Construct the Minimum Bitwise Array I

The core idea of this approach is to exhaustively search for the smallest possible value for ans[i].

1. For each number num in the input array nums, we need to find the smallest non-negative integer x such that x OR (x + 1) == num.
2. We know that x OR (x + 1) will always be greater than or equal to x + 1. Therefore, num >= x + 1, which implies x <= num - 1.
3. This gives us a finite search space for x: from 0 to num - 1.
4. We can iterate through this range. The first value of x that satisfies the condition x OR (x + 1) == num will be the smallest, because we are iterating in increasing order.
5. If the loop completes without finding any such x, it means no solution exists for that num, and we should assign -1.

The algorithm proceeds as follows:

• Initialize an integer array ans with the same length as nums to store the results.
• Iterate through the nums array from i = 0 to n-1.
• For each nums[i], start an inner loop with a variable x from 0 up to nums[i] - 1.
• In the inner loop, compute x | (x + 1) and check if it equals nums[i].
• If the condition is met, we have found the smallest x. We set ans[i] = x and break out of the inner loop to proceed to the next number in nums.
• To handle cases where no solution is found, we can use a flag or initialize ans[i] to -1 before the inner loop. If the inner loop finishes without finding a solution, ans[i] will remain -1.
• After iterating through all numbers in nums, return the ans array.

1class Solution {2    public int[] constructMinimizingBitwiseArray(int[] nums) {3        int n = nums.length;4        int[] ans = new int[n];5 6        for (int i = 0; i < n; i++) {7            int num = nums[i];8            int result = -1;9            // Search space for x is [0, num - 1]10            for (int x = 0; x < num; x++) {11                if ((x | (x + 1)) == num) {12                    result = x;13                    break; // Found the smallest x, no need to check further14                }15            }16            ans[i] = result;17        }18 19        return ans;20    }21}

This method is based on a mathematical derivation from the properties of bitwise operations.

1. First, handle the special case where num is the prime 2. There is no non-negative integer x for which x | (x+1) = 2. So, if num = 2, the answer is -1.
2. For any odd prime num, a solution x can be found. It can be proven that any solution x must be of the form x = num - 2^k for some integer k >= 0.
3. To find the minimum x, we must subtract the largest possible valid power of two, 2^k, from num.
4. The correct value of k is related to the binary representation of num + 1. Let p be the number of trailing zeros in num + 1. The largest valid 2^k we can subtract is 2^(p-1).
5. Thus, the smallest solution x is given by the formula: x = num - (1 << (p - 1)), where p is the number of trailing zeros in num + 1.
6. The number of trailing zeros can be efficiently calculated using built-in functions like Integer.numberOfTrailingZeros().

The core of this approach is a formula derived from the structure of x | (x+1).

Let y = nums[i]. We are solving x | (x+1) = y for the smallest x.

• Case 1: y is even. Since nums[i] are prime, the only even case is y = 2. We can check that 0|1=1, 1|2=3. No non-negative x satisfies x | (x+1) = 2. So, ans[i] = -1.

• Case 2: y is odd. For any odd y, a solution exists. The relationship y = (x+1) + lsb(x+1) - 1 holds, where lsb(v) is the value of the least significant bit of v (e.g., lsb(12) = 4). This can be rearranged to y+1 = (x+1) + lsb(x+1). This equation implies that solutions for x must be of the form y - 2^k.

To minimize x = y - 2^k, we need to maximize k. The largest k that yields a valid solution is determined by the number of trailing zeros in y+1.

Let p = Integer.numberOfTrailingZeros(y + 1). The formula for the minimal x is x = y - (1 << (p - 1)).

This single formula covers all odd y values:

• If y = 5, y+1 = 6 (binary 110). p=1. x = 5 - (1 << 0) = 4.
• If y = 7, y+1 = 8 (binary 1000). p=3. x = 7 - (1 << 2) = 3.
• If y = 11, y+1 = 12 (binary 1100). p=2. x = 11 - (1 << 1) = 9.

1class Solution {2    public int[] constructMinimizingBitwiseArray(int[] nums) {3        int n = nums.length;4        int[] ans = new int[n];5 6        for (int i = 0; i < n; i++) {7            int num = nums[i];8 9            // Since nums[i] is prime, the only even number is 2.10            if (num == 2) {11                ans[i] = -1;12            } else {13                // For any odd prime, a solution exists.14                // The smallest solution x is given by the formula:15                // x = num - 2^(p-1), where p is the number of trailing zeros in num + 1.16                int numPlusOne = num + 1;17                int p = Integer.numberOfTrailingZeros(numPlusOne);18                int powerOf2ToSubtract = 1 << (p - 1);19                ans[i] = num - powerOf2ToSubtract;20            }21        }22 23        return ans;24    }25}