problem per se, but if you consider letters and cases as bits (0/1 for lowercase/uppercase), generating all permutations is analogous to iterating over subsets of bit positions – similar in spirit to how we considered subsets of free bits of x.

• Custom challenge: Try writing a function to generate the entire array given n and x for small cases. Verify that the AND is x and the list is increasing. Then check how the last element relates to the formula. This can solidify your understanding by connecting the sequence to the final result.

• Binary/Bit Manipulation practice: Look for any problems labeled with "Bit Manipulation" on LeetCode. Many of them (like counting bits, flipping bits, etc.) will hone the skills needed to come up with optimized solutions like we did here.