Problem Description

You are given an array nums containing n prime integers. Your task is to construct a new array ans of the same length n that satisfies a specific bitwise condition.

For each index i, you need to find a value for ans[i] such that when you perform a bitwise OR operation between ans[i] and ans[i] + 1, the result equals nums[i]. In other words, the condition ans[i] OR (ans[i] + 1) == nums[i] must hold true.

Additionally, among all possible values that satisfy this condition, you must choose the minimum value for each ans[i].

If no such value exists for a particular position that satisfies the condition, set ans[i] = -1.

The key insight is that for any integer a, the operation a OR (a + 1) flips the rightmost 0 bit in a to 1 (along with setting all bits to its right to 0 in a and 1 in a+1). This means the result is always odd. Therefore, if nums[i] is even (which only happens when it equals 2 since all inputs are prime), no valid ans[i] exists.

For odd values of nums[i], you need to find the smallest a such that performing a OR (a + 1) gives you nums[i]. This involves identifying which bit in a should be the rightmost 0 that gets flipped to produce the desired result.

Intuition

Let's first understand what happens when we perform a OR (a + 1). When we add 1 to a number a, we're essentially flipping the rightmost 0 bit to 1 and all bits to its right become 0. For example:

• If a = 5 (0b101), then a + 1 = 6 (0b110)
• If a = 11 (0b1011), then a + 1 = 12 (0b1100)

When we OR these two consecutive numbers, the result has a specific pattern: it sets the rightmost 0 bit of a to 1 and keeps all other bits to the left unchanged, while all bits to the right of that position become 1.

This means a OR (a + 1) always produces an odd number (since the least significant bit is always 1). Therefore, if our target nums[i] is even, it's impossible to find such an a. Since we're dealing with prime numbers, the only even prime is 2, so we return -1 for this case.

For odd targets, we need to work backwards. Given that nums[i] is the result of a OR (a + 1), we need to figure out what a could be. The key observation is that nums[i] has all 1s up to a certain position (from the right), then continues with some pattern. The transition point tells us where the rightmost 0 bit in a was located.

To find the minimum a, we look for the first 0 bit in nums[i] starting from position 1 (second least significant bit). Once we find this 0 at position i, we know that the rightmost 0 in a was at position i - 1. To construct a from nums[i], we need to flip the bit at position i - 1 from 1 to 0. This can be done using XOR: a = nums[i] XOR (1 << (i - 1)).

This approach guarantees we find the minimum valid a because we're identifying the rightmost possible position for the 0 bit in a, which gives us the smallest value that satisfies the condition.

Solution Approach

The implementation follows a straightforward approach based on our bit manipulation understanding:

1. Initialize Result Array: Create an empty array ans to store our results.

2. Process Each Number: Iterate through each number x in the input array nums.

3. Handle Even Prime Case: First check if x == 2. Since 2 is the only even prime and a OR (a + 1) always produces odd results, we append -1 to indicate no valid answer exists.

4. Find the First Zero Bit: For odd primes, we need to find the position of the first 0 bit starting from index 1 (the second least significant bit). We iterate through bit positions from 1 to 31:

   • Check each bit using x >> i & 1 ^ 1, which evaluates to true when bit at position i is 0
   • The expression works by:
     • x >> i shifts x right by i positions
     • & 1 isolates the least significant bit
     • ^ 1 flips the bit (XOR with 1), so 0 becomes 1 (true) and 1 becomes 0 (false)

5. Calculate the Answer: Once we find the first 0 bit at position i, we calculate ans[i] by flipping the bit at position i - 1 in x:

   • Use the formula: x ^ (1 << (i - 1))
   • 1 << (i - 1) creates a mask with only the (i-1)-th bit set to 1
   • XOR operation flips that specific bit in x from 1 to 0
   • Break out of the loop once we find and process the first 0 bit

6. Return Result: After processing all numbers, return the completed ans array.

The algorithm runs in O(n * log(max(nums))) time, where n is the length of the input array and the log factor comes from checking up to 32 bits for each number. The space complexity is O(1) extra space (not counting the output array).

Example Walkthrough

Let's walk through the solution with nums = [3, 5, 7]:

For nums[0] = 3 (binary: 0b011):

• Since 3 is odd, we look for a value a where a OR (a+1) = 3
• Binary of 3 is 011 - we need to find the first 0 bit starting from position 1
• Position 0: bit is 1 (skip)
• Position 1: bit is 1 (skip)
• Position 2: bit is 0 (found!)
• Since the first 0 is at position 2, we flip the bit at position 2-1 = 1
• Calculate: 3 XOR (1 << 1) = 011 XOR 010 = 001 = 1
• Verify: 1 OR 2 = 001 OR 010 = 011 = 3 ✓
• So ans[0] = 1

For nums[1] = 5 (binary: 0b101):

• 5 is odd, so we proceed
• Binary of 5 is 101 - find first 0 bit from position 1
• Position 0: bit is 1 (skip)
• Position 1: bit is 0 (found!)
• Flip bit at position 1-1 = 0
• Calculate: 5 XOR (1 << 0) = 101 XOR 001 = 100 = 4
• Verify: 4 OR 5 = 100 OR 101 = 101 = 5 ✓
• So ans[1] = 4

For nums[2] = 7 (binary: 0b111):

• 7 is odd, so we proceed
• Binary of 7 is 111 - find first 0 bit from position 1
• Position 0: bit is 1 (skip)
• Position 1: bit is 1 (skip)
• Position 2: bit is 1 (skip)
• Position 3: bit is 0 (found!)
• Flip bit at position 3-1 = 2
• Calculate: 7 XOR (1 << 2) = 111 XOR 100 = 011 = 3
• Verify: 3 OR 4 = 011 OR 100 = 111 = 7 ✓
• So ans[2] = 3

Final result: ans = [1, 4, 3]

Each answer represents the minimum value that, when OR'd with its successor, produces the corresponding prime in the input array.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log M)

The algorithm iterates through each element in the nums array (length n). For each element x, it performs a bit-checking loop that runs from bit position 1 up to at most 31 (since we're checking 32-bit integers). The loop searches for the first bit position i where the i-th bit of x is 0 (using x >> i & 1 ^ 1). This bit-checking operation takes at most O(log M) iterations, where M is the maximum value in the array, as the number of bits needed to represent M is proportional to log M. Therefore, the overall time complexity is O(n × log M).

Space Complexity: O(1) (excluding the output array)

The algorithm uses only a constant amount of extra space. It maintains a few variables (i in the loop and temporary values during bit operations) that don't depend on the input size. The ans array is used to store the results, but as stated in the reference, we ignore the space consumption of the answer array when analyzing space complexity. Therefore, the space complexity is O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Bit Position Logic

A common mistake is confusing the relationship between the position of the rightmost 0 bit and which bit needs to be flipped. When we find a 0 bit at position i, we need to flip the bit at position i-1, not position i itself.

Incorrect approach:

if ((num >> bit_position) & 1) == 0:
    answer = num ^ (1 << bit_position)  # Wrong! Flips the 0 bit itself

Correct approach:

if ((num >> bit_position) & 1) == 0:
    answer = num ^ (1 << (bit_position - 1))  # Correct! Flips the bit before the 0

2. Starting Bit Search from Wrong Position

Another pitfall is starting the bit search from position 0 instead of position 1. Since we're looking for a OR (a+1), the least significant bit (position 0) will always be 1 in the result, so the rightmost 0 that matters must be at position 1 or higher.

Incorrect approach:

for bit_position in range(0, 32):  # Wrong! Starts from position 0

Correct approach:

for bit_position in range(1, 32):  # Correct! Starts from position 1

3. Forgetting to Handle Edge Cases

Not handling the special case where num = 2 (the only even prime) will lead to incorrect results or infinite loops, since no value a satisfies a OR (a+1) = 2.

Solution: Always check for even numbers first:

if num == 2:
    result.append(-1)
    continue

4. Not Breaking After Finding First Zero Bit

Continuing to search after finding the first 0 bit will overwrite the correct answer with incorrect values from higher bit positions.

Incorrect approach:

for bit_position in range(1, 32):
    if ((num >> bit_position) & 1) == 0:
        answer = num ^ (1 << (bit_position - 1))
        result.append(answer)
        # Missing break! Will append multiple values

Correct approach:

for bit_position in range(1, 32):
    if ((num >> bit_position) & 1) == 0:
        answer = num ^ (1 << (bit_position - 1))
        result.append(answer)
        break  # Essential to stop after first 0 bit