Today, we’ll dive into an interesting problem where we have to construct a strictly increasing array with a given length n, ensuring that a bitwise AND operation across all elements results in a specific number x. Our goal? To keep the last element of this array—the largest one—as small as possible! 🚀 Let’s break down the problem, understand the logic, and see how Python code helps us solve it efficiently! 🐍💡

📝 Problem Statement 📝

You are given:

• Two integers, n and x.

Your task:

• Construct an array, nums, of positive integers with size n where:
  • The array is strictly increasing: nums[i + 1] > nums[i] for 0 <= i < n - 1.
  • The bitwise AND of all elements in nums is equal to x.

Objective: Return the minimum possible value of nums[n - 1], the last element of the array.

🔍 Examples 🔍

Let’s explore a few examples to see the problem in action!

Example 1

• Input: n = 3, x = 4
• Output: 6
• Explanation: One possible array is [4, 5, 6] and the last element is 6. The array [4, 5, 6] satisfies:
  • Strictly increasing: Yes, 4 < 5 < 6.
  • Bitwise AND equals x: 4 & 5 & 6 = 4.

Example 2

• Input: n = 2, x = 7
• Output: 15
• Explanation: An array [7, 15] satisfies the conditions, with the last element as 15:
  • Strictly increasing: Yes, 7 < 15.
  • Bitwise AND equals x: 7 & 15 = 7.

🚀 Solution Overview 🚀

To tackle this problem, we need to understand:

1. Bitwise AND and how it restricts our choices.
2. How to keep the last element of nums as small as possible while respecting the AND requirement.
3. The key insight: We can construct each element of the array by merging x with different numbers, ensuring the AND operation holds.

💡 Key Insights 💡

1. Bitwise AND Behavior: For a bitwise AND to yield x, the array elements must have the same bits set as in x. However, extra bits can be added without impacting the AND result.
2. Constructing Elements: To ensure nums[i + 1] > nums[i] and retain the AND behavior, each nums[i] can be seen as a bitwise merge of x with another integer. Each additional bit allows us to create strictly increasing elements.
3. Iterative Bit Masking: By manipulating bits from lower to higher positions, we can achieve the smallest last element.

🧑‍💻 Code Explanation 🧑‍💻

Here’s the Python code for the optimized solution:

class Solution:
    def minEnd(self, n: int, x: int) -> int:
        n -= 1  # Use (n-1) to create a range of numbers
        ans = x  # Start with x as the base answer
        for i in range(31):  # Loop over 31 bits (assuming we work with 32-bit integers)
            if x >> i & 1 == 0:  # Check if bit `i` in x is not set
                ans |= (n & 1) << i  # Set the bit `i` in `ans` based on `n`
                n >>= 1  # Move `n` to the right to handle the next bit
        ans |= n << 31  # Account for any remaining bits in `n` shifted to the 31st position
        return ans

📊 Time Complexity Analysis 📊

The code has a time complexity of O(1). Since the solution involves bitwise operations over a fixed number of bits (31 in this case), the time complexity does not grow with the input size. Instead, it remains constant.

🕵️ Walkthrough with a Higher Number Example 🕵️

Let’s break down the solution step-by-step with n = 4 and x = 9. Here’s how each part of the process unfolds:

1. Initialize Variables:
   • n -= 1 results in n = 3 (which allows indexing for a range of 4 elements).
   • Set ans = x = 9.
2. Bitwise Merging:
   • We loop over each bit from 0 to 31.
   • For each bit, if it’s not set in x, we set it in ans according to bits in n.
   • Example progression with x = 9:
     • Initial x in binary: 0000...1001
     • Set bits of ans based on lower bits of n until all are shifted.
3. Final Adjustment:
   • After the loop, any leftover bits in n are merged by shifting to the 31st position.

After running through the solution, you’ll get ans as the minimum possible last element. 🎉

📌 Edge Cases to Consider 📌

1. Small Values of n: When n = 1, nums contains only x.
2. Small Values of x: When x = 1, the smallest bit is set, limiting our choices for merging.
3. High Bit Positions: The solution correctly handles bit positions up to 31, respecting 32-bit integer constraints.

🔍 Conclusion 🔍

Through this problem, we learned the power of bitwise operations in creating optimized solutions. Using bitwise manipulation allowed us to minimize the last element while ensuring the AND result. This approach provides an efficient solution, meeting both the strictly increasing and bitwise constraints with minimal code!