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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)
using namespace std;
int minimumArrayEnd(int n, int x) {
return x + n - 1;
}
int main() {
int n = 3, x = 4;
cout << minimumArrayEnd(n, x) << endl;
return 0;
}Similar to the C implementation, this C++ function incrementally builds from x, with n-1 increments to the last element.
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)
1using System;
2
3class Program {
4 public static int MinimumArrayEnd(int n, int x) {
5 int highPossibleVal = x;
6 for (int i = 1; i < n; i++) {
7 highPossibleVal |= (1 << i);
8 }
9 return highPossibleVal;
10 }
11
12 static void Main() {
13 int n = 3, x = 4;
14 Console.WriteLine(MinimumArrayEnd(n, x));
15 }
16}In C#, by operating bitwise OR iteratively, each formation of higher numbers adds guaranteeing with enough overrides the functionalities converging tightly bit assumptions requiring modestly straightforward adjustments with O(n) manipulation.