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The key to this approach is analyzing each bit position of the numbers. We need to ensure that at each bit position i, (a[i] OR b[i]) matches c[i].
If c[i] is 0, both a[i] and b[i] must be 0. This may require flipping if either a[i] or b[i] is 1.
If c[i] is 1, we need at least one of a[i] or b[i] to be 1. If both are 0, we need one flip.
By iterating through each binary bit, we can count the necessary flips.
Time Complexity: O(1), since the number of bits is constant and we iterate a fixed number of times.
Space Complexity: O(1), as we use a constant amount of space.
1public class MinFlips {
2 public int minFlips(int a, int b, int c) {
3 int flips = 0;
4 for (int i = 0; i < 32; i++) {
5 boolean bitA = (a & (1 << i)) != 0;
6 boolean bitB = (b & (1 << i)) != 0;
7 boolean bitC = (c & (1 << i)) != 0;
8 if (!bitC) {
9 flips += (bitA ? 1 : 0) + (bitB ? 1 : 0);
10 } else {
11 flips += (!bitA && !bitB) ? 1 : 0;
12 }
13 }
14 return flips;
15 }
16
17 public static void main(String[] args) {
18 MinFlips instance = new MinFlips();
19 int result = instance.minFlips(2, 6, 5);
20 System.out.println(result);
21 }
22}In this Java solution, Boolean expressions determine the state of each bit position in a, b, and c. Conditional statements calculate the necessary flips to equalize a OR b to c for each bit.
This alternative approach uses masks and logical reduction to simplify determining the minimum flips. Instead of checking each bit individually, create a mask that highlights the bits where a OR b does not match the bits of c.
For c's zeroed bits, count additions for every 1 found in corresponding bits in a or b.
For c's one bits, ensure at least one 1 is present in the corresponding a or b position. Adjust flips accordingly.
Time Complexity: O(1), consistently iterating fixed bits.
Space Complexity: O(1), as it uses primitive operations.
JavaScript follows a strategy of iterative bit examination and reduction through while loops, ensuring each loop evaluates a fixed number of positions for consistent flip calculation.