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In this approach, we begin by constructing an array where each element is initialized to the given value of x. We then increment each subsequent element by 1 ensuring that it is greater than the previous element. We stop when we've constructed an array of size n.
This works because the bitwise AND of all numbers will still be x, since all elements are derived from x and only the last element needs to change significantly to ensure uniqueness and satisfy the increasing property.
Time Complexity: O(1)
Space Complexity: O(1)
class Program {
public static int MinimumArrayEnd(int n, int x) {
return x + n - 1;
}
static void Main() {
int n = 3, x = 4;
Console.WriteLine(MinimumArrayEnd(n, x));
}
}In C#, the method uses the same logic for computing the minimal end-element value by direct computation from inputs.
In this approach, we consider constructing the array by leveraging the bitwise properties to ensure the AND operation returns x. We create the first n-1 elements as numbers starting from x incrementally, and ensure the last element is the smallest number fitting the need for bitwise AND to be x.
To optimize the last number's structure, we can explore modifying bits in consideration of the upcoming elements and control zeroed bits strategically to adjust value and ensure valid returns upon AND calculation.
Time Complexity: O(n)
Space Complexity: O(1)
1public class Main {
2 public static int minimumArrayEnd(int n, int x) {
3 int highPossibleVal = x;
4 for (int i = 1; i < n; i++) {
5 highPossibleVal |= (1 << i);
6 }
7 return highPossibleVal;
8 }
9
10 public static void main(String[] args) {
11 int n = 3, x = 4;
12 System.out.println(minimumArrayEnd(n, x));
13 }
14}Using Java, by shifting bits, we dynamically align components increasing separately from initially fixed values of x step by step dynamically, ensuring minimal growth due to AND characteristic of non-detracting additions.