Problem Description

You have an array of integers nums (0-indexed) and an integer k.

You can perform an operation where you:

1. Pick any index i where 0 <= i < nums.length - 1
2. Replace nums[i] and nums[i + 1] with a single value: nums[i] & nums[i + 1] (using bitwise AND)
3. This reduces the array size by 1

The goal is to find the minimum possible value of the bitwise OR of all remaining elements after performing at most k operations.

For example, if you have nums = [3, 5, 7] and perform one operation on indices 0 and 1:

• 3 & 5 = 1 (in binary: 011 & 101 = 001)
• The array becomes [1, 7]
• The bitwise OR of remaining elements is 1 | 7 = 7

The challenge is to strategically choose which adjacent pairs to combine (and in what order) to minimize the final OR value, while being limited to at most k operations.

The solution uses a greedy approach with bit manipulation, checking from the most significant bit to the least significant bit to determine which bits can be eliminated from the final OR result.

Intuition

To minimize the bitwise OR of the final array, we need to minimize the number of 1-bits in the result. The key insight is that the bitwise OR will have a 1-bit at position i if at least one element in the final array has a 1-bit at that position.

Since we can only perform AND operations (which never create new 1-bits, only preserve or eliminate existing ones), our strategy should focus on eliminating as many high-value bits as possible.

The crucial observation is that we should work greedily from the most significant bit to the least significant bit. Why? Because eliminating a higher bit has more impact on reducing the final value than eliminating a lower bit. For example, eliminating bit 29 (worth 2^29) is more valuable than eliminating all bits from 0 to 28 combined.

For each bit position starting from the highest, we ask: "Can we eliminate this bit from the final OR result using at most k operations?"

To check if we can eliminate a specific bit pattern (represented by a mask), we need to verify if we can merge the array elements such that none of the final elements have those bits set. We simulate this by:

1. Creating a test mask that includes all bits we want to eliminate
2. Counting how many "groups" we'd end up with if we performed AND operations optimally
3. If the number of groups minus 1 (which equals the number of merges needed) is at most k, we can eliminate those bits

The counting process works by going through the array and greedily merging adjacent elements when their AND result (masked by our test pattern) becomes 0, which means those bits would be eliminated. If the AND result is non-zero, we must start a new group.

This greedy bit-by-bit approach ensures we get the minimum possible OR value by prioritizing the elimination of the most significant bits first.

Learn more about Greedy patterns.

Solution Approach

The implementation uses a greedy bit manipulation strategy, examining bits from position 29 down to 0 (covering all possible bits in a 32-bit integer).

Key Variables:

• ans: Accumulates the bits we want to try eliminating (our test mask)
• rans: Accumulates the bits that must remain in the final result
• test: The current mask being tested (ans + (1 << i) adds the current bit to our mask)

Algorithm Steps:

1. Iterate through bits from most to least significant (29 to 0):

   for i in range(29, -1, -1):

2. Create a test mask by adding the current bit to our accumulated mask:

   test = ans + (1 << i)

3. Count minimum groups needed when applying the test mask:

   • Initialize cnt = 0 (group counter) and val = 0 (current group's AND accumulator)
   • For each number in the array:

     if val == 0:
         val = test & num  # Start new group
     else:
         val &= test & num  # Continue merging with current group

   • If val becomes non-zero after processing a number, it means we need to keep this as a separate group, so increment cnt

4. Decision making:

   • If cnt > k: We need more than k operations to eliminate this bit pattern
     • Add this bit to the result: rans += 1 << i
     • Don't update ans (we can't eliminate this bit)
   • If cnt <= k: We can eliminate this bit pattern with at most k operations
     • Update our test mask: ans += 1 << i
     • This bit won't appear in the final result

5. Return rans, which contains only the bits that couldn't be eliminated.

Why this works:

• When val & (test & num) = 0, it means we can merge this element with the previous group and eliminate the tested bits
• The number of groups minus 1 equals the number of merge operations needed
• By greedily trying to eliminate the highest bits first, we ensure the minimum possible OR value

Example Walkthrough

Let's walk through the solution with nums = [3, 5, 7] and k = 1.

In binary: 3 = 011, 5 = 101, 7 = 111

Initial state:

• ans = 0 (bits we're trying to eliminate)
• rans = 0 (bits that must remain in result)

Bit 2 (value 4):

• Test mask: test = 0 + 4 = 4 (binary: 100)
• Count groups needed:
  • Process 3: 3 & 4 = 0, so val = 0 (can merge)
  • Process 5: 5 & 4 = 4, so val = 4, increment cnt = 1
  • Process 7: 7 & 4 = 4, so val = 4 & 4 = 4 (still non-zero), keep cnt = 1
• Since cnt = 1 ≤ k = 1, we can eliminate bit 2
• Update: ans = 4

Bit 1 (value 2):

• Test mask: test = 4 + 2 = 6 (binary: 110)
• Count groups needed:
  • Process 3: 3 & 6 = 2, so val = 2, increment cnt = 1
  • Process 5: 5 & 6 = 4, so val = 2 & 4 = 0 (can merge)
  • Process 7: 7 & 6 = 6, so val = 6, increment cnt = 2
• Since cnt = 2 > k = 1, we cannot eliminate bit 1
• Update: rans = 0 + 2 = 2

Bit 0 (value 1):

• Test mask: test = 4 + 1 = 5 (binary: 101)
• Count groups needed:
  • Process 3: 3 & 5 = 1, so val = 1, increment cnt = 1
  • Process 5: 5 & 5 = 5, so val = 1 & 5 = 1 (still non-zero), keep cnt = 1
  • Process 7: 7 & 5 = 5, so val = 1 & 5 = 1 (still non-zero), keep cnt = 1
• Since cnt = 1 ≤ k = 1, we can eliminate bit 0
• Update: ans = 5

Final result: rans = 2

This means with 1 operation, we can achieve a minimum OR of 2. We'd merge indices 0 and 1 (3 & 5 = 1), giving us [1, 7], and 1 | 7 = 7. Wait, that gives 7, not 2...

Let me recalculate more carefully:

Bit 2 (value 4):

• Test mask: test = 4
• Process elements with mask applied:
  • 3 & 4 = 0: Start with val = 0 (empty group)
  • 5 & 4 = 4: Since val was 0, set val = 4, then since val ≠ 0, increment cnt = 1
  • 7 & 4 = 4: val = 4 & 4 = 4 (still non-zero)
• Final cnt = 1, so we need 2 groups total (1 merge operation)
• Since 1 ≤ k, we can eliminate this bit pattern: ans = 4

Bit 1 (value 2):

• Test mask: test = 4 + 2 = 6
• Process elements:
  • 3 & 6 = 2: Start with val = 2, since val ≠ 0, increment cnt = 1
  • 5 & 6 = 4: val = 2 & 4 = 0 (merged!)
  • 7 & 6 = 6: Since val is 0, set val = 6, since val ≠ 0, increment cnt = 2
• Final cnt = 2 means 3 groups (2 merge operations needed)
• Since 2 > k, we cannot eliminate this bit: rans = 2

Bit 0 (value 1):

• Test mask: test = 4 + 1 = 5
• Process elements:
  • 3 & 5 = 1: Start with val = 1, since val ≠ 0, increment cnt = 1
  • 5 & 5 = 5: val = 1 & 5 = 1 (still non-zero)
  • 7 & 5 = 5: val = 1 & 5 = 1 (still non-zero)
• Final cnt = 1, so we can eliminate this bit: ans = 5

The algorithm returns rans = 2, indicating the minimum possible OR is 2.

To achieve this, we'd perform operations to ensure no element has bits 0 and 2 set, leaving only bit 1 possible in the final OR.

Solution Implementation

Time and Space Complexity

Time Complexity: O(30 * n) or O(n) where n is the length of the input array nums.

The algorithm iterates through 30 bit positions (from bit 29 to bit 0) in the outer loop. For each bit position, it performs a complete traversal of the nums array to count valid operations. Since the number of bits (30) is a constant, the overall time complexity simplifies to O(n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space. The variables ans, rans, i, test, cnt, and val are all scalar values that don't depend on the input size. No additional data structures are created that scale with the input, making the space complexity constant.

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

Common Pitfalls

1. Misunderstanding the Group Counting Logic

Pitfall: A common mistake is incorrectly counting the number of groups or operations needed. Developers might think that operations_count directly represents the number of merge operations, but it actually represents the number of groups that remain after merging.

Issue in the code:

# This counts groups, not operations!
if current_and_value:
    operations_count += 1

Why it's confusing:

• The variable name operations_count is misleading - it's actually counting segments/groups
• The number of operations needed is groups - 1, not groups
• When we have n groups, we need n-1 merge operations to combine them all

Solution: Either rename the variable for clarity or adjust the counting logic:

# Option 1: Rename variable to be clearer
segments_count = 0
current_and_value = 0

for num in nums:
    if current_and_value == 0:
        current_and_value = test_mask & num
    else:
        current_and_value &= test_mask & num
  
    if current_and_value:
        segments_count += 1
        current_and_value = 0  # Reset for next segment

# Check if we can merge these segments with k operations
if segments_count - 1 > k:  # segments - 1 = operations needed
    result += 1 << bit_position

2. Incorrect AND Value Reset

Pitfall: Forgetting to reset current_and_value to 0 after counting a group, which causes incorrect group counting in subsequent iterations.

Problematic pattern:

if current_and_value:
    operations_count += 1
    # Missing: current_and_value = 0

Why it matters: Without resetting, the algorithm continues to AND with the previous group's value, leading to incorrect results. Each group should start fresh after being counted.

Correct implementation:

for num in nums:
    if current_and_value == 0:
        current_and_value = test_mask & num
    else:
        current_and_value &= test_mask & num
  
    if current_and_value:
        operations_count += 1
        current_and_value = 0  # Critical: Reset for next group

3. Off-by-One Error in Operation Counting

Pitfall: The code counts groups but compares directly with k, which can lead to allowing too many operations.

The issue: If we have 3 groups, we need 2 operations to merge them into 1. But the code checks if operations_count > k, which for k=2 would incorrectly allow 3 groups.

Fixed logic:

# Count actual groups that would remain
groups_needed = 1  # Start with 1 for the final group
current_and_value = 0

for num in nums:
    masked_value = test_mask & num
    if current_and_value == 0:
        current_and_value = masked_value
    else:
        current_and_value &= masked_value
  
    # When we can't merge anymore, we need a new group
    if current_and_value == 0 and masked_value != 0:
        groups_needed += 1
        current_and_value = masked_value

# Operations needed = groups - 1
operations_needed = groups_needed - 1
if operations_needed > k:
    result += 1 << bit_position
else:
    mask += 1 << bit_position

These pitfalls can cause the algorithm to either incorrectly eliminate bits that shouldn't be eliminated or fail to eliminate bits that could be eliminated, resulting in a suboptimal answer.