Problem Description

You are given a mysterious function func that takes an array and two indices l and r as parameters. The function returns the bitwise AND of all elements from index l to index r (inclusive) in the array. In other words, func(arr, l, r) = arr[l] & arr[l+1] & ... & arr[r].

Given an integer array arr and an integer target, you need to find indices l and r such that the absolute difference |func(arr, l, r) - target| is minimized.

Return the minimum possible value of |func(arr, l, r) - target|.

Constraints:

• 0 <= l, r < arr.length
• Both l and r must be valid indices in the array

Example: If you have an array and a target value, you need to find a subarray (defined by indices l and r) where the bitwise AND of all elements in that subarray is as close as possible to the target value. The answer is the minimum absolute difference you can achieve.

The key insight is that for any subarray [l, r], the function computes the cumulative bitwise AND operation on all elements within that range, and you want this result to be as close to the target value as possible.

Intuition

The key observation is that the bitwise AND operation has a monotonic property: when we extend a subarray by including more elements, the result either stays the same or decreases (it never increases). This is because AND operation can only turn bits from 1 to 0, never from 0 to 1.

If we fix the right endpoint r and vary the left endpoint l from r down to 0, the values of func(arr, l, r) form a decreasing sequence. Moreover, since each element in the array is at most 10^6 (which requires about 20 bits to represent), there can be at most 20 distinct values as we vary l. This is because each bit position can only change from 1 to 0 once as we extend the range leftward.

This suggests we can efficiently track all possible AND values ending at each position. For each position r, instead of computing func(arr, l, r) for all possible l values (which would be O(n²)), we can maintain a set of all distinct AND values that can be obtained with subarrays ending at position r.

When we move from position r to r+1, we can update our set of possible values by:

1. Taking each value in the current set and performing AND with arr[r+1] (extending previous subarrays)
2. Adding arr[r+1] itself (starting a new subarray at position r+1)

Since the set contains at most 20 values at any point (due to the bit limit), this update process is very efficient. For each position, we check all values in our set against the target to find the minimum absolute difference.

This approach cleverly exploits the properties of bitwise AND and the bounded nature of the values to reduce what could be an O(n²) problem to O(n × log M) where M is the maximum value in the array.

Learn more about Segment Tree and Binary Search patterns.

Solution Approach

The implementation uses a hash table (set) to maintain all possible bitwise AND values ending at the current position.

Algorithm Steps:

1. Initialize: Start with the answer as abs(arr[0] - target) and create a set s containing just arr[0].

2. Iterate through the array: For each element x in arr:

   • Update the set: Create a new set containing:
     • All values from the previous set ANDed with x: {x & y for y in s}
     • The current element itself: {x}
     • This is done with the expression: s = {x & y for y in s} | {x}

3. Update the minimum: For each value y in the updated set s, calculate abs(y - target) and update the answer with the minimum value found.

Implementation Details:

class Solution:
    def closestToTarget(self, arr: List[int], target: int) -> int:
        ans = abs(arr[0] - target)  # Initialize with first element
        s = {arr[0]}  # Set to store all possible AND values ending at current position
      
        for x in arr:
            # Update set: AND all previous values with x, and add x itself
            s = {x & y for y in s} | {x}
            # Find minimum difference with target among all values in set
            ans = min(ans, min(abs(y - target) for y in s))
      
        return ans

Why this works:

• The set s at each iteration contains all possible values of func(arr, l, i) where i is the current index and l ranges from 0 to i.
• When we process element arr[i], we extend all subarrays ending at i-1 by including arr[i] (hence x & y for y in s).
• We also start a new subarray at position i (hence adding {x} to the set).
• The set automatically handles duplicates, keeping only unique values.
• Since the set size is bounded by approximately 20 (the number of bits), each iteration is efficient.

Time Complexity: O(n × log M) where n is the array length and M is the maximum value in the array. The log M factor comes from the maximum number of distinct AND values possible.

Space Complexity: O(log M) for storing the set of distinct AND values.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: arr = [9, 12, 3, 7, 15], target = 5

We need to find a subarray where the bitwise AND of all elements is closest to 5.

Initial State:

• ans = |9 - 5| = 4
• s = {9} (only considering subarray [9])

Iteration 1: Process arr[1] = 12

• Calculate new set: s = {12 & 9} | {12} = {8} | {12} = {8, 12}
  • 12 & 9 = 1100 & 1001 = 1000 = 8 (extending subarray [9] to [9,12])
  • 12 itself (new subarray starting at index 1)
• Update answer:
  • |8 - 5| = 3
  • |12 - 5| = 7
  • ans = min(4, 3) = 3

Iteration 2: Process arr[2] = 3

• Calculate new set: s = {3 & 8, 3 & 12} | {3} = {0, 0} | {3} = {0, 3}
  • 3 & 8 = 0011 & 1000 = 0000 = 0 (extending [9,12] to [9,12,3])
  • 3 & 12 = 0011 & 1100 = 0000 = 0 (extending [12] to [12,3])
  • 3 itself (new subarray [3])
• Update answer:
  • |0 - 5| = 5
  • |3 - 5| = 2
  • ans = min(3, 2) = 2

Iteration 3: Process arr[3] = 7

• Calculate new set: s = {7 & 0, 7 & 3} | {7} = {0, 3} | {7} = {0, 3, 7}
  • 7 & 0 = 0 (extending subarrays ending with AND value 0)
  • 7 & 3 = 0111 & 0011 = 0011 = 3 (extending [3] to [3,7])
  • 7 itself (new subarray [7])
• Update answer:
  • |0 - 5| = 5
  • |3 - 5| = 2
  • |7 - 5| = 2
  • ans = min(2, 2) = 2

Iteration 4: Process arr[4] = 15

• Calculate new set: s = {15 & 0, 15 & 3, 15 & 7} | {15} = {0, 3, 7} | {15} = {0, 3, 7, 15}
  • 15 & 0 = 0
  • 15 & 3 = 1111 & 0011 = 0011 = 3
  • 15 & 7 = 1111 & 0111 = 0111 = 7 (extending [7] to [7,15])
  • 15 itself
• Update answer:
  • |0 - 5| = 5
  • |3 - 5| = 2
  • |7 - 5| = 2
  • |15 - 5| = 10
  • ans = min(2, 2) = 2

Final Answer: 2

The closest we can get to target 5 is either:

• Subarray [3] with AND value = 3 (difference = 2)
• Subarray [7] with AND value = 7 (difference = 2)
• Subarray [3,7] with AND value = 3 (difference = 2)

This example demonstrates how the set maintains all possible AND values ending at each position, efficiently tracking only distinct values (notice how duplicate 0s are automatically handled), and how we continuously update our minimum difference.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * log(max(arr)))

The algorithm iterates through each element in the array once. For each element x, it performs bitwise AND operations with all elements in set s. The key insight is that the set s contains at most O(log(max(arr))) distinct values at any point.

This is because when we perform bitwise AND operations starting from any number, the result can only decrease or stay the same (never increase). Since bitwise AND can only turn bits from 1 to 0 (never from 0 to 1), and each number has at most log(max(arr)) bits, there can be at most O(log(max(arr))) distinct values in the set after consecutive AND operations.

Therefore:

• Outer loop: O(n) iterations
• Inner set comprehension: O(log(max(arr))) operations per iteration
• Total: O(n * log(max(arr)))

Space Complexity: O(log(max(arr)))

The space is primarily used by the set s. As explained above, the set s can contain at most O(log(max(arr))) distinct values at any point during the execution, since we're tracking the results of cumulative bitwise AND operations which can only produce a limited number of distinct values based on the bit representation of the numbers.

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

Common Pitfalls

Pitfall 1: Forgetting to Initialize with the First Element

Issue: A common mistake is to start with an empty set or not properly initialize the answer with the first element's difference from the target.

Wrong Implementation:

def closestToTarget(self, arr: List[int], target: int) -> int:
    min_difference = float('inf')  # Wrong: might miss single-element subarrays
    current_and_values = set()     # Wrong: empty set loses first element
  
    for num in arr:
        current_and_values = {num & prev_value for prev_value in current_and_values} | {num}
        min_difference = min(min_difference, min(abs(and_value - target) for and_value in current_and_values))
  
    return min_difference

Correct Approach:

def closestToTarget(self, arr: List[int], target: int) -> int:
    min_difference = abs(arr[0] - target)  # Initialize with first element
    current_and_values = {arr[0]}          # Start with first element in set
  
    for num in arr[1:]:  # Process remaining elements
        current_and_values = {num & prev_value for prev_value in current_and_values} | {num}
        min_difference = min(min_difference, min(abs(and_value - target) for and_value in current_and_values))
  
    return min_difference

Pitfall 2: Not Including Single Elements as Valid Subarrays

Issue: Forgetting to add {num} when updating the set, which would miss all single-element subarrays starting at the current position.

Wrong Implementation:

def closestToTarget(self, arr: List[int], target: int) -> int:
    min_difference = abs(arr[0] - target)
    current_and_values = {arr[0]}
  
    for num in arr:
        # Wrong: only extends previous subarrays, doesn't start new ones
        current_and_values = {num & prev_value for prev_value in current_and_values}
        min_difference = min(min_difference, min(abs(and_value - target) for and_value in current_and_values))
  
    return min_difference

Correct Approach:

def closestToTarget(self, arr: List[int], target: int) -> int:
    min_difference = abs(arr[0] - target)
    current_and_values = {arr[0]}
  
    for num in arr:
        # Correct: includes both extended subarrays AND new subarray starting at current position
        current_and_values = {num & prev_value for prev_value in current_and_values} | {num}
        min_difference = min(min_difference, min(abs(and_value - target) for and_value in current_and_values))
  
    return min_difference

Pitfall 3: Processing the First Element Twice

Issue: If you initialize the set with arr[0] and then iterate through the entire array starting from index 0, you'll process the first element twice, which could lead to incorrect results.

Wrong Implementation:

def closestToTarget(self, arr: List[int], target: int) -> int:
    min_difference = abs(arr[0] - target)
    current_and_values = {arr[0]}
  
    # Wrong: starts from index 0, processing arr[0] again
    for num in arr:  
        current_and_values = {num & prev_value for prev_value in current_and_values} | {num}
        min_difference = min(min_difference, min(abs(and_value - target) for and_value in current_and_values))
  
    return min_difference

Correct Approaches:

Option 1: Start iteration from index 1

def closestToTarget(self, arr: List[int], target: int) -> int:
    min_difference = abs(arr[0] - target)
    current_and_values = {arr[0]}
  
    for num in arr[1:]:  # Start from index 1
        current_and_values = {num & prev_value for prev_value in current_and_values} | {num}
        min_difference = min(min_difference, min(abs(and_value - target) for and_value in current_and_values))
  
    return min_difference

Option 2: Don't initialize, process everything in loop

def closestToTarget(self, arr: List[int], target: int) -> int:
    min_difference = float('inf')
    current_and_values = set()
  
    for num in arr:
        current_and_values = {num & prev_value for prev_value in current_and_values} | {num}
        min_difference = min(min_difference, min(abs(and_value - target) for and_value in current_and_values))
  
    return min_difference

Note: While the original solution processes the first element twice, it still works correctly because arr[0] & arr[0] = arr[0], making it idempotent. However, it's cleaner to avoid the redundancy.